CONTENTS
Unit No. Title Page No.
SEMESTER - II
1. Functions, Derivatives and theirApplications 01
2. Simple Interest and Compound Interest 34
3. Annuities and EMI 47
4. Correlation and Regression 62
5. Time Series 84
6. Index Numbers 110
7. Probability Distributions 129
munotes.in
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UNIT I
Unit -1
FUNCTIONS, DERIVATIVES AND
THEIR APPLICATIONS
Unit Structure :
1.0 Objectives
1.1 Introduction
1.2 Derivatives
1.3 Second Order Derivatives
1.4 Applications of Derivatives
1.5 Maxima and Minima
1.0OBJECTIVE S
After reading this chapter you will be able to recognize.
1)Defin ition of function.
2)Standard Mathematical function .
3)Definition of derivative.
4)Derivatives of standard function s.
5)Second order derivatives.
6)Application of derivatives.
7)Maxim a and Minima .
1.1 FUNCTION S
If y= f(x) is a function then the set of all values of x for which this
function is defined is called the domain of the function ƒ. Here xis called
an independent variable and y is called the dependent variable. The set of
all co rresponding values of y for xin the domain is called the range of the
functionƒ.
The function ƒis defined from the domain to the range.
We shall discuss only those functions where the domain and the
range are subsets of real numbers. Such fun ctions are called 'real valued
functions'.munotes.in
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1.1.1 Standard Mathematical Functions:
(1) Constant function:
The constant function is defined by
y=ƒ(x)= C where C is a constant.
The c onstants are denoted by real numbers or alphabets. The graph
of a constant function is a straight line parallel to x -axis.
Examples:
y=ƒ(x)= 5
y=ƒ(x)=-10
y=ƒ(x)=K
y=ƒ(x)=a
(2) Linear function :
The linear function is defined by y =ƒ(x)= ax+ b where a and b are
constants.
Examples: y= ƒ(x)=2x+5
y=ƒ(x)=-3x+10
y=ƒ(x)= 5x-7
(3) Functions with power of x:
A function ƒ(x)=xnis called power function or function with
power of x. Here xis called base and n is called power .
Examples : ƒ(x) =x2
ƒ(x)=x-5
ƒ(x)=x-4/3ƒ(x)=x3/2
(4) Exponential functions :
These functions are of the type .
ƒ(x) = exand
ƒ(x) = ax, a >0
(5) Lo garithmic function : The logarithmic function is defined by
y=ƒ(x) = log ex,x>0
1.1.2. Standard functions from Economics :
(1) Demand : It refers to the quantity of a product is desired by the
buyers . The demand depends on the pric e. T herefore , there is a
relationship between the price and the quantity demanded. Such
relationship is called a demand function.
Hence the demand function is defined as
D=g(p) where D= demand and p= price .munotes.in
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Here demand is a dependent variable and the price is an
independent variable.
For example, D= 50 + 4 p -3p2
(2) Supply : It refers to the quantity of a product , the market can offer .
The supply depends on the price. Therefore, There is a relationship
between the price and the quantity supplied. Such relationship is called a
supply function.
Hence the supply function is defined as S = ƒ(p) where S = supply and
p = price.
Here supply is a dependent variable and price is an independent
variable.
For example , S = 2p2-6p + 25
(3) Break -even Point : Equilibrium point.
(i) By the law of demand, the demand decreases when the price increases,
the demand curve is a decreasing curve as shown in the figure :
D
D=ƒ(p)
The demand curve
P
(ii)By the law of supply, the supply increases when the price increases,
the supply curve is an increasing curve as shown in the figure.
S = g(p)
The supply curve
(iii) The demand and supply curves D= ƒ(p) and S= g(p) are intersecting
at a point . The point of intersection of the demand and supply curves
represents that specific price at which the demand and supply are equal.
This point is called the Break -even point or equ ilibrium point. The
corresponding price at which this point occu rs is called an equilibrium
price and is denoted by p e
At equilibrium price, the amount of goods supplied is equal to the
amount of goods demanded.munotes.in
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q D = f(p) S = (p)
Break –even point
q=ƒ(p) =D
q= g(p) = S
o pe p
equilibrium price -Break -even point
(4) The total cost function :
The total cost function or cost function is denoted by C and it is
expressed in terms of x. If C is the cost of producing xunits of a p roduct ,
then C is generally a function of xand is called the total cost function .
i.e.C=ƒ(x)
For example , C= 2 x2-5x+ 10
(5) Average cost function :
The ratio between the cost function a nd the number of units produced is
called average cost function . i.e. AC =C
x
For example , AC = x2+2x+5
x
(6) Total Revenue function :
The total revenue function is defin ed as in terms of the demand
and the price per item. If D units are demanded with the selling price of
p per unit , then the total revenue function R is given by
R = p x D where p = price a nd D= demand
For example ,
If D = p2+2p +3 th en R = p x (p2+2p +3) = p3+2p2+3p
(7) Average revenue :
Average revenue is defined as the ratio between the revenue and the
demand and is denoted by AR.
i.e. , AR =R/D AR = p x D/D (as R= p x D)
AR = p
Average revenue is nothing but the selling price per unit.
(8) The Profit function :
The profit function or the total profit function is denoted by P and
is defined by the difference between the total revenue and the total cost.
Total Profit = Total Revenue -Total cost
i.e. P = R -C
Example 1:
Find the total profit function if the cost function C= 40 + 15 x-x2,
x= number of items produced and the demand function is p= 200 -x2munotes.in
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Solution :
Given C= 40 + 15 x-x2
p = 200 -x2
R = p x D (D= x)
= (200 -x2)x
R = 200 x-x3
Profit = Revenue -Cost
P = R -C
= (200 x-x3)-( 40 + 15 x-x2)
= 200 x-x3-40-15x+x2
P= 185 x+x2-x3-40
Example 2:The total cost function is C= 20 -3x2and the demand
function is p= 5 + 6 x. Find the profit when x=100.
Solution :
Given : C= 20 -3x2
R= p x D (D= x)
= p x x
= (5 + 6 x)x
= 5x+ 6x2
Profit = Revenue -Cost
=R-C
= (5x+ 6x2)-( 20-3x2)
= 5x+ 6x2-20 + 3 x2
= 5x+ 9x2-20
When x= 100 , P =5(100) + 9 (100)2-20
=500 + 90000 -20
= 90480.
1.2DERIVATIVES
1.2.1 Derivativ e as rate measure :
Definition : Let y= ƒ(x) be the given function .
If lim [ƒ(x+h)-ƒ(x)]exists ,
h→0 h
then we say that the function ƒ(x) has derivative at xand is denoted by ƒ'
(x).
i.e. ,ƒ' (x) = lim [ƒ(x+h)-ƒ(x) ]
h→0 h
The rate of change is called the " derivative " of y= ƒ(x) with
respect to xand is denoted by dyorƒ' (x).
dx
dy= the rate of change of y with respect to xor the derivative of y
dx with respect to x.
Note : (1) Derivative means " rate of change "
(2)The process of finding the derivative of a fu nction is called "
differentiation".munotes.in
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dC= the rate of change cost with respect to x.
dx
For example dD=therateof change of demand with respect to p .
dp
1.2.2 Derivatives of Standard functions :
(1)Ify=xn, where n is a real number , then
dy= nxn-1
dx
i.e. , dy=d (xn)= nxn-1
dx dx
(2)If y = C , where C is a constant ,
then dy=0
dx
i.e. , dy=d (C) = 0
dxdx
(3) If y = ex, then dy=ex
dx
i.e. , dy=d (ex)=ex
dxdx
(4)If y = ax, where a isapositive real number , then
dy= axlog a
dx
i.e. , dy=d (ax)= axlog a
dx dx
(5) If y= log x, then dy=1
dxx
where x> 0
i.e. , dy=d (log x)=1
dx dx x
Examples :
(1) y= x=x'dy= 1.x1-1= 1.x0=1
dx
(2) y= x4dy= 4.x4-1= 4.x3
dx
(3) y= x10dy= 10. x10-1= 10x9
dx
(4) y= 1=x-1dy=-1.x-1-1=-1x-2=-1
x dx x2munotes.in
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(5) y= 1=x-3dy=-3x-3-1=-3x-4=-3
x3dx x4
(6) y=√x=x1/2dy=1x1/2-1=1x-1/2=1 =1
dx2 2 2 x1/22√x
(7) y= x5/2dy=5x5/2-1=5x3/2
dx2 2
(8) y= x-3/2
dy=-3x-3/2-1=-3x-5/2
dx 2 2
(9) y= x-7/2
dy=-7x-7/2-1
dx2
=-7x-9/2
2
(10) y=5 , 5 is a constant
dy= 0
dx
(11) y=K , dy=0
dx
(12) y= log2 , dy=0
dx
(13) y= -10dy= 0
dx
(14) y= exdy= ex
dx
(15) y= 2xdy= 2xlog 2
dx
(16) y= 10xdy= 10xlog 10
dx
(17) y= log xdy=1
dxx
Exercise: 1.1
Find dyfor the following :
dx
(1) y = x6
(2) y = 1
x2
(3) y = x7/2munotes.in
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(4) y = x-5/2
(5) y = 3
(6) y = log 10
(7) y = -8
(8) y = 4x
(9) y = 9x
(10)y = 15x
Answers :
(1) 6 x5(2)-2(3)7x5/2(4)-5x-7/2(5) 0 (6) 0 (7) 0 (8) 4xlog 4
x32 2
(9) 9xlog 9 (10) 15xlog 15
1.2.3 Rules of derivatives :
Rule : 1 Addit ion Rule (or) Sum rule :
If y= u + v where u and v are differentiable functions of xthen
dy=du+dv
dx dxdx
i.e,dy=d(u+v) =du+du
dx dx dxdx
Examples:
(1) If y = x2+ ex, find dy
dx
Solution : Given : y = x2+ ex
dy=d (x2+ ex)
dx dx
=d(x2)+d (ex)
dx dx
dy= 2x+ ex
dx
(2) If y= x10+ log x, find dy
dx
Solution: Given : y= x10+ log x
dy=d[x10+logx]
dxdx
=d(x10) +d(logx)
dx dx
dy= 10 x9+1
dx x
Rule:2 Subtraction Rule (or) Difference rule:
If y= u -v where u and v are differentiable functions of xthenmunotes.in
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dy=du-dv
dxdxdx
i.e,dy=d(u-v) = du-dv
dxdx dxdx
Examples :
(1) If y= x5-2x, find dy
dx
Solution : Given : y = x5-2x
dy=d (x5-2x)
dx dx
=d(x5)-d(2x)
dx dx
dy= 5x4-2xlog2
dx
(2) If y = 100 -logx, find dy
dx
Solution : Given : y = 100 -logx
dy=d(100 -logx)
dx dx
=d(100) -d(logx)
dx dx
= 0 -1
x
dy=-1
dx x
Rule : 3 Product Rule :
If y = uv where u and v are differentiable functions of x,
then dy= udv+ vdu
dx dx dx
i.e. , dy=d(uv) = u dv+ vdu
dx dx dx dx
Examples :
(1) If y = x4logx, find dy
dx
Solution : Given y = x4logx
u =x4, v = log x
dy=d(x4logx)
dxdxmunotes.in
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=x4d(logx)+logxd(x4)
dx dx
=x4(1/x) + log x(4x3)
=x3+ 4x3logx
dy=x3[ 1+ 4 log x]
dx
(2) If y = x2ex, find dy
dx
Solution : Given y = x2ex
dy=x2d(ex) + exd(x2)
dx dx dx
=x2ex+ex(2x)
dy= ex[x2+ 2x]
dx
Rule :4 Quotient Rule :
If y = u, v≠0where u and v are differentiable functions ofx, then
2du dvv udy dx dx
dx v
Examples :
(1) If y = x+ 4, find dy
logx dx
Solution: Given y= x+ 4
logx
Here u= x+ 4
v= log x
dy= log xd(x+ 4 ) -(x+ 4 ) d( logx)
dx dx dx
(logx)2
=(logx)(1+0) -(x+ 4) ( ¼)
(logx)2
dy=( logx)-(x+ 4) ( ¼)
dx (logx)2
(2) If y = ex+5, find dy
x6-10 d x
Solution :
Given : y = ex+5
x6-10munotes.in
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Here u = ex+ 5 ; v= x6-10
dy= (x6-10)d(ex+5)-(ex+5)d(x6-10)
dx dx dx
(x6-10 )2
dy=(x6-10 ) (ex)-( ex+5) (6 x5)
dx (x6-10)2
(3) If y = x3-1, find dy
x3+1 dx
Solution :
Given : y =x3-1
x3+1
Here u = x3-1
v =x3+1
dy= (x3+1)d(x3-1)-(x3-1)d(x3+1)
dx dx dx
(x3+1)2
=(x3+ 1) (3 x2)-(x3-1) (3 x2)
(x3+ 1)2
=3x2[ (x3+ 1) -(x3-1)]
(x3+1)2
=3x2[x3+ 1-x3+ 1]
(x3+1)2
=3x2[2]
(x3+ 1)2
dy
dx=2
26
( 1)x
x
Rule 5 : Scalar multiplication rule or constant mul tiplied by a function
rule.
If y = cu , c is a constant , where u is a differentiable function of x,then
dy= cdu
dx dx
i.e.,dy=d(cu) = c du
dxdx dx
Examples :
(1) If y= 5 x3, find dy
dx
Solutio n : Given : y= 5 x3
dy=d(5x3)munotes.in
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dxdx
= 5d(x3) = 5(3 x2)= 15 x2
dx
(2)If y= 10 log x, find dy
dx
Solution : Given : y = 10 log x
dy=d(10 log x)
dxdx
= 10 d(logx)
dx
= 10 (1/ x)
=10
x
1.2.4 List of formulae :
y =ƒ(x) dy=ƒ' (x)
dx
1. xnnxn-1
2. C , C = constant 0
3. exex
4. axaxlog a
5. log x 1/x
6. x 1
7. √x 1/ 2√x
8. 1/ x -1/x2
1.2.5 List of Rules :
(1) d(u+v) = du+dv
dx dx dx
(2) d(u-v) = du-dv
dx dxdx
(3) d(uv) =u dv+vdu
dx dx dx
(4) d(u/v) =v du-udv
dx dx dx
v2
(5) d(cu) = c du, c = constant.
dx dx
1.2.6 Examples :
Find dyfor each of the following :
dx
Ex: (1) y= x6+ 4 ex+ log x+10
Solution:
dy=d(x6+ 4 ex+ log x+10)munotes.in
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dxdx
=d(x6) + d(4ex) +d(logx) +d(10)
dx dx dx dx
=d(x6) + 4 d(ex) +d(logx) +d(10)
dx dx dx dx
=6x5+4ex+ 1/x+ 0
dy=6x5+4ex+ 1/x
dx
Ex : ( 2) y= 5 x4-3ex+4√x+ 2x
Solution :
dy=d(5x4-3ex+4√x+ 2x)
dxdx
=d(5x4)-d(3ex) +d(4√x) +d(2x)
dx dx dx dx
=5d(x4)-3d(ex) + 4 d(√x) +d(2x)
dx dx dx dx
= 5(4 x3)-3ex+ 4(1/2√x) + 2Xlog 2
= 20 x3-3ex+ 2/√x+2Xlog 2
Ex: (3) y= x3/2+ 4 log x-10x2+ 15
Solution:
dy=d(x3/2+ 4 log x-10x2+ 15)
dxdx
=d(x3/2) +d(4 log x)-d(10x2)+d(15)
dx dx dx dx
=3x3/2-1+ 4 ( 1)-10(2x) + 0
2 x
dy=3x1/2+4-20x.
dx2 x
Ex: (4) y= ( x+ ex) (5+log x)
Solution: Here u= x+ ex
v= 5+ log x
dy= (x+ ex)d(5+log x) + (5+log x)d(x+ ex)
dx dx dx
= (x+ ex) (0+1/ x) + (5+ log x) (1+ex)
dy= (x+ ex) (1/x) + (5+ log x) (1+ex)
dxmunotes.in
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Ex: (5) y= ( x10) (10x)
Solution : Here u= x10, v = 10x
dy = x10d(10x) + 10xd(x10)=x10(10xlog 10) + 10X(10x9)
dx dx dx
Ex:(6) y= (√x+ex)(2x3+7)
Solution: Here u=√x+ex, v= 2x3+7
dy=(√x+ex)d(2x3+7) + (2 x3+7)d(√x+ex)
dx dx dx
dy = (√x+ex) (6x2) + (2 x3+7) ( 1/2√x+ ex)
dx
Ex: (7) y = x2+ 5x+6
x+ 7
Solution : Here u= x2+ 5x+ 6
v=x+ 7
dy=(x+ 7) d(x2+ 5x+ 6) -(x2+ 5x+ 6) d(x+ 7)
dx dx dx
(x+ 7)2
=(x+ 7) (2 x+5)-(x2+ 5x+ 6) (1)
(x+ 7)2
=2x2+ 19x+ 35 -x2-5x-6
(x+ 7)2
dy=x2+ 14x+ 29
dx (x+ 7)2
Ex: 8y=10ex+ 5logx
x3+ 12
Solution : Here u= 10ex+ 5 log x
v=x3+ 12
dy=(x3+ 12) d(10 ex+ 5 log x)-(10 ex+ 5 log x)d(x3+ 12)
dx dx dx
(x3+ 12)2
dy=(x3+ 12) (10ex+ 5/x)-(10 ex+ 5 log x) ( 3x2)
dx (x3+ 12)2
Ex:9 y=4x+ 6
2x2+5
Soluti on : Here u = 4x+ 6 , v=2 x2+5munotes.in
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dy= (2x2+5)d(4x+ 6) -(4x+ 6)d(2x2+5)
dx dx dx
(2x2+5)2
=(2x2+5)(4xlog 4) -(4x+6)(4 x)
(2x2+5)2
Ex: 10 y= 2 √x+ 16ex+(6x)+ 20x
Solution:
dy=d(2√x+16ex+ 6x+ 20x)
dxdx
= 2 ( 1/2√x) + 16ex+ 6xlog 6 + 20
dy= 1/√x+ 16ex+ 6xlog 6 + 20
dx
Exercise : 1.2
Differentiate the f ollowing with respect to x.
(1) y = x8-6ex+ 4x3/2-3x2+ 5
(2) y = 6 log x-3x+ 2ex+ 10√x+2
(3) y = 5 x4-12x3+ 18 ex+ 10x-25
(4) y = 8x(5x3+3x+1)
(5) y = (10 x2+ 2x+5) (√x+ex)
(6) y = (2 x3+3x2) (5log x+ 14)
(7) y = (x+ log x) (x5-4x2+ 10)
(8) y = (8 x5-6x5/2+ 1) (40√x+ 2ex)
(9) y = (ex+ 2 log x+ 2) (6x+ 2x2+5)
(10)y = x2+ 1
x4-1
(11)y = 2x+4√x
2ex+ 5
(12)y = x3-x2+ 2
x2-4
(13)y = x+√x
√x-1
(14)y= 3log x+5
x5+ 2x
(15)y = ex-√x
2√x+1munotes.in
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Answers :
(1) 8 x7-6ex+ 6x1/2-6x
(2) 6/ x-3xlog3 + 2ex+5/√x
(3) 20 x3-36x2 + 18ex+ 10xlog 10
(4) 8x(15x2+3) + (5 x3+ 3x+1) (8xlog8)
(5) (10 x2+2x+5)(1/2√x+ ex) +(√x+ex)(20x+2)
(6) (2 x3+3x2) (5/x) + (5log x+14) (6 x2+ 6x)
(7) (x+logx) (5x4-8x) + (x5-4x2+10) (1+1/ x)
(8) (8 x5-6x5/2+1) (20/√x+2ex) + (40√x+2ex) (40 x4-15x3/2)
(9)(ex+ 2logx+2) (6xlog6+4x) + (6x+ 2x2+5) ( ex+2/x)
(10) -2x5-4x3-2x
(x4-1)2
(11) (2ex+5) (2xlog2+2/√x)-(2x+ 4√x) (2ex)
(2ex+5)2
(12) x4-12x2+ 4x
(x2-4)2
(13)√x-1-(1+x)/2√x
(√x-1)2
(14) (x5+2x) (3/x)-(3logx+5) (5 x4+2)
(x5+2x)2
(15) (2√x+1) (ex-1/2√x)-(ex-√x) (1/√x)
(2√x+1)2
1.3 SECOND ORDER DERIVATIVES
If y=ƒ(x) is differentiable function with respect to x, then
dy=ƒ'(x) is called first order derivative of y with respect to xand
dx
d2y=ddy is called the second order derivative of y with
dX2 dx dx
respect to X
Notation :
(i) First order derivative :
dy=ƒ'(x) = y 1= y'
dx
(ii) Second order derivative :
d2y=ƒ"(x) = y 2=y"
dx2munotes.in
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1.3.1 Examples :
(1) If y = x3-6x2+ 19 x+ 100 , find d2y
dx2
Solution :
Given : y = x3-6x2+ 19 x+ 100
dy=3x2-12x+19
dx
d2y=d(3x2-12x+19)
dx2dx
d2y= 6x-12
dx2
(2) If y = ex+2x3+5x2+4 , find d2y
dx2
Solution :
Given : y = ex+2x3+5x2+4
dy= ex+ 6x2+10x
dx
d2y=d(dy/d x)
dx2dx
=d(ex+ 6x2+10x)
dx
d2y= ex+ 12x+10
dx2
(3) If y= ƒ(x)=x5-6x3+ 2x2+10x+5, findƒ"(x)
Solution:
Givenƒ(x) =x5-6x3+2x2+ 10x+5
ƒ'(x) = 5 x4-18x2+4x+10
ƒ"(x)= 20 x3-36x+ 4
(4) If y=x+1/x,x≠ 0, find d2y/dx2
Solution: Given: y= x+1/x
dy/dx= 1+( -1/x2) = 1-1/x2
d2y=0-(-2/x3) =2
dx2x3
Exercise:9.3
Find d2yfor each of the following:
dx2
(1) y= 8 x5-16x4+4x3+x+2
(2) y= 2 x3-5x2+12x+15
(3) y= x+25/x
(4) y= 2 x2+ex+5x+12
(5) y= x2+x+logxmunotes.in
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Answers:
1) 160 x3-192x2+24x
2) 12 x-10
3)50
x3
4) 4+ex
5) 2-1
x2
1.4 APPLICATIONS OF DERIVATIVES
Applications to Economics:
1.4.1 The Total Cost function:
(i) Total cost function C= ƒ(x)
(ii) Average cost function AC= C
x
(iii) Marginal cost function:
The rate of cha nge of cost with respect to the number of units
produced is called the Marginal cost and is denoted by MC.
i.e, MC= dC
dx
1.4.2 The Total Revenue Function:
(i) Total Revenue Function R = p x D
(ii) Average Revenue Function AR= p
(iii) Marginal Revenue function:
The rate of change of total revenue with respect to the
demand D is called
the Marginal revenue function and is denoted by MR.
i.e MR = d R
d D
1.4.3 Elasticity:
Let D be the demand and p be the price. The quantity -PdDis
D dP
calledelasticity of demand with respect to the price and is denoted by η
(η = eta : Greek alphabet).
i.e, η = -pdD
D d pmunotes.in
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If η = 0, demand D is Constant and the demand is said to be perfectly
elastic.
If 0< η < 1, the demand is said to be inelastic
If η = 1, the demand is directly proportional to the price.
If η > 1, the demand is said to be elastic.
1.4.4 Relation between the Marginal Revenu e and elasticity of
demand.
Let R = Total revenue
p = price
D = demand
R = pD
MR = dR
dD
=d(pD)
dD
= p d(D) + D. d(p) (by product rule)
dD dD
= p (1) + D dp
dD
MR = p + D dp
dD
η = -pdD
D dp
Ddp= -p
dD η
MR = p + ( -p)
η
MR = p [ 1 -1]
η
MR = AR [ 1 -1]
η (AR = p )
1.4.5 Examples:
Ex: (1) The cost of producing ҳ items is given bymunotes.in
20
2 x2+ 5x + 20. Find the total cost, average cost and marginal cost
when x= 10.
Solution:
Let C = ƒ(ҳ) = 2x2+ 5x+20
AC = C=2x2+ 5x+20
x x
MC = dC
dx
=d(2x2+ 5x+20)
dx
MC = 4x + 5
when x = 10
C = 2 ( 10)2+ 5 (10 ) + 20
C = 270
AC = C=270 = 27
x 10
MC = 4 ( 10 ) + 5 = 45
Ex: ( 2)
The demand function is given by ρ = 50 + 6 D + 4D2.
Find t he total revenue, average revenue and the marginal revenue when
the demand is 5 units.
Solution:
Given : ρ = 50 + 6 D + 4 D2
R = ρ x D
= ( 50 + 6 D + 4 D2) ( D)
R = 50 D + 6 D2+ 4 D3
AR= ρ = 50 + 6 D + 4 D2
MR =dR
dDmunotes.in
21
=d( 50 D + 6 D2+ 4D3)
dD
MR = 50 + 12 D + 12 D2
When D = 5
R = 50 ( 5) + 6 ( 5 )2+ 4 ( 5 )3
= 250 + 150 + 500
R = 900
AR = 50 + 6 (5) + 4 (5)2
= 50 + 30 + 100
= 180
MR = 50 + 12 ( 5 ) + 12 ( 5 )2
=50 + 60 + 300
= 410
Ex : (3)
The total revenue function is given by R = 20 D -D2, D =
Demand. Find the demand function. Also find AR when MR = 0 .
Solution :
Given R = 20 D -D2
ρD = 20 D -D2( R = ρD )
ρ = 20D -D2
D
ρ = 20 -D
The demand function is ρ = 20 -D .
Now, MR = dR
dD
=d( 20D -D2)
dD
MR = 20 -2Dmunotes.in
22
Given that MR = 0.
0 = 20 -2D
2D = 20
D = 10
AR = ρ = 20 -D
AR = 20 -10
AR = 10
Ex : ( 4 )
The demand function is given by
D = 25 - 2 ρ - ρ2
Find the elasticity of demand when the price is 4.
Solution :
Given : D = 25 - 2 ρ - ρ2
dD= 0-2- 2 ρ
dp
dD=-2- 2 ρ
dp
η = - ρ dD
D dρ
=(- ρ) (-2- 2 ρ )
25- 2ρ - ρ2
η = p( 2 + 2 p)
25-2p-p2
When p = 4,
η = 4 ( 2 + 8 )
25-8-16
=4 ( 10 )
1
η = 40munotes.in
23
Ex: 5
The demand function is given by D = p + 3
2p-1
where D = Demand and p = price. Find the elasticity of demand whe n the
price is 8.
Solution:
Given D = p + 3
2p-1
( 2p-1 )d( p + 3 ) -( p + 3 ) d( 2p-1 )
dD dp dp
dp =
(2p-1)2
= ( 2p-1) (1) -(p +3) (2)
(2p-1)2
= 2p-1-2p-6
(2p-1)2
dD= -7
dp (2p-1)2
η = -p dD
D dp
(-p) [ -7 ]
= (2p-1)2
[(p +3)/(2p -1)]
=7 p ( 2 p -1 )
( p + 3 ) ( 2 p -1 )2
η = 7 p
( p + 3 ) ( 2 p -1 )
When p = 8
η = 7 ( 8)
( 8 + 3 ) [ 2 ( 8 ) -1 ]
= 56
( 11) (15)munotes.in
24
=56
165
η = 0.34
Ex : (6) If MR = 45, AR = 75, Find η
Solution:
Given MR = 45
AR = 75
η = ?
MR =AR [ 1- 1/η ]
45 = 75 [ 1 - 1/ η ]
45= 1 - 1/ η
75
3= 1-1 / η
5
0.6 = 1 - 1 / η
1 / η = 1 -0.6
1 / η = 0.4
η = 1 / 0.4
= 2.5
Ex : ( 7)
If AR = 95 and η = 7/ 2 , Find MR.
Solution : Given AR = 95
η = 7/2 = 3.5
MR = ?
MR = AR [ 1 - 1 / η ]
= 95 [ 1-1 / 3.5 ]
= 95 [ 1 -0.29 ]
= 95 [ 0.71]
MR = 67.45munotes.in
25
Exercise : 1.4
(1) The cost of producing x items is given by x3+ 4x +15. Find the total
cost, average cost and marginal cost when x = 6.
(2) The total cost function is given by C= x3+2 x2+ 5 x + 30. Find the
total cost, average cost and marginal cost when x = 10.
(3) The demand function is given by p= 20 -8 D + 3 D2. Find the total
revenue, average revenue and marginal revenue when the demand is
4 units.
(4) The total revenue f unction is given by R= 30 D -2 D2+ D3.Find
the demand function. Also find total revenue, average revenue and
marginal revenue when the demand is 5 units.
(5) The demand function is given by D = -28-5p + 2 p2Find the
elasticity of demand when t he price is 3.
(6) The demand function is given by D = 2p + 5 Where D = Demand
and p-3
p = price. Find the elasticity of demand when price is 6.
(7) IfAR = 65 and η = 3, find MR.
(8)If MR = 85 and η = 4.5 , find AR.
(9)If MR = 55 and AR = 98 , find η .
(10) If the price is 65 and the elasticity of demand is 5.2, find the
marginal revenue.
Answers:
(1) C = 255 ; AC = 42.5 ; MC = 112
(2) C = 1280 ; AC = 128 ; MC = 345
(3) R = 144 ; AR = 36 ; MR = 100
(4) p = 30 -2D + D2
R = 225 ; AR = 45 ; MR = 85
(5)η = 0.64
(6)η = 1.29
(7) MR = 43.3munotes.in
26
(8) AR = 108.97
(9)η = 2.28
(10) MR = 52.5
1.5 MAXIMA AND MINIMA
Let y =ƒ(x) be th e given function. A curve ƒ(x) is said to have a
maximum or minimum point (extreme point), if ƒ(x) attains either a
maximum or minimum of that point.
Y
y = f(x)
f(a)
0 x=a X
Maximum at x = a
Y
y = f (x)
(f(c) b
0 x=b X
Maximum at x = b
In the first figure, x = a is the point where the curve ƒ(x) attains a
maximum. In the second figure, x = b is the point where the curve ƒ(x)
attains a minimum
1.5.1 Conditions for Maximum & Minimum:
1. Condition for Maximum:
(i)ƒ' (x) = 0
(ii)ƒ" (x) < 0 at x = a
2. Condition for Minimum:
(i) ƒ' (x) = 0
(ii)ƒ" (x) > 0 at x = bmunotes.in
27
1.5.2. To find the Maximum and Minimum v alues of ƒ(x) :
Steps:
(i) Findƒ'(x) andƒ"(x) .
(ii) putƒ'(x) = 0 , solve and get the values of x.
(iii) Substitute the values of x in ƒ"(x) .
Ifƒ"(x) < 0 , then ƒ(x) has maximum value at x = a. If ƒ"(x) > 0, then
ƒ(x) has minimum valu e at x= b.
(iv) To find the maximum and minimum values, put the points
x = a and x = b in ƒ(x).
Note: Extreme values of ƒ(x) = Maximum and minimum values of ƒ(x).
1.5.3. Examples :
Ex: (1) Find the extreme values of ƒ(x) = x3-3x2-45x + 25.
Solution:
Given : ƒ(x) = x3-3x2-45x + 25
ƒ'(x) = 3x2-6x-45
ƒ"(x) = 6x -6.
Sinceƒ(x) has maximum or minimum value,
ƒ'(x) = 0
3x2-6x-45 = 0
3 (x2-2x-15) = 0
x2-2x-15 = 0
x2-5x + 3x -15 = 0
x ( x -5 ) + 3 (x -5 ) = 0
( x-5 ) ( x + 3 ) = 0
x-5 = 0 or x + 3 = 0
x = 5 or x = -3.munotes.in
28
When x = 5, ƒ"(5)=6(5) -6=24>0
ƒ(x) has minimum at x = 5.
When x = -3,ƒ"(-3 )= 6 ( -3 )-6
=-24 < 0
ƒ(x) has maximum at x = -3.
To find the maximum and minimum values of ƒ(x) :
putx = 5 and x = -3 inƒ(x) .
ƒ(5) = 53-3 ( 5 )2-45(5) + 25
= 125 -75-225 + 25
=-150
ƒ(-3 ) = ( -3 )3-3 (-3 )2-45(-3 ) + 25
=-27-27 + 135 + 25
= 106
Maximum value = 106 at x = -3
Minimum value = -150 at x = 5.
Ex: ( 2 ) Find the maximum and minimum values of ƒ(x) = x +16
x,x≠ 0.
Solution : Given :ƒ( x ) = x + (16 / x )
ƒ'( x ) = 1 + 16 ( -1 / x2)
ƒ'( x ) = 1 -16
x2
ƒ"( x ) = 0 -16 ( -2)
x3
ƒ"( x ) = 32
x3
Sinceƒ( x ) has maximum or minimum value, ƒ'( x ) = 0.
1-16= 0
x2
x2-16= 0
x2munotes.in
29
x2-16 = 0
x2-42= 0
( x-4 ) ( x + 4 ) = 0
x = 4 or x = -4
when x = 4 , ƒ"( 4 ) = 32=32=1> 0
( 4 )364 2
ƒ( x ) has minimum at x = 4.
when x = -4 ,ƒ"(-4 ) = 32 =32=-1< 0
(-4 )3-64 2
ƒ( x ) has maximum at x = -4.
Now to find the extreme values of ƒ( x ) : put x = 4 and x = -4 inƒ( x )
ƒ( 4 ) = 4 + 16 = 4 + 4 = 8
4
ƒ(-4 ) = -4 + 16=-4-4 = -8
(-4)
Maximum value = -8 at x = -4
Minimum value = 8 at x = 4.
Ex: ( 3) Divide 80 into two parts such that the sum of their squares is a
minimum.
Solutio n:
Let x and 80 -x be the two required numbers.
By the given condition,
ƒ(x) = x2+ ( 80 -x)2
ƒ(x)= x2+ 802-2(80) (x) + x2
ƒ(x) = 2x2-160x + 6400
ƒ'(x)= 4x -160
ƒ"(x) = 4
Sinceƒ(x) has minimum,
ƒ'(x)= 0
4x-160= 0
4x=160munotes.in
30
x=160=40.
4
ƒ" (x) = 4 > ƒ" (40)= 4>0.
ƒ(x) has minimum at x= 40.
The required numbers are
40 and 80 -40=40
The required parts of 80 are
40 and 40.
Ex: (4)
A manufacturer can sell x items at a price of Rs. (330 -x) each. The
cost of p roducing x items is Rs. (x2+ 10x+12). Find x for which the profit
is maximum.
Solution:
Given that the total cost function is
C= x2+ 10x+ 12.
Selling price p= 330 -x
Revenue is R= p x D
= p x x (D= x)
= (330 -x) x
= 330x -x2
Profit= Revenue -Cost
P= R -C
= (330x -x2)-(x2+10x+ 12)
= 330x -x2-x2-10x-12
P= 320x -2x2-12
dp= 320 -4x
dx
d2p=-4 < 0
dx2
The profit is maximum.
Since the profit is maximum,
dp= 0
dxmunotes.in
31
320-4x= 0
4x=320
x=80.
Hence the profit is maximum when 80 items are sold.
Ex: (5) The total cost function is
C= x3-9x2+ 24x+ 70.
Find x for whic h the total cost is minimum.
Solution:
Let C =ƒ(x)= x3-9x2+ 24x+ 70
C'=ƒ' (x) = 3x2-18x+24
C"=ƒ" (x) = 6x -18
Sinceƒ(x) has minimum, ƒ' (x) = 0
3x2= 18x+ 24=0
3(x2-6x+8) = 0
x2-6x +8=0
x2-2x-4x+8=0
x(x-2)-4(x-2)=0
(x-2)(x-4)= 0
x-2=0 or x -4=0
x=2 or x=4
When x=4
ƒ"(x) = 6x -18
ƒ" (4) = 6(4) -18
= 6>0
ƒ(x) has minimum at x=4.
The t otal cost is minimum at x=4.
Ex: (6) The total revenue function is given by
R = 4x3-72x2+ 420x+ 800.
Find x for which the total revenue is maximum.
Solution:
Let R =ƒ(x)= 4x3-72x2+ 420x+800
R'=ƒ'(x) = 12x2-144x+420
R" =ƒ" (x) = 24x -144
Sinceƒ(x) has maximum,munotes.in
32
ƒ'(x)= 0
12x2-144x+420=0
12(x2-12x+35)=0
x2-12x+35=0
x2-5x-7x+35=0
x(x-5)-7(x-5)=0
(x-5)(x-7)=0
x-5=0 or x -7=0
x=5 or x=7
When x=5
R"=ƒ"(x) = 24x -144
= 24(5) -144
=-24<0
ƒ(x) has maximum at x=5
The total revenue is maximum at x=5
Exercise : 1.5
(1) Find the extreme values of ƒ(x)=2x3-6x2-48x+90.
(2) Find the maximum and minimum values of ƒ(x) = x+ (9/x), x0
(3) Find the extreme values of ƒ(x)= 4x3-12x2-36x+25
(4) Find the extreme values of ƒ(x) =x + (36/x), x0.
(5) Divide 120 into two parts such that their product is maximum.
(6) Divide 70 into two parts such that the sum of their squares is a
minimum.
(7) A manufacturer sells x items at a price of Rs. (400 -x) each. The cost of
producing x items is Rs (x2+ 40x+52). Find x for which the profit is
maximum.
(8) The cost function is given by C= x3-24x2+ 189x+120. Find x for
which the cost is minimum,
(9) The total revenue function is given by
R= 2x3-63 x2+ 648x+250.
Find x for which the total revenue is maximum.
(10) The total cost of producing x units is Rs.(x2+ 2x+5) and the price is
Rs.(30-x) per unit. Find x for which the profit is maximum.munotes.in
33
Answers:
(1) Maximum value= 146 at x= -2
Minimum value = -70 at x=4
(2) Maximum value= -6 at x= -3
Minimum value = 6 at x= 3
(3) Maximum value= 45 at x= -1
Minimum value= -83 at x =3
(4) Maximum value = -12 at x= -6
Minimum value = 12 at x=6
(5) 60,60
(6) 35,35
(7) 90
(8) 9
(9) 9
(10) 7
munotes.in
34
Unit II
Unit -2
SIMPLE INTEREST AND
COMPOUND INTEREST
Unit Structure :
2.0 Objectives
2.1 Introduction
2.2 Definitions of Terms Used In This Chapter
2.3 Simple Interest
2.4 Compound Interest
2.0OBJECTIVES
After reading this chapter you will be abl e to:
Define interest, principal, rate of interest, period.
Find simple interest (SI), rate of S.I., period of investment.
Find Compound Interest (CI), rate of C.I., Amount accumulated at the
end of a period.
Compound interest compounded yearly, half -yearl y, quarterly or
monthly.
2.1INTRODUCTION
In every day life individuals and business firms borrow money
from various sources for different reasons. This amount of money
borrowed has to be returned from the lender in a stipulated time by paying
some interest on the amount borrowed. In this chapter we are going to
study the two types of interests viz. simple and compound interest. We
start with some definitions and then proceed with the formula related to
both the types of interests.
2.2DEFINITIONS OF T ERMS USED IN THIS CHAPTER
Principal : The sum borrowed by a person is called its principal . It is
denoted by P.
Period : The time span for which money is lent is called period . It is
denoted by n.munotes.in
35
Interest : The amount paid by a borrower to the lender for the use of
money borrowed for a certain period of time is called Interest . It is
denoted by I.
Rate of Interest : This is the interest to be paid on the amount of Rs. 100
per annum (i.e. per year). This is denoted by r.
Total Amount : The sum of the princ ipal and interest is called as the total
amount (maturity value )and is denoted by A. Thus, A=P + I .
i.e. Interest paid I=A–P.
2.3SIMPLE INTEREST
The interest which is payable on the principal only is called as
simple interest (S.I.). For ex ample the interest on Rs. 100 at 11% after one
year is Rs.11 and the amount is 100 + 11 = Rs. 111.
It is calculated by the formula: I =100Pnr=Pxnx100r
Amount at the end of nthyear = A =P + I =P +100Pnr=P1100nr
Remark : The period nis always taken in ‘years’. If the period is given in
months/days, it has to be converted into years and used in the above
formula. For example, if period is 4 months then we take n=4/12 = 1/3 or
if period is 60 days then n= 60/365.
Example 1:If Mr. Sagar borrows Rs. 500 for 2years at 10% rate of
interest, find (i) simple interest and (ii) total amount.
Ans: Given P= Rs. 500,n=2andr=10 %
(i)I=100Pnr=500 x 2 x 10
100= Rs. 100
(ii)A = P + I = 500 + 100 = Rs. 600
3.3.1Problems involving unknown factors in the formula I =100Pnr
The formula I =100Pnrremaining the same, the unknown factor in
the formula is taken to the LHS and its value is computed. For example, if
rate of interest is unknown then the formula is rewritten asx 100
P xIrn .
Example 2:If Mr. Prashant borrows Rs. 10 00 for 5 years and pays an
interest of Rs. 300, find rate of interest.
Ans: Given P=1000,n= 5 and I= Rs. 300Simple Interest = Prinicpal x period x rate of interest
munotes.in
36
Now, I=100Pnr x 100
P xIrn =300 x 100
1000 x 5=6
Thus, the rate of interest is 6%.
Example 3:Find the period for Rs. 2500 to yield Rs. 900 in simple
interest at 12 %.
Ans: Given P= Rs. 2500, I= 900, r= 12%
Now, I=100Pnr x 100
xInP r =900 x 100
2500 x 12= 3
Thus, the period is 3 years.
Example 4:Find the period for Rs. 1000 to yield Rs. 50 in simple interest
at 10%.
Ans:Given P= Rs. 1000, I= 50, r= 10%
Now, I=100Pnr x 100
xInP r =50 x 100
1000 x 10= 0.5
Thus, the period is 0.5 years i.e. 6 months.
Example 5:Mr. Ak ashlent Rs. 5000 to Mr. Prashantand Rs. 4000 to Mr.
Sagarfor 5 years and received total simple interest of Rs. 4950. Find (i)
the rate of interest and (ii) simple interest of each.
Ans: Let the rate of interest be r.
S.I. for Prashant=5000 x 5 x
100r=250r … (1)
and S.I. for S agar =4000 x 5 x
100r= 200 r … (2)
from (1) and (2), we have,
total interest from both = 250 r+ 200 r
= 450 r
But total interest received be Mr. Ak ash= Rs. 4950
450r= 4950 r=4950
450= 11
the rate of interest = 11%
Example 6:The S.I. on a sum of money is one -fourth the principal. If the
period is same as that of the rate of interest then find the rate of interest.
Ans: Given I=4Pandn = r
Now, we know that I=100Pnr
4P=x x
100P r r100
4=r2
r2= 25 r= 5.
the rate of interest = 5%munotes.in
37
Example 7:If Rs. 8400 amoun t to Rs. 11088 in 4 years, what will Rs.
10500 amount to in 5 years at the same rate of interest?
Ans:
(i) Given n= 4,P= Rs. 8400, A= Rs. 11088
I=A–P= 11088 –8400 = Rs. 2688
Letrbe the rate of interest.
Now, I=100Pnr2688 =8400 x 4 x
100r
r= 8%
(ii) To find Awhen n= 5,P= Rs. 10500, r= 8
A = P 1100nr= 10500 x5 x 81100 = 10500 x140
100= 14700
the required amount = Rs. 14,700
Example 8: Mr. Shirish borrowed Rs. 12,000 at 9% interest from Mr.
Girish on January 25, 2007. The interest and principal is due on August
10, 2007. Find the interest and total amount paid by Mr. Shirish.
Ans: Since the period is to be taken in years, we first count number of
days from 25thJanuary t o 10thAugust, which is 197days.
Now, I=100Pnr= 12000 x197
365x9
100
I= Rs. 582.9
Total amount = P + I = 12000 + 582.9
A= Rs. 12 ,582.9
Check your progress 10.1
1.Find the SI and amount for the following data giving principal, rate
of interest and number of years:
(i) 1800, 6%, 4 years. (ii) 4500, 8%, 5 years
(iii) 7650, 5.5%, 3 years. (iv) 6000, 7.5%, 6 years
(v) 25000, 8%, 5 years (vi) 20000, 9.5%, 10 years.
Ans: (i) 432, 2232 (ii) 1800, 6300, (iii) 1262.25, 8912.25
(iv) 2700, 8700 (v) 10000, 35000 (vi) 19000, 39000
2.Find the S.I. and the total amount for a principal of Rs. 6000 for 3
years at 6% rate of interest.
Ans: 1080, 7080
3.Find the S.I. and the total amount for a principal of Rs. 3300 for 6
years at 3½ % rate of interest.
Ans:693, 3993January 6
February 28
March 31
April 30
May 31
June 30
July 31
August 10
Total 197
munotes.in
38
4.Find the S.I. and the total amount for a principal of Rs. 10550 for 2
years at 10¼ % rate of interest.
Ans:2162.75, 1271 2.75
5.Find the rate of interest if a person invests Rs. 1000 for 3 years and
receives a S.I. of Rs. 150.
Ans: 5%
6.Find the rate of interest if a person invests Rs. 1200 for 2 years and
receives a S.I. of Rs. 168.
Ans: 7%
7.A person invests Rs. 4050 in a bank which pays 7% S.I. What is the
balance of amount of his savings after ( i) six months, ( ii) one year?
Ans: 141.75, 283.5
8.A person invests Rs. 3000 in a bank which offers 9% S.I. After how
many years will his balance of amount will be Rs. 3135?
Ans: 6 mont hs
9.Find the principal for which the SI for 4 years at 8% is 585 less than
the SI for 3½ years at 11%.
Ans: 9000
10.Find the principal for which the SI for 5 years at 7% is 250 less than
the SI for 4 years at 10%.
Ans: 5000
11.Find the principal for which t he SI for 8 years at 7.5% is 825 less
than the SI for 6½ years at 10.5%.
Ans: 10000
12.Find the principal for which the SI for 3 years at 6% is 230 more
than the SI for 3½ years at 5%.
Ans: 46000
13.After what period of investment would a principal of Rs. 12,3 50
amount to Rs. 17,043 at 9.5% rate of interest?
Ans: 4 years
14.A person lent Rs. 4000 to Mr. Xand Rs. 6000 to Mr. Yfor a period
of 10 years and received total of Rs. 3500 as S.I. Find ( i) rate of
interest, ( ii) S.I. from Mr. X, Mr. Y.
Ans: 3.5%, 1400, 2 100
15.Miss Pankaj Kansra lent Rs. 2560 to Mr. Abhishek and Rs. 3650 to
Mr. Ashwin at 6% rate of interest. After how many years should he
receive a total interest of Rs. 3726?
Ans: 10 yearsmunotes.in
39
16.If the rate of S.I. on a certain principal is same as that of the p eriod
of investment yields same interest as that of the principal, find the
rate of interest.
Ans: 10%
17.If the rate of S.I. on a certain principal is same as that of the period
of investment yields interest equal to one -ninth of the principal, find
the rate of interest.
Ans: 31
3years
18.Find the principal and rate of interest if a certain principal amounts
to Rs. 2250 in 1 year and to Rs. 3750 in 3 years.
Ans: 1500, 50%
19.Find the principal and rate of interest if a certain principal amounts
to Rs. 3340 in 2 years and to Rs. 4175 in 3 years. Ans: 1670, 50%
20.If Rs. 2700 amount Rs. 3078 in 2 years at a certain rate of interest,
what will Rs. 7200 amount to in 4 years at the same rate on interest?
Ans: 7%, 9216
21.At what rate on interest will certain sum of money amount to three
times the principal in 20 years?
Ans: 15%
22.Mr. Chintan earns as interest Rs. 1020 after 3 years by lending Rs.
3000 to Mr. Bhavesh at a certain rate on interest and Rs. 2000 to Mr.
Pratik at a rate on interest 2% m ore than that of Mr. Bhavesh. Find
the rates on interest.
Ans: 6%, 8%
23.Mr. Chaitanya invested a certain principal for 3 years at 8% and
received an interest of Rs. 2640. Mr. Chihar also invested the same
amount for 6 years at 6%. Find the principal of Mr. Chaitanya and
the interest received by Mr. Chihar after 6 years.
Ans: 11000, 3960
24.Mr. Ashfaque Khan invested some amount in a bank giving 8.5%
rate of interest for 5 years and some amount in another bank at 9%
for 4 years. Find the amounts invested in bot h the banks if his total
investment was Rs. 75,000 and his total interest was Rs. 29,925.
Ans: 45000, 30000
25.Mrs. Prabhu lent a total of Rs. 48,000 to Mr. Diwakar at 9.5% for 5
years and to Mr. Ratnakar at 9% for 7 years. If she receives a total
interest of Rs. 25,590 find the amount she lent to both.
Ans: 18000, 30000munotes.in
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2.4COMPOUND INTEREST
The interest which is calculated on the amount in the previous year
is called compound interest .
For example, the compound interest on Rs. 100 at 8% after one
year is Rs. 8 and after two years is 108 + 8% of 108 = Rs. 116.64
IfPis the principal, ris the rate of interest p.a. then the amount at the end
ofnthyear called as compound amount is given by the formula:
Thecompound interest is given by the form ula:
Note :
1.The interest may be compounded annually (yearly), semi -annually
(half yearly), quarterly or monthly. Thus, the general formula to calculate
the amount at the end of nyears is as follows:
Here p: number of times the interest is compou nded in a year.
p= 1 if interest is compounded annually
p= 2 if interest is compounded semi -annually (half-yearly )
p= 4 if interest is compounded quarterly
p= 12 if interest is compounded monthly
2.It is easy to calculate amount first and then the compo und interest as
compared with finding interest first and then the total amount in case of
simple interest.
Example 9:Find the compound amount and compound interest of Rs.
1000 invested for 10 years at 8% if the interest is compounded annually.
Ans: Give nP=1000,r =8,n= 10.
Since the interest is compounded annually, we have
A= 1100nrP=1000 x1081100= 1000 x 2.1589 = Rs.2158.9
Example 10:Find the principal which will amount to Rs. 11,236 in 2
years at 6% co mpound interest compounded annually.
Ans: Given A= Rs. 11236, n= 2,r= 6 and P= ?A= P 1100nr
CI = A–P
A = P 1x 100npr
p
munotes.in
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Now, A= 1100nrP
11236 = P261100=Px 1.1236
P=11236
1.1236= 10,000
the required principal is Rs. 10,000.
Exam ple11:Find the compound amount and compound interest of Rs.
1200 invested for 5 years at 5% if the interest is compounded ( i) annually,
(ii) semi annually, ( iii) quarterly and ( iv) monthly.
Ans: Given P= Rs. 1 200,r=5,n= 5
Let us recollect the formu laA=P1x 100npr
p
(i)If the interest is compounded annually ,p= 1:
A= 1100nrP= 1200 x551100= 1200 x 1. 2763 =Rs. 1 531.56
CI= A–P= 1531.56 –1200 = Rs. 33 1.56
(ii)If the interest is compounde d semi -annually ,p= 2:
A=2
12 x 100nrP = 1200 x1051200= 1200 x 1. 28=Rs. 1 536
CI= A–P= 1536–1200 = Rs. 3 36.
(iii)If the interest is compounded quarterly ,p= 4:
A=4
14 x 100nrP = 1200 x2051400= 1200 x 1. 2820 =
Rs. 1 538.4
CI= A–P= 1538.4 –1200 = Rs. 3 38.4
(iv)If the interest is compounded monthly ,p= 12:
A=12
112 x 100nrP = 1200 x60511200 = 1200 x 1. 2834 =Rs. 1 540
CI= A–P= 1540–1200 = Rs. 340
Example 12:Mr. Santosh wants to invest some amount for 4 years in a
bank. Bank Xoffers 8% interest if compounded half yearly while bank Y
offers 6% interest if compounded monthly. Which bank should Mr.
Santosh select for better benefits?
Ans: Givenn= 4.
Let the principal Mr. Santosh wants to invest be P= Rs. 100
From Bank X:r= 8and interest is compounded half -yearly, so p= 2.
A=2
12 x 100nrP = 100 x481200= 116.9858 … (1)munotes.in
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From Bank Y:r= 6,p= 12
A=12
112 x 100nrP = 100 x48611200 = 127.0489 … (2)
Comparing (1) and ( 2), Dr. Ashwinikumar should invest his amount in
bank Yas it gives more interest at the end of the period.
Example 13:In how many years would Rs. 75,000 amount to Rs.
1,05,794.907 at 7% compound interest compounded semi -annually?
Ans: Given A= Rs. 1057 94.907, P= Rs. 75000, r= 7,p= 2
A=2
12 x 100nrP
105794.907 = 75000 x271200n
105794.907
75000=(1.035)2n
1.41059876 = (1.035)2n
(1.035)10= (1.035)2n2n= 10
Thus, n= 5
Example 14:A certa in principal amounts to Rs. 4410 after 2 years and to
Rs.4630.50 after 3 years at a certain rate of interest compounded annually.
Find the principal and the rate of interest.
Ans: Let the principal be Pand rate of interest be r.
Now, we know that A=P1100nr
From the given data we have,
4410 = P2
1100rand 4630.5 = P3
1100r
4410 =P(1 + 0.01 r)2… (1)
4630.5 = P(1 + 0.01 r)3… (2)
Dividing (2) by (1), we haveDo not write ‘1 + 0.01 r’
as 1.01 rmunotes.in
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4630.5
4410=3
2(1 0.01 )
(1 0.01 )P r
P r
1.05 = 1 + 0.01 r
0.05 = 0.01 r
Thus, r= 5%
Example 15:Find the rate of interest at which a sum of Rs. 2000 amounts
to Rs. 2690 in 3 years given that the interest is compounded half yearly.
(61.345 = 1.05)
Ans: Given P= Rs. 2000, A= Rs. 2680, n= 3,p= 2
Now, A=2
12 x 100nrP
2690 = 2000 x6
1200r
2690
2000=6
1200r1.345 =6
1200r
61.345 = 1+200r1.05 = 1 +200r
r= 0.05 x 200 = 10%
Thus, the rate of compound interest is 10 % .
Example 16:If the interest compounded half yearly on a certain principal
at the e nd of one year at 8% is Rs. 3264, find the principal.
Ans: Given CI= Rs. 3264, n= 1,p= 2 and r= 8
Now, CI=A–P=P281200–P
i.e. 3264 = P[ (1.04)2–1] = 0.0816 P
P=3264
0.0816= 40000
Thus, the principal is Rs. 40,000.
Check your progress 10.2
1.Compute the compound amount and interest on a principal of Rs.
21,000 at 9% p.a. after 5 years.
Ans:32,311.10, 11,311.10
2.Compute the compound amount and interest on a principal of Rs.
6000 at 11% p.a. after 8 years.
Ans:13827.23, 7827.23
3.Compute the compound amount and compound interest of Rs. 5000
if invested at 11% for 3 years and the interest compounded i)
annually, ( ii) semi annually, ( iii) quarterly and ( iv) monthly.
Ans: (i)6838.16, 1838.16 (ii) 6894.21, 18942 1
(iii) 6923.92, 1923.92 (iv) 6944.39, 1944.39munotes.in
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4.Compute the compound amount and compound interest of Rs. 1200
if invested at 9% for 2 years and the interest compounded i)
annually, ( ii) semi annually, ( iii) quarterly and ( iv) monthly.
Ans: (i) 1425.7 2, 225.72 (ii) 1431.02, 231.02
(iii) 1433.8, 233.8 (iv) 1435.7, 235.7
5.Miss Daizy invested Rs. 25,000 for 5 years at 7.5% with the interest
compounded semi -annually. Find the compound interest at the end
of 5 years. Ans:
11,126.10
6.Mr.Dayanand borrowed a sum of Rs. 6500 from his friend at 9%
interest compounded quarterly. Find the interest he has to pay at the
end of 4 years? Ans: 2779.54
7.Mr. Deepak borrowed a sum of Rs. 8000 from his friend at 8%
interest compounded annually. Fi nd the interest he has to pay at the
end of 3 years? Ans:
2077.70
8.Mr. Deshraj borrowed Rs. 1,25,000 for his business for 3 years at
25% interest compounded half yearly. Find the compound amount
and interest after 3 years. Ans: 2,53,410.82;
12,841 0.82
9.Mrs. Hemlata bought a Sony Digital Camera for Rs. 15,800 from
Vijay Electronics by paying a part payment of Rs. 2,800. The
remaining amount was to be paid in 3 months with an interest of 9%
compounded monthly on the due amount. How much amount did
Mrs. Hemlata paid and also find the interest.
Ans: 13294.70, 294.70
10.Mr. Irshad bought a Kisan Vikas Patra for Rs. 10000, whose
maturing value is Rs. 21,000 in 4½ years. Calculate the rate of
interest if the compound interest is compounded quarterly.
Ans: 16.8%
11.What sum of money will amount to Rs. 11236 in 2 years at 6% p.a.
compound interest? Ans: 10,000
12.Find the principal which will amount to Rs. 13468.55 in 5 years at
6% interest compounded quarterly. [ (1.015)20= 1.346855]
Ans: 10000
13.Find the principal which will amount to Rs. 30626.075 in 3 years at
7% interest compounded yearly. Ans: 25000
14.Find the principal if the compound interest payable annually at 8%
p.a. for 2 years is Rs. 1 664. Ans: 10000munotes.in
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15.If Mr. Sagar wants to earn Rs. 50000 after 4years by investing a
certain amount in a company at 10% rate of interest compounded
annually, how much should he invest? Ans:
34150.67
16.Find after how many years will Rs. 4000 amount to Rs. 4494.40 at
6% rate of interest compounded yearly. Ans: n= 2
17.Find after how many years Rs. 10,000 amount to Rs. 12,155 at 10%
rate of interest compounded half -yearly . Ans: n= 1
18.Find the rate of interest at which a principal of Rs.10000 amounts to
Rs. 11236 after 2 years. Ans:6%
19.Find the rate of interest at which a principal of Rs.50000 amounts to
Rs. 62985.6 after 3 years. (31.259712 = 1.08) Ans: 8%
20.Mrs. Manisha Lokhande deposited Rs. 20,000 in a bank for 5 years.
If she received Rs.3112.50 as in terest at the end of 2 years, find the
rate of interest p.a. compounded annually. Ans: 7.5%
21.A bank Xannounces a super fixed deposit scheme for its customers
offering 10% interest compounded half yearly for 6 years. Another
bank Yoffers 12% simple interest for the same period. Which bank’s
scheme is more beneficial for the customers? Ans: Bank X
22.ABC bank offers 9% interest compounded yearly while XYZ bank
offers 7% interest compounded quarterly. If Mr. Arunachalam wants
to invest Rs. 18000 for 5 years, which bank should he choose?
Ans: Bank ABC
23.Mangesh borrowed a certain amount from Manish at a rate of 9% for
4 years. He paid Rs. 360 as simple interest to Manish. This amount
he invested in a bank for 3 years at 11% rate of interest compounded
quarterly. Find the compo und interest Mangesh received from the
bank after 3 years. Ans:
1384.78
24.On a certain principal for 3 years the compound interest
compounded annually is Rs. 11 25.215 while the simple interest is
Rs. 1050, find the principal an d the rate of interest.
Ans: 5000, 7%
25.On a certain principal for 4 years the compound interest
compounded annually is Rs. 13923 while the simple interest is Rs.
12000, find the principal and the rate of interest.
Ans: 30000, 10%.munotes.in
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26.Which investment is better for Mr. Hariom Sharma (i) 6%
compounded half yearly or (ii) 6.2% compounded quarterly?
Ans:
27.Which investment is better for Mr. Suyog Apte (i) 9% compounded
yearly or (ii) 8.8% compounded quarterly?
Ans:
28.A bank Xoffers 7% interest compounded s emi-annually while
another bank offers 7.2% interest compounded monthly. Which bank
gives more interest at the end of the year?
Ans:
29.Mr. Nitin Tare has Rs. 10000 to be deposited in a bank. One bank
offers 8% interest p.a. compounded half yearly, while the other
offers 9% p.a. compounded annually. Calculate the returns in both
banks after 3 years. Which bank offers maximum return after 3
years?
Ans:
munotes.in
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Unit II
Unit -3
ANNUITIES AND EMI
Unit Structure :
3.0 Objectives
3.1 Introduction
3.2 Annuity
3.3 Types of Annuities
3.4 Sinking Fund
3.5 Equated Monthly Installment (Emi)
3.0OBJECTIVES
After reading this chapter you will be able to:
Define annuit y, future value, present value, EMI, Sinking Fund.
Compute Future Value of annuity due, Present Value of an ordinary
annuity.
Compute EMI of a loan using reducing balance method and flat
interest method.
Compute Sinking Fund (periodic payments).
3.1INTRO DUCTION
In the previous chapter we have seen how to compute compound
interest when a lump sum amount is invested at the beginning of the
investment. But many a time we pay (or are paid) a certain amount not in
lump sum but in periodic installments. This series of equal payments done
at periodic intervals is called as annuity .
Let us start the chapter with the definition of an annuity .
3.2ANNUITY
A series of equal periodic payments is called annuity . The
payments are of equal size andat equal t ime interval .
The common examples of annuity are: monthly recurring deposit
schemes, premiums of insurance policies, loan installments, pension
installments etc. Let us understand the terms related to annuities and then
begin with the chapter.munotes.in
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Periodic Payment:
The amount of payment made is called as periodic payment .
Period of Payment:
The time interval between two payments of an annuity is called as the
period of payment .
Term of an annuity:
The time of the beginning of the first payment period to t he end of the last
payment period is called as term of annuity . An annuity gets matured at
the end of its term.
3.3TYPES OF ANNUITIES
Though we will be discussing two types of annuities in detail, let
us understand different types of annuities based on the duration of the term
or on the time when the periodic payments are made. On the basis of the
closing of an annuity, there are three types of annuities:
1.Certain Annuity :
Here the duration of the annuity is fixed (or certain), hence called
certain an nuity . We will be learning such annuities in detail.
2.Perpetual Annuity :
Here the annuity has no closing duration, i.e. it has indefinite duration.
Practically there are rarely any perpetuities.
3.Contingent Annuity :
Here the duration of the annuity depen ds on an event taking place. An
example of contingent annuity is life annuity . Here the payments are to
be done till a person is alive, like, pension, life insurance policies for
children (maturing on the child turning 18 years) etc.
Onthe basis of whe n the periodic payments are made we have two
types of annuities: ordinary annuity and annuity due.
3.3.1 Immediate (Ordinary) Annuity:
The annuity which is paid at the end of each period is called as
immediate (ordinary )annuity . The period can be mont hly, quarterly or
yearly etc. For example, stock dividends, salaries etc.
Let us consider an example of an investment of Rs. 5000 each year
is to be made for four years. If the investment is done at the end of each
year then we have the following diagr ammatic explanation for it:munotes.in
49
3.3.2 Present Value :
The sum of all periodic payments of an annuity is called its present
value . In simple words, it is that sum which if paid now will give the same
amount which the periodic payments would have given at the end of the
decided period. It is the one time payment of an annuity.
The formula to find the present value ( PV) is as follows:
Leti=x 100r
p, the rate per period, then the above formula can be
rewritten as f ollows:
3.3.3 Future Value (Accumulated value) :
The sum of all periodic payments along with the interest is called
thefuture value (accumulated amount ) of the annuity.
The formula to find the future value ( A) of an immediate annuity is as
follow s:PV=11
1x 100 x 100npP
r r
p p
Where
P: periodic equal payment
r: rate of interest p.a.
p: period of annuity
PV=
11
1npP
i i
munotes.in
50
Example 1:Find the future value after 2 years of an immediate annuity of
Rs. 5000, the rate of interest being 6% p.a compounded annually.
Ans: Given n= 2,P= Rs. 5000, r= 6 and p= 1i=6
100= 0.06
A=P
i1 1npi =5000
0.0621 0.06 1 = 50001.1236 1
0.06
A= 5000 x 2.06 = Rs. 10300
Example 2: Find the amount for an ordinary annuity with periodic
payment of Rs. 3000, at 9% p. a. compounded semi -annually for 4 years.
Ans: Given n= 4,P= Rs. 3000, r= 9 and p= 2i=9
2 x 100= 0.045
Now, A =P
i1 1npi =3000
0.0452 x 4[(1 0.045) 1] =3000
0.045x 0.4221
Thus, A= Rs. 28,140
Example 3: Mr. Ravi invested Rs. 5000 in an annuity with quarterly
payments for a period of 2 years at the rate of interest of 10%. Find the
accumulated value of the annuity at the end of 2ndyear.
Ans: Given n= 2,P= Rs.5000, r= 10 and p= 4i=10
4 x 100= 0.025
Now, A =P
i1 1npi =5000
0.0252 x 4(1.025) 1 =5000
0.025x 0.2184
Thus, A= Rs. 43,680
Example 4:Mr. Ashok Rane borrowed Rs. 20,000 at 4% p.a. compounded
annually for 10 years. Find the periodic payment he has to make.
Ans: Given PV= Rs. 20,000; n= 10; p =1 and r= 4i= 0.04
Now to find the periodic payments Pwe use the following formula:
PV=
11
1npP
i i
20000 =
10110.04 1 0.04P
=0.04Px 0.3244
P=20000 x 0.04
0.3244= 2466.09A= 1 1x 100
x 100npP r
r p
p
A=P
i1 1npi Here,
P: periodic equal payment
r: rate of interest p.a.
p: period of annuity i.e. yearly,
half yearly, quarterly or
monthly
and i=x 100r
p
munotes.in
51
Thus, Mr. Rane has to make the periodic payments of Rs. 2466.09
Example 5:Find the future value of an immediate annuity after 3 years
with the periodic payment of Rs. 12000 at 5% p.a. if the period of
payments is (i) yearly, (ii) half -yearly, (iii) quarterly and (iv) monthly.
Ans: Given P= Rs. 1200, n= 3,r= 5
(i) period p= 1 then i=5
100= 0.05
A=P
i1 1ni =12000
0.0531 0.05 1 =12000
0.05 1.1576 1
A= 12000 x 3.1525 = Rs. 37,830
(ii) period p= 2 then i=5
2 x 100= 0.025
A=P
i21 1ni =12000
0.0256(1 0.025) 1 =12000
0.025x 0.1597
A= 12000 x 6.388 = Rs. 76,656
(iii) period p=4 then i=5
4 x 100= 0.0125
A=P
i41 1ni =12000
0.0125[(1 + 0.0125)12–1] =12000
0.0125x 0.16075
A= 12000 x 12.86 = Rs. 1,54,320
(iv) period p= 12 then i=5
12 x 100= 0.00417
A=P
i121 1ni =12000
0.00417[(1 + 0.00417)36–1] =12000
0.00417x 0.1615
A= 1200 x 38.729 = Rs. 4,64,748
Example 6:Mr. Nagori invested certain princ ipal for 3 years at 8%
interest compounded half yearly. If he received Rs.72957.5 at the end of
3rdyear, find the periodic payment he made. [(1.04)6= 1.2653]
Ans: Given n= 3,r= 8,p= 2i=8
2 x 100= 0.04
Now, A=P
i1 1npi
72957.5 =0.04P[(1 + 0.04)6–1]=0.04Px 0.2653
72957.5 = P[6.6325]munotes.in
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P=72957.5
6.6325= 11000
Thus, the periodic payment is Rs. 11,000
3.4SINKING FUND
The fund (money) which is kept aside to accumulate a certain sum
in a fixed period through periodic equal payments is called as sinking
fund .
We can consider an example of a machine in a factory which needs
to be replaced after say 10 years. The amount for buying a new machine
10 ye ars from now may be very large, so a proportionate amount is
accumulated every year so that it amounts to the required sum in 10 years.
This annual amount is called as sinking fund . Another common example is
of the maintenance tax collected by any Society from its members.
A sinking fund being same as an annuity, the formula to compute
the terms is same as that we have learnt in section 2.3.3
Example 7:A company sets aside a sum of Rs. 15,000 annually to enable
it to pay off a debenture issue of Rs. 1 ,80,000 at the end of 10 years.
Assuming that the sum accumulates at 6% p.a., find the surplus after
paying off the debenture stock.
Ans: Given P= Rs. 15000, n= 10, r= 6i= 0.06
A=P
i1 1ni =15000
0.06x [(1 + 0.06)10–1] =15000
0.06x 0.7908
A= Rs. 1,97,700
Thus, the surplus amount after paying off the debenture stock is
= 197712 –180000 = Rs. 17712.
Example 8:Shriniketan Co -op Hsg. Society has 8 members and colle cts
Rs. 2500 as maintenance charges from every member per year. The rate of
compound interest is 8% p.a. If after 4 years the society needs to do a
work worth Rs. 100000, are the annual charges enough to bear the cost?
Ans: Since we want to verify whether Rs. 2500 yearly charges are enough
or not we assume it to be Pand find its value using the formula:
A=P
i1 1ni
Here A= Rs. 100000, n= 4,r= 8i= 0.08munotes.in
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P=
x
1 1nA i
i=
4100000 x 0.08
1 0.08 1= 22192
Thus, the annual payment of all the members i.e. 8 members should be Rs.
22192.
the annual payment per member =22192
8= Rs. 2774
This payment is less than Rs. 2500 which the society has decided to take
presently. Th us, the society should increase the annual sinking fund.
3.5EQUATED MONTHLY INSTALLMENT (EMI)
Suppose a person takes a loan from a bank at a certain rate of
interest for a fixed period. The equal payments which the person has to
make to the bank per m onth are called as equated monthly installments in
short EMI.
Thus, EMI is a kind of annuity with period of payment being
monthly and the present value being the sum borrowed .
We will now study the method of finding the EMI using reducing
balance method and flat interest method.
(a)Reducing balance method :
Let us recall the formula of finding the present value of an annuity.
PV=
11
1npP
i i
The equal periodic payment ( P) is our EMI which is denoted it by E.
The present value ( PV) is s ame as the sum ( S) borrowed.
Also the period being monthly p= 12 ,i=1200ras we are interested in
finding monthly installments and nis period in years .
Substituting this in the above formula we have:
S=
1211
1nE
i i
Thus, if Sis the sum borrowed for nyears with rate of interest r% p.a.
then the EMI is calculated by the formula:
E=
12x
11(1 )nS i
imunotes.in
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(b) Flat Interest Method:
Here the amount is calculated using Simple Interest for the period and
the EMI i s computed by dividing the amount by total number of monthly
installments.
LetSdenote the sum borrowed, rdenote the rate of interest and n
denote the duration in years, then as we know the amount using simple
interest formula is A=S1100nr.The total number of monthly
installments for duration of nyears is 12n. Hence the EMI is calculated as
E=12A
n
Example 9:Mr. Sudhir Joshi has taken a loan of Rs. 10,00,000 from a
bank for 10 years at 11% p.a. Find his EMI using (a) reducing balance
method and (b) Flat interest method.
Ans: Given S= Rs. 1000000, n= 10, r= 11i=11
1200= 0.0092
(a)Using flat interest method :
A=S1100nr=10000001101100=2100000
Thus, E=12A
n=2100000
120= 17,500 … (1)
(b)Using reducing balance method :
Now, E=
12x
11(1 )nS i
i=
1201000000 x 0.0092
11(1 0.0092)= 13797.65
E= Rs. 13,798 approximately … (2)
Comparing (1)and (2), we can see that the EMI using flat interest method
is higher than by reducing balance method .
Example 10:Mr. Prabhakar Naik has borrowed a sum of Rs. 60,000 from
a person at 6% p.a. and is due to return it back in 4 monthly installments.
Find th eEMI he has to pay and also prepare the amortization table of
repayment.
Ans: Given S= Rs. 60,000; n= 4 months;
r= 6%i=6
1200= 0.005
Now, E=x
11(1 )nS i
i=
460000 x 0.005
11(1 0.005)=300
0.01975
E= Rs. 15,187.97munotes.in
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Now, we will prepare the amortization table i.e. the table of repayment of
the sum borrowed using reducing balance method.
In the beginning of the 1stmonth the outstanding principal is the sum
borrowed i.e. Rs. 60000 and the EMI paid is Rs. 15187.97
The interest on the outstanding principal is 0.005 x 60000 = Rs. 300 … (1)
Thus, the principal repayment is 15187.97 –300 = Rs. 14887.97 … (2)
The outstanding principal ( O/P) in the beginning of the 2ndmonth is now
60000 –14887.97 = 45112.03.
Note :
(1) is called the interest part of the EMI and (2) is called as the
principal part of the EMI.
As the tenure increases the interest part reduces and the principal part
increases.
This calculation can be tabulated as follows:
O/P EMI Interest PartPrincipal
Part Month
(a) (b) (c) = (a) xi (b)-(c)
1 60000 15187.97 300 14887.97
2 45112.03 15187.97 225.56 14962.45
3 30141.02 15187.97 150.75 15037.22
4 15111.80 15187.97 75.56 15112.41
In the beginning of the 4thmonth t he outstanding principal is Rs. 15111.80
but the actual principal repayment in that month is Rs. 15112.41. This
difference is due to rounding off the values to two decimals, which leads
the borrower to pay 61 paise more!!
Example 11:Mr. Shyam Rane has bo rrowed a sum of Rs. 100000 from a
bank at 12% p.a. and is due to return it back in 5 monthly installments.
Find the EMI he has to pay and also prepare the amortization table of
repayment.
Ans: Given S= Rs. 100000; n= 5 months;
r= 12% p.a. =12
12= 1% p.m i= 0.01
Now, E=x
11(1 )nS i
i=
5100000 x 0.01
11(1 0.01)=1000
0.0485343= 20603.98munotes.in
56
The amortization table is as follows:
O/P EMI Interest PartPrincipal
Part Month
(a) (b) (c) = (a) xi (b)-(c)
1 100000 20603.98 1000 19603.98
2 80396.02 20603.98 803.96 19800.02
3 60596 20603.98 605.96 19998.02
4 40597.98 20603.98 405.98 20198
5 20399.98 20603.98 204 20399.98
Check your progress
1.An overdraft of Rs. 50,000 is to be paid back in equal annual
installments in 20 years. Find the installments, if the interest is 12%
p.a. compounded annually. [(1.12)20= 9.64629]
2.A man borrows Rs. 30,000 at 6% p.a. compounded semi -annually
for 5 years. Find the periodic payments he has to make.
3.What periodic payments Mr. Narayana n has to make if he has
borrowed Rs. 1,00,000 at 12% p.a. compounded annually for 12
years? [(1.12)12= 3.896]
4.Find the future value of an immediate annuity of Rs. 1200 at 6% p.a.
compounded annually for 3 years.
5.Find the future value of an immediate annui ty of Rs. 500 at 8% p.a.
compounded p.m. for 5 years.
6.Find the accumulated value after 2 years if a sum of Rs. 1500 is
invested at the end of every year at 10% p.a. compounded quarterly.
7.Find the accumulated amount of an immediate annuity of Rs. 1000 at
9%p.a. compounded semi -annually for 4 years.
8.Find the future value of an immediate annuity of Rs. 2800 paid at
10% p.a. compounded quarterly for 2 years. Also find the interest
earned on the annuity.
9.Find the sum invested and the accumulated amount for an ordinary
annuity with periodic payment of Rs. 2500, at the rate of interest of
9% p.a. for 2 years if the period of payment is (a) yearly, (b) half -
yearly, (c) quarterly or (d) monthly.
10.Find the sum invested and the accumulated amount for an ordinary
annui ty with periodic payment of Rs. 1500, at the rate of interest of
10% p.a. for 3 years if the period of payment is (a) yearly, (b) half -
yearly, (c) quarterly or (d) monthly.
11.Mr. Banerjee wants to accumulate Rs. 5,00,000 at the end of 10
years from now. How much amount should he invest every year at
the rate of interest of 9% p.a. compounded annually?munotes.in
57
12.Find the periodic payment to be made so that Rs. 25000 gets
accumulated at the end of 4 years at 6% p.a. compounded annually.
13.Find the periodic payment to be ma de so that Rs. 30,000 gets
accumulated at the end of 5 years at 8% p.a. compounded half
yearly.
14.Find the periodic payment to be made so that Rs. 2000 gets
accumulated at the end of 2 years at 12% p.a. compounded quarterly.
15.Find the rate of interest if a pe rson depositing Rs. 1000 annually for
2 years receives Rs. 2070.
16.Find the rate of interest compounded p.a. if an immediate annuity of
Rs. 50,000 amounts to Rs. 1,03,000 in 2 years.
17.Find the rate of interest compounded p.a. if an immediate annuity of
Rs. 50 00 amounts to Rs. 10400 in 2 years.
18.What is the value of the annuity at the end of 5 years, if Rs. 1000
p.m. is deposited into an account earning interest 9% p.a.
compounded monthly? What is the interest paid in this amount?
19.What is the value of the annuit y at the end of 3 years, if Rs. 500 p.m.
is deposited into an account earning interest 6% p.a. compounded
monthly? What is the interest paid in this amount?
20.Mr. Ashish Gokhale borrows Rs. 5000 from a bank at 8% compound
interest. If he makes an annual paym ent of Rs. 1500 for 4 years,
what is his remaining loan amount after 4 years?
(Hint : find the amount using compound interest formula for 4 years and
then find the accumulated amount of annuity, the difference is the
remaining amount.)
21.Find the present val ue of an immediate annuity of Rs. 10,000 for 3
years at 6% p.a. compounded annually.
22.Find the present value of an immediate annuity of Rs. 100000 for 4
years at 8% p.a. compounded half yearly.
23.Find the present value of an immediate annuity of Rs. 1600 for 2
years at 7% p.a. compounded half yearly.
24.A loan is repaid fully with interest in 5 annual installments of Rs.
15,000 at 8% p.a. Find the present value of the loan.
25.Mr. Suman borrows Rs. 50,000 from Mr. Juman and agreed to pay
Rs. 14000 annually for 4 yea rs at 10% p.a. Is this business profitable
to Mr. Juman?
(Hint: Find the PVof the annuity and compare with Rs. 50000)
26.Mr. Paradkar is interested in saving a certain sum which will amount
to Rs. 3,50,000 in 5 years. If the rate of interest is 12% p.a., ho w
much should he save yearly to achieve his target?
27.Mr. Kedar Pethkar invests Rs. 10000 per year for his daughter from
her first birthday onwards. If he receives an interest of 8.5% p.a.,
what is the amount accumulated when his daughter turns 18?munotes.in
58
28.Dr. Wakan kar, a dentist has started his own dispensary. He wants to
install a machine chair which costs Rs. 3,25,000. The machine chair
is also available on monthly rent of Rs. 9000 at 9% p.a. for 3 years.
Should Dr. Wakankar buy it in cash or rent it?
29.A sum of Rs. 50,000 is required to buy a new machine in a factory.
What sinking fund should the factory accumulate at 8% p.a.
compounded annually if the machine is to be replaced after 5 years?
30.The present cost of a machine is Rs. 80,000. Find the sinking fund
the com pany has to generate so that it could buy a new machine after
10 years, whose value then would be 25% more than of today’s
price. The rate of compound interest being 12% p.a. compounded
annually.
31.Mr. Mistry has to options while buying a German wheel alignm ent
machine for his garage: (a) either buy it at Rs. 1,26,000 or (b) take it
on lease for 5 years at an annual rent of Rs. 30,000 at the rate of
interest of 12% p.a.. Assuming no scrap value for the machine which
option should Mr. Mistry exercise?
32.Regency Co-op. Hsg. Society which has 50 members require Rs.
12,60,000 at the end of 3 years from now for the society repairs. If
the rate of compound interest is 10% p.a., how much fund the society
should collect from every member to meet the necessary sum?
33.Mr. L alwaney is of 40 years now and wants to create a fund of Rs.
15,00,000 when he is 60. What sum of money should he save
annually so that at 13% p.a. he would achieve his target?
34.If a society accumulates Rs. 1000 p.a. from its 200 members for 5
years and rec eives 12% interest then find the sum accumulated at the
end of the fifth year. If the society wants Rs. 13,00,000 for society
maintenance after 5 years, is the annual fund of Rs. 1000 per
member sufficient?
35.How much amount should a factory owner invest eve ry year at 6%
p.a. for 6 years, so that he can replace a mixture -drum (machine)
costing Rs. 60,000, if the scrap value of the mixture -drum is Rs.
8,000 at the end of 6 years.
36.If a society accumulates Rs. 800 p.a. from its 100 members for 3
years and receiv es 9% interest then find the sum accumulated at the
end of the third year. If the society wants Rs. 2,50,000 for society
maintenance after 3 years, is the annual fund of Rs. 800 per member
sufficient?
37.Mr. Kanishk wants clear his loan of Rs. 10,00,000 taken at 12% p.a.
in 240 monthly installments. Find his EMI using reducing balance
method.
38.Using the reducing balance method find the EMI for the following:munotes.in
59
Loan amount
(in Rs.)Rate of Interest
(in % p.a.)Period of Loan
(in years)
i) 1000 6 5
ii) 50000 6 10
iii) 8000 7 6
iv) 12000 9 10
v) 1000 9.5 10
vi) 1100000 12.5 20
39.Mr. Vilas Khopkar has taken a loan of Rs. 90,000 at 11% p.a. Find
the EMI using (a) reducing balance method and (b) Flat interest
method, if he has to return the loan in 4 years.
40.Find the EMI using reducing balance method on a sum of Rs. 36,000
at 9%, to be returned in 6 monthly installments.
41.Find the EMI using reducing balance method on a sum of Rs. 72,000
at 12%, to be returned in 12 installments.
42.Mr. Sachin Andhale has borrowed Rs. 1 0,000 from his friend at 9%
p.a. and has agreed to return the amount with interest in 4 months.
Find his EMI and also prepare the amortization table.
43.Mr. Arvind Kamble has borrowed Rs. 30,000 from his friend at 14%
p.a. If he is to return this amount in 5 monthly installments, find the
installment amount, the interest paid and prepare the amortization
table for repayment.
44.Mrs. Chaphekar has taken a loan of Rs. 1,25,000 from a bank at 12%
p.a. If the loan has to be returned in 3 years, find the EMI, Mrs.
Chaphekar has to pay. Prepare the amortization table of repayment
of loan and find the interest she has to pay.
45.A loan of Rs. 75,000 is to be returned with interest in 4 installments
at 15% p.a. Find the value of the installments.
46.A loan of Rs. 60,000 is to b e returned in 6 equal installments at 12%
p.a. Find the amount of the installments.
47.Find the sum accumulated by paying an EMI of Rs. 11,800 for 2
years at 10% p.a.
48.Find the sum accumulated by paying an EMI of Rs. 1,800 for 2 years
at 12% p.a.
49.Find the sum accumulated by paying an EMI of Rs. 12,000 for 3
years at 9% p.a.
50.Find the sum accumulated by paying an EMI of Rs. 11,000 for 8
years at 9.5% p.a.munotes.in
60
Hints & Solutions to Check your progress
(1)6694 (2)3517 (3)16,144 (4)3820.32
(5)36555.65 (6)13104 (7)9380 (8)24461
(9)
Period Sum Invested Accumulated
Amount
Yearly 5000 5225
Half-yearly 10000 10695.5
Quarterly 20000 21648
Monthly 60000 65471
(10)
Period Sum Invested Accumulated
Amount
Yearly 4500 4965
Half-yearly 9000 10203
Quarterly 18000 20693
Monthly 54000 62635
(11) 32910 (12) 5715 (13) 2498.72 (14) 225
(15) 7% (16) 6% (17) 8% (18) 75424, 15424
(19) 19688 , 1688 (20) 4719 (21) 26730
(22) 673274.5 (23) 5877 (24) 59890.65 (25) 44378, Yes
(26) 97093.4 (27) 393229.95 (28) 283021.25, take it on rent
(29)12523 (30) 17698.42 (31) 108143.28 < 126000, Mr. Mistry
should use the second option. (32) 16245 (33) 18530
(34) 1270569.47, not sufficient (35) 7454.86 (36) 2,62,248; yes
(37) 11,011
(38)
Loan amount
(in Rs.)Rate of inte rest
(in % p.a.)Period of Loan
(in yrs.)EMI
(in Rs.)
i) 1000 6 5 19
ii) 50000 6 10 555
iii) 8000 7 6 136
iv) 12000 9 10 152
v) 1000 9.5 10 13
vi) 1100000 12.5 20 12498
(39) 2326, 2700 (40) 6158.48 (41) 6397.11munotes.in
61
(42)
O/P EMI Intere st PartPrincipal
Part Month
(a) (b) (c) = (a) xi (b)-(c)
1 10000 2547.05 75 2472.05
2 7527.95 2547.05 56.45 2490.6
3 5037.35 2547.05 37.78 2509.27
4 2528.08 2547.05 18.96 2528.09
(43)
O/P EMI Interest PartPrincipal
Part Month
(a) (b) (c) = (a) xi (b)-(c)
1 30000 6212.23 351 5861.23
2 24138.77 6212.23 282.42 5929.81
3 18208.96 6212.23 213.04 5999.19
4 12209.77 6212.23 142.85 6069.38
5 6140.39 6212.23 71.84 6140.39
(45)19339.57 (46) 16353 (47) 3,12,673.60
(48) 48552.24 (49) 4,93,832.6 (50)15,72,727
munotes.in
62
UNIT -III
Unit -4
CORRELATION AND REGRESSION
Unit Structure :
4.0 Objectives
4.1 Introduction
4.2 Types of Correlation
4.3 Measurement of Correlation
4.4 Rank Correlation
4.5 Regression Analysis
4.0OBJECTIVE S
To understand the relationship between two relevant characteristics of
a statistical unit.
Learn to obtain the numerical me asure of the relationship between two
variables.
Use the mathematical relationship between two variables in order to
estimate value of one variable from the other.
Use the mathematical relationship to obtain the statistical constants
line means andS.D.’s
4.1 INTRODUCTION
Inthe statistical analysis we come across the study of two or more
relevant characteristics together in terms of their interrelations or
interdependence. e.g. Interrelationship among production, sales and
profits of a company. Inter relationship among rainfall, fertilizers, yield
and profits to the farmers .
Relationship between price and demand of a commodity When we
collect the information (data) on two of such characteristics it is called
bivariate data. It is generally denoted by (X,Y) where X and Y are the
variables representing the values on the characteristics.
Following are some examples of bivariate data.
a)Income and Expenditure of workers.
b)Marks of students in the two subjects of Maths and Accounts.
c)Height of Husband and Wif e in a couple.
d)Sales and profits of a company.munotes.in
63
Between these variables we can note that there exist some sort of
interrelationship or cause and effect relationship. i.e. change in the value
of one variable brings out the change in the value of other vari able also.
Such relationship is called as correlation.
Therefore, correlation analysis gives the idea about the nature and
extent of relationship between two variables in the bivariate data.
4.2 TYPES OF CORRELATION:
There are two types of correlation .
a)Positive correlation. and b) Negative correlation.
4.2.1Positive correlation: When the relationship between the variables X
andYis such that increase or decrease in Xbrings out the increase or
decrease in Yalso, i.e. there is direct relation betwe enXand Y, the
correlation is said to be positive. In particular when the ‘change in X
equals to change in Y’ the correlation is perfect and positive. e.g. Sales and
Profits have positive correlation.
4.2.2Negative correlation: When the relationship betw een the variables X
andYis such that increase or decrease in Xbrings out the decrease or
increase in Y, i.e. there is aninverse relation between Xand Y, the
correlation is said to be negative. In particular when the ‘change in X
equals to change in Y’but in opposite direction the correlation is perfect
and negative. e.g. Price and Demand have negative correlation.
4.3 MEASUREMENT OF CORRELATION
The extent of correlation can be measured by any of the following
methods:
Scatter diagrams
Karl Pearson’ s co-efficient of correlation
Spearman’s Rank correlation
4.3.1 Scatter Diagram: The Scatter diagram is a chart prepared by
plotting the values of X and Y as the points (X,Y) on the graph. The
pattern of the points is used to explain the nature of correla tion as follows.
The following figures and the explanations would make it clearer.
(i)Perfect Positive Correlation :
If the graph of the values of the variables is a
straight line with positive slope as shown in
Figure 4.1, we say there is aperfect positive
correlation between XandY. Here r =1.
(ii)Imperfect Positive Correlation :
If the graph of the values of XandYshow a band
of points from lower left corner to upper right corner
as shown in Figure 4.2, we say that ther e is an imperfectO Fig4.1 X
Y
O Fig4.2 X
munotes.in
64
positive correlation . Here 0 < r< 1.
(iii)Perfect Negative Correlation :
If the graph of the values of the variables is a
straight line with negative slope as shown in
Figure 4.3, we say there is a perfect negative
correlation between XandY. Here r=–1.
(iv)Imperfect Nega tive Correlation :
If the graph of the values of XandYshow a band
of points from upper left corner to the lower right
corner as shown in Figure 4.4, then we sa y that
there is an imperfect negative correlation . Here –1
If the graph of the values of XandYdo not show any of the above trend
then we say that there is a zero correlation between XandY. The graph of
such type can be a straight line perpendicular to the axis, as shown in
Figure 4.5and4.6,or may be completely scattered as shown in Figure
4.7. Here r= 0.
The Figure 4.5show that the increase in the values of Yhas no effect on
the value of X, it remains the same, hence zero correlation. The Figure 4.6
show that the increase in the values of Xhas no effect on the value of Y, it
remains the same, hence zero correlation. The Figure 4.7show that the
points are completely scattered on the graph and show no particular trend,
hence there is no correlation or zero correlation between XandY.
4.3.2Karl Pearson’s co -efficient of correlation.
This co -efficient provides the numerical measure of the correlation
between the variables X and Y. It is suggested by Prof. Karl Pearson and
calculated by the formula
( , )
.x yCov x yr
Where, Cov(x,y) : Covariance between x&y
x.: Standard deviation of x&y: Standard deviation of y
Also, Cov(x,y) =1
n(x-x) (y-y) =1
nxy-xyY
O Fig4.5 XY
O Fig4.6XY
O Fig4.7 XY
O X
Fig.4.3
Y
O Fig.4.3 X
munotes.in
65
S.D.(x) = x =2X)1x-n =2 2X1xn and
S.D.(y) =y =2)1y-Yn2 21y Yn
Remark : We can also calculate this co -efficient by using the formula
given by
2 22 22 2X) ) X
X) )1 xyx- y- -
1 1 x yx- y- XY Y
Y Yn nr
n n n n
The Pearson’s Correlation co -efficient is also called as the ‘product
mom ent correlation co -efficient’
Properties of correlation co -efficient ‘r’
The value of ‘r’ can be positive (+) or negative( -)
The value of ‘r’ always lies between –1 & +1, i.e. –1< r<+1]
Significance of ‘r’ equals to –1, +1 & 0
When ‘r’= +1; the correl ation is perfect and positive.
‘r’=-1; the correlation is perfect and negative.
andwhen there is no correlation ‘r’= 0
SOLVED EXAMPLES :
Example .1:Calculate the Karl Pearson’s correlation coefficient from the
following.
X: 12 10 20 13 15
Y: 7 14 6 12 11
Solution: Table of calculation,
And n= 5
The Pearson’s correlation coefficient r is gi ven by,
( , )
.x yCov x yr
Where,X Y XxY X2Y2
12 7 84 144 49
10 14 140 100 196
20 6 120 400 36
13 12 156 169 144
15 11 165 225 121
x =70y= 50xy=665x2=1038y2=546munotes.in
66
x=x
n=70
5= 14 y=y
n=50
5=10
Cov(x,y) = Xxy-Ynx=2 2X1xn y=2 21y Yn
=665
5-14x10 =2 1038145 =2 546105
= 133 -140 =11.6 =3.40 9.2 =3.03 =-07
... Cov(x,y) =-7x=3.40 and y= 3.03
Substituting the values in the formula ofrwe get
7
3.40x3.03r =-0.68
.
... r = -0.68
Example 2:Let us calculate co -efficient of correlation between Marks of
students in the Subjects of Maths & Accounts. in a certain test conducted.
Table of calculation:
n=10
Now Pearson’s co -efficient of correlation is given by the fomula,
( , )
.x yCov x yr
Where,
x=x
n=242
10= 24.2 y=y
n=287
10=28.7Marks
InMaths
XMarks In
Accounts
Y XY X2Y2
28 30 840 784 900
25 40 1000 625 1600
32 50 1600 1024 2500
16 18 288 256 324
20 25 500 400 625
15 12 180 225 144
19 11 209 361 121
17 21 357 289 441
40 45 1800 1600 2025
30 35 1050 900 1225
x= 242y=287xy=7824x2= 6464y29905munotes.in
67
Cov(x,y) = Xxy-Ynx=2 2X1xny=2 21y Yn
=7824
10-24.2x 28.7x=2 646424.210y=2 990528.710
=782.4 -694.54 =60.76 =166.81
Cov(x,y) = 87.86 ,x=7.79 andy= 12.91
... Cov(x,y) = 87.86 x= 7.79 and y= 12.91
Substituting the values in the formula ofrwe get
87.86
7.79 12.91rx =0.87
... r = 0.87
4.4RANK CORRELATION
In many practical situations, we do not have the scores on the
characteristics, but the rank s (preference order) decided by two or more
observers. Suppose, a singing competition of 10 participants is judged by
two judges A and B who rank or assign scores to the participants on the
basis of their performance. Then it is quite possible that the ran ks or scores
assigned may not be equal for all the participants. Now the difference in
the ranks or scores assigned indicates that there is a difference of openion
between the judges on deciding the ranks. The rank correlation studies the
association in th is ranking of the observations by two or more observers.
The measure of the extent of association in rank allocation by the two
judges is calculated by the co -efficient of Rank correlation ‘R’. This co -
efficient was developed by the British psychologist Ed ward Spearman in
1904.
Mathematically, Spearman’s rank correlation co -efficient is defined as,
R= 1 -2
2d
( 1)n n
Where d= rank difference and n= no of pairs.
Remarks: We can note that, the value of ‘R’ always lies between –1 and
+1
Thepositive value of ‘R’ indicates the positive correlation (association) in
the rank allocation. Whereas ,the negative value of ‘R’ indicates the
negative correlation (association) in the rank allocation.munotes.in
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SOLVED EXAMPLES:
Example 3
a)When ranks are g iven: -
Data given below read the ranks assigned by two judeges to 8 participants .
Calculate the co -efficient of Rank correlation.
Ranks by Judge Participant
No. A BRank diff
Square d2
1 5 4 (5-4)2= 1
2 6 8 4
3 7 ` 1 36
4 1 7 36
5 8 5 9
6 2 6 16
7 3 2 1
8 4 3 1
N = 8 Total 104 =d2
Spearman’s rank correlation co -efficient is given b y
R=1-2
26 d
( 1)n n
Substituting the values from the table we get,
R= 1 -26x 104
8(8 1)=-0.23
The value of correlation co -efficient is -0.23. This indicates that there is
negative association in rank allocation by the two judges A and B
b)When scores are given: -
Example 4
The data given below are the marks given by two Examiners to a
set of 10 students in a aptitude test. Calculate the Spearman’s Rank
correlation co -efficient, ‘R’
Now the Spearman’s rank correlation co-efficient is given by
R= 1 -2
26 d
( 1)n n
Substituting the values from the table we get,
R= 1 -26x5
10(10 1)
= 1-0.03
= 0.97munotes.in
69
The value of correlation co -efficient is + 0. 97. This indicates that
there is positive association in assessment of two examiners, A and B.
c)Case of repeated values: -
It is quite possible that the two participants may be assigned the
same score by the judges. In such cases Rank allocation and cal culation of
rank correlation can be explained as follows.
Example : The data given below scores assigned by two judges for 10
participants in the singing competition. Calculate the Spearman’s Rank
correlation co -efficient.
Score assigned B y
JudgesRanks Rank
difference
squareParticipant
No.
A B RA RB D2
1 28 35 9 (8.5) 6 (8.5-6)2
=6.25
2 40 26 3 10(9.5) 42.25
3 35 42 5 (4.5) 3 2.25
4 25 26 10 9 (9.5) 0.25
5 28 33 8 (8.5) 7 2.25
6 35 45 4 (4.5) 2 6.25
7 50 32 1 8 49
8 48 51 2 1 1
9 32 39 6 4 4
10 30 36 7 5 4
N = 10 Total d2=117.5
Marka ByExaminer Ranks Rank
difference
squareStudent No.
A B RA RB D2
1 85 80 2 2 0
2 56 60 8 7 1
3 45 50 10 10 0
4 65 62 6 6 0
5 96 90 1 1 0
6 52 55 9 8 1
7 80 75 3 4 1
8 75 68 5 5 0
9 78 77 4 3 1
10 60 53 7 9 1
N = 10 Total 5 =d2munotes.in
70
Explanation: -In the column of A and B there is repeatation of scores so
while assigning the ranks we first assign the ranks by treating them as
different values and then for rereated scores we assign the average rank.
e.g In col A the score 35 appears 2 times at number 4 and 5 in the order of
ranking so we calculate the average rank as ( 4+5)/2 = 4.5.
Hence the ranks assigned are 4.5 each. The other repeated scores can be
ranked in the same manner.
Note: In this example we can note that the ranks are in fraction e.g. 4.5,
which is logically incorrect or meaningless. Therefore in the calculation of
‘R’ we add a correction factor (C.F.) to d2calculated as follows.
Table of correction factor (C.F.)
Value
Repeate dFrequency
M m(m2-1)
35 2 2x(22-1)=6
28 2 6
26 2 6
Totalm(m2-1)=18
Now3m ). .12mC F = 18/12 =1.5
...d2= 117.5+1.5= 119
We use this value in the calculation of ‘R’
Now the Spearman’s rank correlation co -efficient is g iven by
R= 1 -2
26 d
( 1)n n
Substituting the values we get, R= 1 -26x 119
10(10 1)= 1-0.72 = 0.28
EXERCISE I
1.What is mean by correlation? Explain the types of correlation with
suitable examples.
2.What is a scatter diagram? Draw different scattered diagrams to
explain the correlation between two variables x and y.
3.State the significance of ‘r’ = +1, –1 and 0.
4.Calculate the coefficient of correlation r from the following data.
X: 18 12 16 14 10 15 17 13
Y: 9 13 20 15 11 24 26 22munotes.in
71
5.The following table gives the price and demand of a certain
commodity over the period of 8 months. Calculate the Pearson’s
coefficient of correlation.
Price: 15 12 23 25 18 17 11 19
Demand 45 30 60 65 48 45 28 50
6.Following results are obtained on a certain bivariate data.
(i) n = 10 x= 75 y = 70 x2= 480
y2= 600xy = 540
(ii) n = 15 x= 60 y = 85 x2= 520
y2=1200xy = -340
Calculate the Pears on’s correlation coefficient in each case.
7.Following data are available on a certain bi -variate data :
(i)(x-x) (y-y)=120,(x-x)2= 150(y-y)2= 145
(ii)(x-x) (y-y)=-122,(x-x)2= 136(y-y)2= 148
Find the correlation coefficient.
8. Calculate the Pearson’s coefficient of correlation fr om the given
information on a bivariate series:
No of pairs: 25
Sum of x values:300
Sum of y values:375
Sum of squares of x values: 9000
Sum of squares of y values:6500
Sum of the product of x and y values:4000.
9.The ranks assigned to 8 participants by two judges are as
followes.Calculate the Spearman’s Rank correlation coefficient ‘R’.
Participant No: 1 2 3 4 5 6 7 8
Ranks by JudgeI: 5 3 4 6 1 8 7 2
JudgeII : 6 8 3 7 1 5 4 2
10.Calculate the coefficient of rank correlation from the data given below.
X: 40 33 60 59 50 55 48
Y: 70 60 85 75 72 82 69
11. Marks given by two Judges to a group of 10 participants are as
follows. Calculate the coefficient of rank correlation.
Marks by Judge
A: 52 53 42 60 45 41 37 38 25 27
Judge B: 65 68 43 38 77 48 35 30 25 50.
12. An examination of 8 applicants for a clerical post was by a bank. The
marks obtained by the applicants in the subjects of Mathematics and
Accountancy were as f ollows. Calculate the rank correlation coefficient.
Applicant: A B C D E F G Hmunotes.in
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Marks in
Maths: 15 20 28 12 40 60 20 80
Marks in
Accounts: 40 30 50 30 20 10 25 60
4.5REGRESSION ANALYSIS
As the correlation analysis studies the nature and extent of
interrelationship between the two variables X and Y, regression analysis
helps us to estimate or approximate the value of one variable when we
know the value of other var iable. Therefore we can define the
‘Regression’ as the estimation (prediction) of one variable from the other
variable when they are correlated to each other. e.g. We can estimate the
Demand of the commodity if we know it’s Price.
Why are there two regres sions?
When the variables XandYare correlated there are two possibilities,
(i)Variable Xdepends on variable y. in this case we can find the value of
x if know the value of y. This is called regression of xon.
(ii)Variabledepends on variable X. we can find the value of yif know
the value of X. This is called regression of yonx. Hence there are two
regressions,
(a)Regression of X on Y; (b) Regression of XonY.
4.5.1 Formulas on Regression equation,
Regression of XonY Regression of XonY
Assumption: X depends on Y Y depends on X
The regression equation is The regression equation is
(x-x) = b xy(y-y) (y-y) = b yx(x-x)
bxy= Re gression co -efficient of b yx= Regression co -efficient
X on Y =(x, )
( )Cov y
V yof Y on X =(x, )
(x)Cov y
V
Where,
Cov(x,y) =1
n(x-x) (y-y) =1
nxy-xy
V(x) =1
n(x-x)2and V(y) =1
n(y-y)2
V(x) =1
nx2-x2and V(y) =1
ny2-y2
Use: To find X Use: To find munotes.in
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SOLVED EXAMPLES
Example 1:
Obtain the two regression equatio ne and hence find the value of x when
y=25
Data: -
X Y X2Y2XxY
8 15 64 225 120
10 20 100 400 200
12 30 144 900 360
15 40 225 1600 600
20 45 400 2025 900
x=65y=150x2=933y2=
5150xy=
2180
And n= 5
Now the two regr ession equations are,
(x-x) = b xy(y-y) ------- x on y (i)
(y-y) = b yx(x-x)------- y on x (ii)
Where,
x=1
nx =65
5=13 and y=1
ny =150
5=30
Also,
Cov(x,y,) =1
nxy-xy V(x) =1
nx2-x2V(y) =1
ny2-y2
=2180
5-13x30 =933
5-132=5150
5-302
= 436 -390 = 186.6 -169 = 1030 –900
... Cov(x,y) = 46 V(x) = 17.6 V(y) =130
Now we find,
Regression co -efficient of X on Y Regression co -efficient of X on Y
bxy=(x, )
( )Cov y
V ybyx=(x, )
(x)Cov y
V
=46
130=46
17.6
...bxy= 0.35 and byx= 2.61
Now substituting the values of x,y,bxyand b yxin the regression
equations we get,
(x-13) = 0.35(y -30)-------x on y (i)
(y-30) =2.61(x -13)------- y on x (ii)
as the two regression equations.munotes.in
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Now to estimate x when y =25, we use the regression equation of x on y
...(x-13) = 0.35(25 -30)
... x = 13 -1.75 = 11.25
Remark:
From the above example we can note some points about Regression
coefficients.
Both the regression coefficients carry the same sign (+ or -)
Both the regression coefficients can not be greater than 1 in
number
(e.g. -1.25 and -1.32) is not possible.
Product of both the regression coefficients b xyand b yxmust be < 1
i.e. b xy Xbyx< 1 Here 0.35x2.61 = 0.91< 1 (Check this
always)
Example 2 :
Obtain the two regression equation sand hence find the value of y when
x=10
Data: -
X Y XxY X2Y2
12 25 300 144 625
20 18 360 400 324
8 17 136 64 289
14 13 182 196 169
16 15 240 256 225
x=70y=88xy=1218x2=1060y2=1632
And n= 5
Now the two regression equations are,
(x-x) = b xy(y-y) ------- x on y ( i)
(y-y) = b yx(x-x)------- y on x (ii)
Where,
x=1
nx =70
5= 14 and y=1
ny =88
5=17.6
Also,
Cov(x,y,) =1
nxy-xy V(x) =1
nx2-x2V(y) =1
ny2-y2
=1218
5–14x17.6 =1060
5-142=1632
5-17.62
= 243.6 -246.4 =212-196 = 326.4 –309.76
... Cov(x,y) =-2.8 V(x) = 16 V(y) = 16.64munotes.in
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Now we find,
Regression co -efficient of X on Y Regression co -efficient of X on Y
bxy=(x, )
( )Cov y
V ybyx=(x, )
(x)Cov y
V
=2.8
16.64=-2.8
16.64
...bxy=-0.168 byx= 0.175
Now substituting the values of x,y,bxyand b yxin the regression
equations we get,
(x-14) = -0.168(y -17.6) ------- x on y (i)
(y-17.6)= -0.175(x -14)------- y on x (ii)
as the two regression equations.
Now to estimate y when x =10, we use the regressi on equation of y on x
...(y-17.6) = -0.175(10 -14)
... y =17.6 +0.7 = 24.3
Example 3 :
The following data give the experience of machine operators and
their performance rating given by the number of good parts turned out per
100 pieces.
Operator: 1 2 3 4 5 6 7 8
Experience: 16 12 18 4 3 10 5 12
(in years)
Performance: 87 88 89 68 78 80 75 83
Rating
Obtain the two regression equations and estimate the permance rating of
an operator who has put 15 years in service.
Solution : We define the varia bles,
X: Experience y: Performance rating
Table of calculations:
X Y Xy x2Y2
16 87 1392 256 7569
12 88 1056 144 7744
18 89 1602 324 7921
4 68 272 16 4624
3 78 234 9 6084
10 80 800 100 6400
5 75 375 25 5625
12 83 996 144 6889
x=80y=648xy=6727x2= 1018 y2=52856munotes.in
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Now the two regression equations are,
(x-x) = b xy(y-y) ------- x on y (i)
(y-y) = b yx(x-x)------- y on x (ii)
Wher e,
x=1
nx =80
8= 10 and y=1
ny =648
8=81
Also,
Cov(x,y,) =1
nxy-xy V(x) =1
nx2-x2V(y) =1
ny2-y2
=6727
8–10x81 =1018
8-102=52856
8-812
= 840.75 -810 = 127.25 -100 = 6607 –6561
... Cov(x,y) = 30.75 V(x) = 27.25 V(y) = 46
Now we find,
Regression co -efficient of X on Y Regression co -efficient of X on Y
bxy=(x, )
( )Cov y
V ybyx=(x, )
(x)Cov y
V
=30.75
46=30.75
27.25
...bxy= 0.67 and byx= 1.13
Now substituting the values of x,y,bxyand b yxin the regression
equations we get,
(x-10) = 0.67(y -81)------- x on y (i)
(y-81)=1.13(x -10)------- y on x (ii)
as the two regression equations.
Now to estimate Performance rating (y) when Experience (x) = 15, we use
the regression equation of y on x
...(y-81) =1.13(15 -10)
...y = 81+ 5.65 = 86.65
Hence the est imated performance rating for the operator with 15 years of
experience is approximately 86.65 i.e approximately 87
4.5.2 Regression coefficients in terms of correlation coefficient.
We can also obtain the regression coefficients b xyand b yxfrom
standa rd deviations, x.,yand correlation coefficient ‘r’ using the
formulas
bxy=x
yr
and byx=y
xr
munotes.in
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Also consider,
bxyXbyx=x
yr
y
xr
= r2i.e. r = xxy yxb b
Hence the correlation coefficient ‘r’ is the geometric mean of the
regression coefficients, bxyandbyx
Example 5:
You are given the information about advertising expe nditure and sales:
---------------------------------------------------------------------------------------
Exp. on Advertisiment Sales (Rs. In Lakh)
(Rs. In Lakh)
----------------------------------------------------------------------------------- -----
Mean 10 90
S.D. 3 12
----------------------------------------------------------------------------------------
Coefficient of correlation between sales and expenditure on Advertisement
is 0.8 .Obtain the two regression equations.
Find the likely sales when advertisement budget is Rs. 15 Lakh.
Solution : We define the variables,
X: Expenditure on advertisement
Y: Sales achieved.
Therefore we have,
x= 10, y=90, 6x= 3, 6y = 12 and r = 0.8
Now, using the above results we can write the two regression equations as
(x-x) =x
yr
(y-y)------- x on y (i)
(y-y) =y
xr
(x-x)------- y on x (ii)
Substituting the values in the equations we get,
(x-10) =30.812(y-90)
i.e x -10 = 0.2 (y -90) ------- x on y (i)
also (y-90) =120.83(x-10)
i.e. y-90 = 3.2 (x -10) ------- y on x (ii)
Now when expenditure on advertisement (x) is 15, we can find the sales
from eqn (ii) as,
y-90 = 3.2 (15 -10)
...y = 90 + 16 = 106munotes.in
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Thus the likely sales are Rs.106 Lakh.
Example 6: Comput the two regression equations on the basis of the
following information:
X Y
Mean 40 45
Standard deviation 10 9
Karl Pearson’s coefficient of correlation between x and y = 0.50.
Also estimate the value of x when y = 48 using the appropriate equation.
Solution: We have,
x= 40, y=45,x= 10,y = 9 and r = 0.5
Now, we can write the two regression equations as
(x-x) =x
yr
(y-y)------- x on y (i)
(y-y) =y
xr
(x-x)------- y on x (ii)
Substituting the values in the equations we get,
(x-40) =100.59(y-45)
i.e x-40 = 0.55 (y -45)------- eqn of x on y (i)
and (y -45) =90.510(x-40)
i.e. y-45 = 0.45(x -40)------- eqn of y on x (ii)
Now when y is 48, we can find x from eqn (i) as,
x-40 = 0.55(48 -45)
...x = 40 +1.65 = 41.65
Example 7:
Find the marks of a student in the Subject of Mathematics who
have scored 65 marks in Accountancy Given,
Average marks in Mathematics 70
Accountancy 80
Standard Devi ation of marks in Mathematics 8
in Accountancy 10
Coefficient of correlation between the marks of Mathematics and marks of
Accountancy is 0.64.munotes.in
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Solution : We define the variables,
X: Marks in Mathematics
Y: Marks in Accountancy
Therefore w e have,
x= 70, y=80,σx= 8,σy=10 and r= 0.64
Now we want to approximate the marks in Mathematics (x), we obtain the
regression equation of x on y, which is given by
(x-x)=x
yr
(y-y)------- x on y (i)
Substituting the values we get,
(x-70) =80.6410(y-80)
i.e x-70 = 0.57 (y -80)
Therefore, when marks in Accountancy (Y) = 65
x-70 = 0.57(65 -80)
...x = 70 -2.85 = 67.15 i.e. 67 appro.
Use of regression equations to find means x,yS.D.sx ,yand
correlation coefficient ‘r’
As we have that, we can obtain t he regression equations from the
values of Means, standard deviations and correlation coefficients ‘r’, we
can get back these values from the regression equations.
Now, we can note that the regression equation is a linear equation
in two variables x and y. Therefore, the linear equation of the type
Ax+By+C = 0 or y = a+bx represents a regression equation.
e.g. 3x+5y -15 = 0 and 2x+7y+10 = 0 represent the two regression
equations.
The values of means x,ycan be obtain by solving the two equations as
the simultaneous equations.
Example 8 :
From the following regression equation, find means x,y,x,yand ‘r’
3x-2y-10 = 0, 24x -25y+145 = 0
Solution : The two regression equations are,
3x-2y-10 = 0 -------- (i)
24x-25y+145 = 0 ---(ii)
Now for xand ywe solve the two equations as the simultaneous
equations.munotes.in
80
Therefore, by (i) x 8 and (ii) x1, we get
24x-16y-80 = 0
24x-25y+145 = 0
-+-
9y-225 = 0 y =225
9= 25
Putting y = 25 in eqn (i), we get
3x-2(25) -10 = 0
3x–60 = 0 x =60
3= 20
Hence x= 20 and y= 25.
Now to find ‘r’ we express the equations in the form y=a+bx
So, from eqns (i) and (ii)
y =3
2x–10
2and y =24
25x+145
25
... b1=3
2= 1.5 ... b 2=24
25= 0.96
Since, b1> b 2(i.e. b 2is smaller in number irrespective of sign + or -)
... Equation (ii) is regression of y on x and b yx= 0.96
Hence eqn (i) is regression of x on y and bxy= 1/1.5 = 0.67
________ _________
Now we find, r = √bxy Xbyx i.e. r =√0.67x0.96 = + 0.84
(The sign of ‘r’ is same as the sign of regression coefficients)
Example 9:
Find the means values of x,y, and r from the two regression equations.
3x+2y -26=0 and x+y-31=0. Also find xwheny= 3.
Solution : The two regression equations are,
3x+2y -26=0 -------- (i)
6x+y -31=0 ---------- (ii)
Now for x and y we solve the two equations as the simultaneous
equations.
Therefore, by (i) x 2 and (ii) x1, we get
6x+4y -52 = 0
6x+ y -31 = 0
--+ ...y =21
3= 7
3y-21 = 0munotes.in
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Putting y = 7 in eqn (i), we get
3x+2(7) -26 = 0
3x–12 = 0 x =12
3= 4.
Hence x = 4 and y = 7.
Now to find ‘r’ we express the equations in the form y= a+bx
So, from eqns (i) and (ii)
y =-3
2x–26
2and y =-6
1x +31
1
... b1=-3
2=-1.5 ... b 2=6
1=-6
since, b1< b 2(i.e. b 1is smaller in number irrespective of sign + or -)
... Equation (i) is regression of y on x and byx=-1.5
Hence, eqn (ii) is regression of x on y and bxy=-1/6 = -0.16
Now we find, . r =bxyxbyx r = 0.16x1.5 = = -0.16
Note: The sign of ‘r’ is same as the sign of regression coefficients
Now to find 6x when 6y = 3, we use the formula,
byx=x
yr
-1.5=-0.16x3
6x
... 6x =0.48
1.5= 0.32
Hence means x= 4,y= 7, r = -0.16 and 6x = 0.32.
EXERCISES
1.What is mean by Regression? Expl ain the use of regression in the
statistical analysis.
2.Why are there two Regressions? Justify.
3.State the difference between Correlation and Regression.
4.Obtain the two regression equations from the data given bellow.
X: 7 4 6 5 8
Y: 6 5 9 8 2
Hence estimate y when x = 10.
5.The data given below are the years of experience (x) and monthly
wages (y) for a group of workers. Obtain the two regression equations
and approximate the monthly wages of a workers who have completed
15 year s of service.munotes.in
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Experience: 11 7 9 5 8 6 10
In years
Monthly wages: 10 8 6 8 9 7 11
(in ‘000Rs.)
6.Following results are obtained for a bivariate data. Obtain the two
regression equa tions and find y when x = 12
n = 15x= 130y = 220 x2= 2288y2= 5506xy = 3467
7.Marks scored by a group of 10 students in the subjects of Maths and
Stats in a class test are given below.Obtain a suitable regression
equation to find th e marks of a student in the subject of Stats who have
scored 25 marks in Maths.
Student no: 1 2 3 4 5 6 7 8 9 10
Marks 13 18 9 6 14 10 20 28 21 16
in Maths
Marks 12 25 11 7 16 12 24 25 22 20
in Stats:
8.The data given below are the price and demand for a certain
commodity over a period of 7 years. Find the regression equation of Price
on Demand and hence obtain the most likely demand for the in the year
2008 when it’s price is Rs.23.
Year: 2001 2002 2003 2004 2005 2006 2007
Price(in RS): 15 12 18 22 19 21 25
Demand 89 86 90 105 100 110 115
(100 units)
9.For a bivariate data the following results were obtained
x= 53.2 , y= 27.9 , 6x = 4.8, y =.4 and r =0.75
Obtain the two regression equations, fi nd the most probable value of x
when y =25.
10. A sample of 50 students in a school gave the following statistics
about Marks of students in Subjects of Mathematics and Science,
------------------------------------------------------------------------------ ---------
Subjects: Mathematics Science
---------------------------------------------------------------------------------------
Mean 58 79
S.D. 12 18
--------------------------------------------- -------------------------------------------munotes.in
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Coefficient of correlation between the marks in Mathematics and marks in
Science is 0.8. Obtain the two regression equations and approximate the
marks of a student in the subject of Mathematics whose score in Sci ence is
65.
11. It is known that the Advertisement promotes the Sales of the
company. The company’s previous records give the following results.
---------------------------------------------------------------------------------------
Expenditure on Advertise ment Sales
(Rs. In Lakh) (Rs. In Lakh)
----------------------------------------------------------------------------------------
Mean 15 190
S.D. 6 20
-------------------------------------------------------------------------- --------------
Coefficient of correlation between sales and expenditure on Advertisement
is 0.6. Using the regression equation find the likely sales when
advertisement budget is Rs.25 Lakh.
12. Find the values of x,y, and r from the two regression equ ations given
bellow. 3x+2y -26=0 and 6x+y -31=0. Also find 6x when y = 3.
13. Two random variables have the regression equations:
5x+7y -22=0 and 6x+2y -20=0. Find the mean values of x and y. Also
find S.D. of x when S.D. of y = 5.
14.The two regression equat ions for a certain data were y = x+5 and 16x
= 9y-94. Find values of x,yand r. Also find the S.D. of y when S.D. of x
is 2.4.
munotes.in
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UNIT I V
Unit -5
TIME SERIES
Unit Structure :
5.0 Objectives
5.1 Introduction
5.2 Importance of Time Series Analysis
5.3 Components of Time Series
5.4 Methods tofind Trend
5.0OBJECTIVE S
From this chapter student should learn analysis of data using
various methods. Methods involve moving average method and least
square method seasonal fluctuations can be studied by business for casting
method.
5.1INTRODUCTION
Every business venture needs to know their performance in the
past and with the help of som e predictions based on that, would like to
decide their strategy for the present By studying the past behavior of the
characteristics, the nature of variation in the value can be determined. The
values in the past can be compared with the present values o f comparisons
at different places during formulation of future plan and policies. This is
applicable to economic policy makers, mete orological department, social
scienti sts, political analysis. Forecasting thus is an important tool in
Statistical analysis .The statistical data, particularly in the field of social
science, are dynamic in nature. Agricultural and Industrial production
increase every year or due to improved medical facilities, there is decline
in the death rate over a period of time. There is increase in sales and
exports of various products over a period of years. Thus, a distinct change
(either increasing or decreasing) can be observed in the value of time -
series.
A time series is a sequence of value of a phenomenon arranged in
order of their occurrence. Mathematically it can expressed as a function,
namely y = f(t) where t represents time and y represents the corresponding
values. That is, the value y 1, y2, y3…… of a phenomenon with respect to
time periods t 1, t2, t3…. F orm a Tim e Series.munotes.in
85
Forecasting tech niques facilities prediction on the basic of a data
available from the past. This data from the past is called a time series. A
set of observations, of a variable, taken at a regular (fixed and equal)
interval of time is calle d time series. A time series is a bivariate data,
with time as the independent variable and the other is the variable under
consideration. There are various forecasting method for time series which
enable us to study the variation or trends and estimat e the same for the
future.
5.2IMPORTANCE OF TIME SERIES ANALYSIS
The analysis of the data in the time series using various forecasting
model is c alled as time analysis. The importance of time series analysis is
due to the following reasons:
Understand ing the past behavior
Planning the future action
Comparative study
5.3COMPONENTS OF TIME SERIES
The fluctuation in a time series are due to one or more of the
following factors which are called “components” of time series.
(a) Secular Trend :
Thegeneral tendency of the data, either to increase, to decrease or
to remain constant is called Secular Trend. It is smooth, long term
movement of the data. The changes in the values are gradual and
continuous. An increasing demand for luxury items like r efrigerators or
colour T.V. sets reflect increasing trend. The production of steel, cement,
vehicles shows a rising trend. On the other hand, decreasing in import of
food grains is an example of decreasing trend. The nature of the trend
may be linear or curvilinear, in practice, curvilinear trend is more
common.
Trend in due to long term tendency. Hence it can be evaluation if
the time series is a available over a long duration.
(b)Seasonal Variation :
The regular, seasonal change in the time series are called
“Seasonal Variation”. It is observed that the demand for umbrellas,
raincoats reaches a peak during monsoon or the advertisement of cold
drinks, ice creams get a boom in summer. The demand for greeting cards,
sweets, increase during festival l ike Diwali, Christmas. In March, there is
maximum withdrawal of bank deposits for adjustment of income -tax
payment, so also variation tax -saving schemes shoot up during this period.
The causes, for these seasonal fluctuations, are thus change in
weather conditions, the traditions and customs of people etc. The seasonalmunotes.in
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component is measured to isolate these change from the trend component
and to study their effect, so that, in any business, future production can be
planned accordingly and necessary adjus tments for seasonal change can be
made
(c)Cyclical Variation :
These are changes in time series, occurring over a period which is
more than a year. They are recurring and periodic in nature. The period
may not be uniform. These fluctuations are due to ch anges in a business
cycle. There are four important phases of any business activity viz.
prosperity, recession, depression and recovery. During prosperity, the
business flourishes and the profit reaches a maximum level. Thereafter, in
recession, the pro fit decreases, reaching a minimum level during
depression. After some time period, the business again recovers
(recovery) and it is followed by period of prosperity. The variation in the
time series due to these phases in a business cycle are called “Cyc lical
Variation:.
Depression RecoveryT im e S c a le
One cycle
The knowledge of cyclic variations is important for a businessman
to plan his activity or design his policy for the phase of recession or
depression. But one should know that the factors affecting the cyclical
variations are quite irregular, difficult to identify and measure. The
cyclical variation are denoted by
(d)Irregular Variation :
The changes in the time series which can not be predicated and are
erratic in nature are called “Irregular Variation ”. Usually, these are short
term changes having signification effect on the time series during that time
interval. These are caused by unforeseen event like wars, floods, strikes,
political charges, etc. During Iran-Iraq war or recent Russian revolution ,
prices of petrol and petroleum product soared very high. In recent budget,
control on capital issued was suddenly removed. As an effect, the all
Indian -Index of share market shooted very high , creates all time records.munotes.in
87
If the effect of other components of the time series is eliminated, the
remaining variation are called “Irregular or Random Variations”. No
forecast of these change can be made as they do not reflect any fixed
pattern.
MODELS FOR ANALYSIS OF TIME SERIES
The purpose of studying time s eries is to estimate or forecast the
value of the variable. As there are four components of the time series,
these are to be studied separately. There are two types of models which
are used to express the relationship of the components of the time series .
They are additive model and multiplicative model.
O = Original Time Series
T = Secular Trend
S = Seasonal Variations
C = Cyclical Variations and
I = Irregular Variations
In Additive model, it is assumed that the effect of the individual
compo nents can be added to get resultant value of the time series, that is
the components are independent of one another. The model can be
expressed as
O = T + S + C + I
In multiplicative model, it is assumed that the multiplication of the
individual effect of the components result in the time series, that is, the
components are due to different causes but they are not necessarily
independent, so that changes in any one of them can affect the other
components. This model is more commonly used. It is expres sed as
O = T × S × C × I
If we want to estimate the value in time series, we have to first
estimate the four components and them combine them to estimate the
value of the time series. The irregular variations can be found. However,
we will restrict ou rselves, to discuss method of estimating the first
components, namely Secular Trend.
5.4METHODS TO FIND TREND
There are various method to find the trend. The major methods are
as mentioned below:
I.Free Hand Curve.
II.Method of Semi –Averages.
III.Method of Moving Averages.
IV.Method of Least Squares.
we will study only the method of moving average and least squares.munotes.in
88
5.4.1.Method of Moving Averages
This is a simple method in which we take the arithmetic average of
the given times series over a certain period o f time. These average move
over period and are hence called as moving averages. The time interval
for the average is taken as 3 years, 4 years or 5 years and so on. The
average are thus called as 3 yearly, 4 yearly and 5 yearly moving average.
The movi ng average sare useful in smoothing the fluctuations caused to
the variable. Obviously larger the time interval of the average more is the
smoothing. We shall study the odd yearly (3 and 5) moving average first
and then the 4 yearly moving average.
Odd Y early Moving Average
In this method the total of the value in the time series is taken for
the given time interval and is written in front of the middle value. The
average so taken is also written in front of this middle value. This average
is the trend for that middle year. The process is continued by replacing the
first value with the next value in the time series and so on till the trend for
the last middle value is calculated. Let us understand this with example:
Example 1:
Find 3 years moving aver ages and draw these on a graph paper.
Also represent the original time series on the graph.
Year 199
9200
0200
1200
22003 200
4200
5200
62007
Production
(in thousand
unit)12 15 20 18 25 32 30 40 44
Solution:
We calculate arithmetic mean of first three observations viz. 12, 15
and 20, then we delete 12 and consider the next one so that now, average
of 15, 20 and 18 is calculated and so on. These averages are placed
against the middle year of each group, viz. the year 2000, 2001 and so on.
Note moving averages are not obtained for the year 1999 and2007.
Year Production
(in thousand
unit)3 Years Total 3yrly. Moving
Average
1999 12
2000 15 12 + 15 + 20 = 47 47 / 3 = 15.6
2001 20 15 + 20 + 18 = 53 53 / 3 = 17.6
2002 18 20 + 18 + 25 = 63 63 / 3 = 21.0
2003 25 18 + 25 + 32 = 75 75 / 3 = 25.0
2004 32 25 + 32 + 30 = 87 87 / 3 = 29.0
2005 30 32 + 30 + 40 = 102 102 / 3 = 34.0
2006 40 30 + 40 + 44 = 114 114 / 3 = 38.0
2007 44munotes.in
89
Example 2:
Find 5 yearly moving average for the following data.
Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006
Sales (in
lakhs of
Rs.)51 53 56 57 60 55 59 62 68 70
Solution:
We find the average of first five values, namely 51, 53, 56 , 57 and 60.
Then we omit the first value 51 and consider the average of next five
values, that is, 53, 56, 57, 60 and 55. This process is continued till we get
the average of the last five values 55, 59, 62, 68 and 70. The following
table is prepared.
Year Sales
(in lakhs of Rs.)5 Years Total Moving
Average
(Total / 5)
1997 51 …. ….
1998 53 …. ….
1999 56 51 + 53 + 56 + 57 +60 = 277 55.4
2000 57 53 + 56 + 57 + 60 + 55 = 281 56.2
2001 60 56 + 57 + 60 + 55 + 59 = 287 57.4
2002 55 57 + 60 + 55 + 5 9 + 62 = 293 58.6
2003 59 60 + 55 + 59 + 62 + 68 = 304 60.8
2004 62 55 + 59 + 62 + 68 + 70 = 314 62.8
2005 68 …. ….
2006 70 …. ….
Example 3:
Determine the trend of the following time series using 5 yearly moving
average s.
Solution :The time series is divided into overlapping groups of five years,
their 5 yearly total and average are calculated as shown in the following
table.
Year Export (Y) 5–yearly total (T) 5–yearly moving
average: (T/5)
1981
1982
1983
1984
1985
198678
84
80
83
86
8978+84+80+83+86 = 411
84+80+83+86+89 = 422
80+83+86+89+88 = 426411 / 5 = 82.2
422 / 5 = 84.4
85.2Year 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991
Exports
in‘000Rs 78 84 80 83 86 89 88 90 94 93 96munotes.in
90
1987
1988
1989
1990
199188
90
94
93
9683+86+89+88+90 = 436
86+89+88+90+94 = 447
89+88+90+94+93 = 454
88+90+94+93+96 = 46187.2
89.4
90.8
92.2
Observation s:
I.In case of the 5 –yearly moving avera ge, the total and average for
the first two and the last two in the time series is not calculated.
Thus, the moving average of the first two and the last two years in
the series cannot be computed.
II.To find the 3 –yearly total (or 5 –yearly total) for a particular
years, you can subtract the first value from the previous year’s
total, and add the next value so as to save your time!
Even yearly moving averages
In case of even yearly moving average the method is slightly
different as here we cannot find the middle year of the four yea rs in
consideration. Here we find the total for the first four years and place it
between the second and the third year value of the variable. These totals
are again sunned into group of two, called as centered total and is placed
between the two totals. The 4 –yearly moving average is found by
dividing these centered totals by 8. Let us understa nd this method with
an example
Example 4:Calculate the 4 yearly moving averages for the following
data.
Ans: The table of calculation is show below. Student should leave one
line blank after every to place the centered total in between two years.
Years(Import)
Y4-Yearly
Total4-Yearly
Centered Total4-Yearly
Moving Averages :
1991 15 - -
1992 18 - -
1993 20 77 77 + 83 = 160 160/8 = 20
1994 24 83 83 + 90 = 173 173/8 = 21.6
1995 21 90 90 + 98 = 188 188/8 = 23.5
1996 25 98 98 + 100 = 198 198/8 = 24.8
1997 28 100 100 + 109 = 209 209/8 = 26.1
1998 26 109 - -
1999 30 - - -Year 1991 1992 1993 1994 1995 1996 1997 1998 1999
Import
in‘000Rs 15 18 20 24 21 25 28 26 30
munotes.in
91
Example 5:
Find the moving average of length 4 for the following data. Represent the
given data and the moving average on a graph paper.
Year 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007
Sales (in
thousand
unit)60 69 81 86 78 93 102 107 100 109
Solution: We prepare the following table.
Year Sale (in
thousand
unit)4 Yearly Totals Centred Total Moving /
Avg. Central
= Total / 8
1998 60
1999 69
60+ 69 + 81 + 86 = 296
2000 81 296 + 314 = 610 76.25
69 + 81 + 86 + 78 = 314
2001 86 314 + 338 = 652 81.5
81 + 86 + 78 + 93 = 338
2002 78 338 + 359 = 697 87.125
86 + 78 + 93 + 102 = 359
2003 93 359 + 380 = 739 92.375
78 + 93 + 102 + 107 = 380
2004 102 380 + 402 = 782 97.75
93 + 102 + 107 + 100 = 402
2005 107 402 + 418 = 820 102.5
102 + 107 + 100 + 109 = 418
2006 100
2007 109
Note that 4 yearly total are written between the years 1999 -2000, 2000 -
01, 2001 -02 etc. and the central total are written against the years 2000,
2001, 2002 etc. so also the moving average are considered w.r.t. years;
2000, 2001 and so on. The moving averages are obtained by dividing the
certain total by 8.
The graph of the given set of values and the moving averages
against time representing the trend component are shown below. Note that
the moving averages are not obtained for the years 1998, 1999, 2006 and
2007. (i.e. first and last two extreme years).
When the values in th e time series are plotted, a rough idea about the
type of trend whether linear or curvilinear can be obtained. Then,
accordingly a linear or second degree equation can be fitted to the values.
In this chapter, we will discuss linear trend only.munotes.in
92
5.4.2.LEAST SQUARES METHOD:
Let y = a + bx be the equation of the straight line trend where a, b are
constant to be determined by solving the following normal equations,
∑ y = na + b∑ x
∑ xy = a∑ x + b∑ x2
where y represents the given time series.
We de fine x from years such that ∑ x = 0. So substituting ∑ x = 0 in the
normal equation and simplifying, we get
b =
x2xyand a =ny
Using the given set of values of the time series, a, b can be calculated and
the s traight line trend can be determined as y = a + bx. This gives the
minimum sum of squares line deviations between the original data and the
estimated trend values. The method provides estimates of trend values for
all the years. The method has mathemati cal basis and so element of
personal bias is not introduced in the calculation. As it is based on all the
values, if any values are added, all the calculations are to be done again.
Odd number of years in the time series
When the number of years in t he given time series is add, for the
middle year we assume the value of x = 0. For the years above the middle
year the value given to x are …, -2,-1 while those after the middle year
are values 1,2, … and so on.
Even number of years in the time series
When the number of years in the time series is even, then for the
upper half the value of x are assumed as…., -5,-3,-1. For the lower half
years, the values of x are assumed as 1, 3, 5, …. And so on.
Example 6:
Fit a straight line trend for the follo wing data giving the annual profits (in
lakhs of Rs.) of a company. Estimate the profit for the year 1999.
Years 1992 1993 1994 1995 1996 1997 1998
Profit 30 34 38 36 39 40 44
Solution : Let y = a + bx be the straight line trend.
The number of years i s seven, which is sold. Thus, the value of x is taken
as 0 for the middle years 1995, for upper three years as -3,-2,-1 and for
lower three years as 1, 2, 3.munotes.in
93
The table of computation is as shown below:
Years Profit (y) x xy x2Trend Value:
Yt= a+ bx
1992
1993
1994
1995
1996
1997
199830
34
38
36
39
40
44-3
-2
-1
0
1
2
3-90
-68
-38
0
39
80
1329
4
1
0
1
4
931.41
33.37
35.33
37.29
39.25
41.21
43.17
Total∑y = 261∑x = 0∑xy = 55∑x2= 28
From the table : n = 7, ∑xy = 55, ∑x2= 28,∑y = 261
There fore b =
x2xy=2855= 1.96 and a =ny=7261= 37.29
Thus, the straight line trend is y = 37.29 + 1.96x.
The trend values in the table for the resp ective years are calculated by
substituting the corresponding value of x in the above trend line equation.
Forthe trend value for 1992: x = -3:
y1992= 37.29 + 1.96 (-3)=37.29 -5.88 = 31.41
Similarly, all the remaining trend values are calculate d.
(A short -cut method in case of odd number of years to find the remaining
trend values once we calculate the first one, is to add the value of b to the
first trend value to get the second trend value, then to the second trend
value to get the third one and so on. This is because the difference in the
values of x is 1.)
To estimate the profit for the years 1999 in the trend line equation, we
substitute the prospective value of x, if the table was extended to 1999. i.e.
we put x = 4, the next value afte r x = 3 for the year 1998.
y1999= 37.29 + 1.96 (4) = 45.13
There fore the estimated profit for the year 1999 is Rs. 45.13 lakhs.
Example 7:
Fit straight line trend by the method of lease squares for the following data
represent ing production in thousand units. Plot the data and the trend line
on a graph paper. Hence or otherwise estimate the trend for the years
2007 .munotes.in
94
Year 1999 2000 2001 2002 2003 2004 2005
Production (in
thousand unit) 14 15 17 16 17 20 23
Solution :
Here, the total number of years is 7, an odd number. So we take the center
as 1986 the middle -most year and define x as year 2002. The values of x
will be -3,-2,-1, 0, 1, 2, 3.
Prepare the following table to calculate the required summations. Note
that the trend values can be written in the table only after calculation of a
and b.
Year Production
(y)x x2x y Trend
Values
1999 14 -3 9 -42 13.47
2000 15 -2 4 -30 14.79
2001 17 -1 1 -17 16.11
2002 16 0 0 0 17.43
2003 17 1 1 17 18.75
2004 20 2 4 40 20.07
2005 23 3 3 69 21.39
122 28 37
Here, n = 7, ∑ y = 122, ∑ x2= 28,∑ x y = 37
Now, a and b are calculated as follows:
a = 43.17 4286.177122
ny
b = 32.1 3214.12837
2
xxy
So, the equation is used to find trend values.
y = a + b x
i.e. y = 17.43 + 1.32x
The equation is used to find trend va lues.
For the year 1999, x = -3, substituting the value od x, we get,
y = 17.43 + 1.32 ( -3) = 17.43 –3.96 = 13.47
to find the remaining trend values we can make use of the property of a
straight line that as all the values of x are equidistant with diff erent of one
unit ( -3,-2,-1,----and so on), the estimated trend value will also be
equidistant with a difference of b unit.
In this case as b = 1.32, the remaining trend values for x = -2,-1, 0, ---
etc. are obtained by adding b = 1.32 to the previo us values. So, the trend
values are 13.47, 14.79, 16.11, 17.43, 18.75, 20.07 and 21.39.munotes.in
95
Now to estimate trend for the year 2007, x = 5, substituting in the equation
y = 17.43 + 1.32x
= 17.43 + 1.32 (5) = 24.03
So, the estimated trend value for the year 2007 is 24,030 unit.
For graph of time series, all point sare plotted. But for the graph of trend
line, any two trend values can be plotted and the line joining these points
represents the straight line trend.
0510152025
1999 2000 2001 2002 2003 2004 2005 2006 2007
Yea rsTimeSeries
Time
S eries
Trend L ine
For the trend line, the trend values 17.43 and 21.39 for the years
2002 and 2005 are plotted and then a straight line joining these two points
is drawn and is extended on both the sides.
The estimate of trend for the year 2007 can also be obtained from
the graph by drawing a perpendicular for the year 2007, from x -axis which
meet the trend line at point P. From P, a perpendicular on y -axis gives the
required ternd estimate as 24.
Now, to find straight line trend, when number of years is even, consider
the follo wing example.
Example 8:
Fit a straight line trend to the following time –series, representing
sales in lakhs of Rs. of a company, for the year 1998 to 2005. Plot the
given data well as the trend line on a graph paper. Hence or otherwise
estimate tre nd for the year 2006.
Year 1998 1999 2000 2001 2002 2003 2004 2005
Sales
(Lakhs of Rs.) 31 33 30 34 38 40 45 49munotes.in
96
Solution:
Here the number of years = 8, an even number, so we define
x =5.05. 2001 year,so that the values of x are -7,-3,-1, 1, 3, 5 and 7, to
get∑ x = 0.
Prepare the following table to obtain the summations ∑ x2,∑ y, ∑ x y.
Year Sales
(in Lakhs of
Rs.)x x2x y Trend
Values
1998 31 -7 49 -217 28.33
1999 33 -5 25 -165 30.95
2000 30 -3 9 -90 33.57
2001 34 -1 1 -34 36.19
2002 38 1 1 38 38.81
2003 40 3 9 120 41.43
2004 45 5 25 225 44.05
2005 49 7 49 343 46.67
300 168 220
Here, n = 8, ∑ y = 300, ∑ x2= 168,∑ x y = 220
Now, a and b are calculated as follows:
a = 5.378300
ny
b = 31.1168220
2
xxy
So, the equation of the straight line trend is y = a + b x
i.e. y = 37.5 + 1.31 x
To obtain the trend values, first calculate y for x = -7, for the year 1998
y = 37.5 + 1.31 ( -7)
= 37.5 –9.17 = 28.33
To find the successive trend values, go on addition 2b = 2 × 1.31 = 2.62,
to the preceding values as in this case the different between x values is of
2 units.
So, the estimated values of trend for x = -5,-3,-1, 1, 3, 5, 7 and 7 are
30.95, 33. 57, 36.19, 38.81, 41.43, 44.05 and 46.67 respectively.
Write down these values in the table.
Hence the estimated trend value for the year 2006 is 49.29 (in lakhs of
Rs.).munotes.in
97
Now, for the graph of trend line, note that only two trend values 30.95 and
46.67 w.r.t. years 1999 and 2005 are considered as point. The line joining
these two points represents trend line.
0102030405060
1998 1999 2000 2001 2002 2003 2004 2005 2006
Yea rsTimeSeries
Time S eries
To estimated the trend for the year 2006, drawn a perpendicular
from x -axis at this point meeting the line in P. then from P, draw another
perpendicular on y -axis which gives estimate of trend as 49.
Example 9:
Fit a straight line trend to the following data. Draw the graph of
the actual time series and the trend line. Estimate the sales for the year
2007.
Year 1998 1999 2000 2001 2002 2003 2004 2005
Sales
in‘000Rs 120 124 126 130 128 132 138 137
Solution : lety = a + bx be the straight line trend.
The number of years in the given time series is eight, which is an even
number. The upper four years are assigned the values of x as 1, 2, 3, and
7. Note that ere the difference between the values of x is 2, but the sum is
zero.
Now, the table of computation is completed as shown below:
Years Profit (y) X Xy X2Trend Value:
Yt= a + bx
1998
1999
2000
2001
2002
2003120
124
126
130
128
132-7
-5
-3
-1
1
3-840
-620
-378
-130
128
39649
25
9
1
1
9120.84
123.28
125.72
128.16
130.06
133.04munotes.in
98
2004
2005138
1375
7390
35925
49135.48
137.92
Total∑y = 1035∑x = 0∑xy = 205∑x2= 168
From the table : n = 8, ∑xy = 205, ∑x2= 168,∑y = 1035
b = 22.1168205
2
xxyand a = 38.12981035
ny
Thus, the straight line trend is y = 129.38 + 1.22x.
The trend values in the table for the respective years are calculated by
substituting the corresponding value of x in the above trend line equation.
For the trend value for 1998: x = -7:
y1998= 129.38 + 1.22 ( -7)= 129.38 -8.54 = 120.84
Similarly, all t he remaining trend values are calculated.
(A short -cut method in case of even number of years to find the remaining
trend values once we calculate the first one, is to add twice the value of b
to the first trend value to get the second trend value, then t o the second
trend value to get the third one and so on. This is because the difference
in the values of x is 2. In this example we add 2 x 1.22 = 2.44)
Estimation:
To estimate the profit for the years 2007 in the trend line equation, we
substitute the prospective value of x, if the table was extended to 2007. i.e.
we put x =11, the next value after x = 9 for the year 2006 and x = 7 for
2005.
y2007= 129.38 + 1.22 (11) = 142.8
There fore the estimated profit for the year 2007 is Rs. 1,42,800.
Now w e draw the graph of actual time series by plotting the sales
against the corresponding year, the period is taken on the X -axis and the
sales on the Y -axis. The points are joined by straight lines. To draw the
trend line it is enough to plot any two poin t (usually we take the first and
the last trend value) and join it by straight line.
To estimate the trend value for the year 2007, we draw a line
parallel to Y -axis from the period 2007 till it meet the trend line at a point
say A. From this point we d raw a line parallel to the X -axis till it meet the
Y-axis at point say B. This point is our estimate value of sales for the year
2007. The graph and its estimate value (graphically) is shown below:munotes.in
99
From the graph, the estima ted value of the sales for the year 2007
is 142 i.e. Rs 1,42,000 (approximately)
Example 10:
Fit a straight line trend to the following data. Draw the graph of
the actual time series and the trend line. Estimate the import for the year
1998.
Solution : Here again the period of years is 6 i.e. even. Proceeding
similarly as in the above problem, the table of calculation and the
estimation is as follows:
Years Import
(y)x xy x2Trend Value:
Yt= a + bx
1991
1992
1993
1994
1995
199640
44
48
50
46
52-5
-3
-1
1
3
5-200
-132
-48
50
138
26025
9
1
1
9
2541.82
43.76
45.7
47.64
49.58
51.52
Total∑y = 280∑x = 0∑xy = 68∑x2= 70
From the table : n = 6, ∑xy = 68, ∑x2= 70,∑y = 280
Their four b = 97.07068
2
xxyand 67.468280
ny
Thus, the straight line trend is y = 46.67 + 0.97x.Year 1991 1992 1993 1994 1995 1996
Import
in‘000Rs 40 44 48 50 46 52
munotes.in
100
All the remaining trend value s are calculated as described in the above
problem.
Estimation:
To estimate the import for the year 1998, we put x = 9 in the tried line
equation. There fore y1997= 46.67 + 0.97 (9) = 55.4
There fore the imports for the year 1997is Rs. 55,400.
The gr aph of the actual time series and the trend values along with the
graphical estimation is an shown below:
From graph the estimated import are Rs. 55,000.
Example 11:
Fit a straight line trend to the following time –series, r epresenting
sales in lakhs of Rs. of a company, for the year 1998 to 2005. Plot the
given data well as the trend line on a graph paper. Hence or otherwise
estimate trend for the year 2006.
Year 1998 1999 2000 2001 2002 2003 2004 2005
Sales
(Lakhs of Rs.) 31 33 30 34 38 40 45 49
Solution:
Here the number of years = 8, an even number, so we define
x =5.05. 2001 year, so that the val ues of x are -7,-3,-1, 1, 3, 5 and 7, to
get∑ x = 0.
Prepare the following table to obtain the summations ∑ x2,∑ y, ∑ x y.munotes.in
101
Year Sales
(in Lakhs of
Rs.)x x2x y Trend
Values
1998 31 -7 49 -217 28.33
1999 33 -5 25 -165 30.95
2000 30 -3 9 -90 33.57
2001 34 -1 1 -34 36.19
2002 38 1 1 38 38.81
2003 40 3 9 120 41.43
2004 45 5 25 225 44.05
2005 49 7 49 343 46.67
300 168 220
Here, n = 8, ∑ y = 300, ∑ x2= 168,∑ x y = 220
Now, a and b are calculated as follows:
a = 5.378300
ny
b = 31.1168220
2
xxy
So, the equation of the straight line trend is y = a + b x
i.e. y = 37.5 + 1.31 x
To obtain the trend values, first calculate y for x = -7, for the year 1998
y = 37.5 + 1.31 ( -7)
= 37.5 –9.17 = 28.33
To fi nd the successive trend values, go on addition 2b = 2 × 1.31 = 2.62,
to the preceding values as in this case the different between x values is of
2 units.
So, the estimated values of trend for x = -5,-3,-1, 1, 3, 5, 7 and 7 are
30.95, 33. 57, 36.1 9, 38.81, 41.43, 44.05 and 46.67 respectively.
Write down these values in the table.
Hence the estimated trend value for the year 2006 is 49.29 (in lakhs of
Rs.).
Now, for the graph of trend line, note that only two trend values 30.95 and
46.67 w.r.t. y ears 1999 and 2005 are considered as point. The line joining
these two points represents trend line.munotes.in
102
0102030405060
1998 1999 2000 2001 2002 2003 2004 2005 2006
Yea rsTimeSeries
Time S eries
To estimate the trend for the year 2006, drawn a perpendicular
from x -axis at this point meeting the line in P. then from P, draw another
perpendicular on y -axis which gives estimate of trend as 49.
MEASURESMENT OF OTHER COMPONENTS
We have studied four method of estimation of Secular Trend. The
following procedure is applied to separate the remaining components of
the time s eries.
Using seasonal indices (s), the seasonal variations in a time series
can be measured. By removing the trend and the seasonal factors, a
combination of cyclical and irregular fluctuations is obtained.
If we assume, multiplicative model, represen ted by the equation
O = T × S × C × I
Then, to depersonalize the data, the original time series (O) divided
by the seasonal indices (S), which can be express as,
ICTSICST
SO
If it is further divided by trend values (T), then we have
ICTICT
Thus a combination of cyclical and irregular variation can be
obtained. Irregular fluctuations, because of their nature, can not be
eliminated completely, but these can be minimized by taking short term
averages and then the estimate of cyclical variation can be obtained.
METHODS TO ESTIMATE SEASONAL FLUCTUATIONS
We have seen methods to separate the trend component of Time
Series. Now, let us see, how to separate the seasonal component of it.
Methods of Seasonal Indexmunotes.in
103
It is used to finds the effect of seasonal variations in a Time Series. The
steps are as follows:
i.Find the totals for each season, as well as the grand total, say G.
ii.Find the arithmetic means of these total, and the grand total by
dividin g the values added.
iii.Find seasonal indices, representing the seasonal component for
each season, using the formula
Seasonal Index =ge GrandAveraSeasonal Averagefor 100
Where, Grand Average =ofValues TotalNoG
.
Example 12:
Find the seasonal component of the time s eries, using method of seasonal
indices.
Seasonal /
YearsI II IV Grand
2003 33 37 32 31
2004 35 40 36 35
2005 34 38 34 32
2006 36 41 35 36
2007 34 39 35 32
Solution:
I II III IV Grand
Total
Average
Seasonal
Index172
34.4 (172 /
5)
25.351004.34
= 97.59195
39
25.3510039
= 110.64172
34.4
25.351004.34
= 97.59166
33.2
25.351002.33
= 94.18705 (G)
35.25(G/20)
The time series can be deseasonalised by removing the effect of seasonal
component from it. It is done using the formula.
Deseasonalised Value =dex SeasonalInlue OriginalVa 100
BUSINESS FORECASTING:
In this chapter, few methods of analyzing the past data and
predicting the future values are already discussed. Analysis of time series
an important ro le in Business Forecasting. One of the aspects of it
estimating future trend values. Now -a-days, any business or industry is
governed by factors like supply of raw material, distribution network,
availability of land, labour and capital and facilitates l ike regular supply of
power, coal, water, etc. a business has to sustain intricate government
regulations, status, everchanging tastes and fashions, the latest technology,
cut throat competition by other manufacturers and many other.munotes.in
104
While making a forec ast, combined effect of above factors should
be considered. Scientific method are used to analyse the past business
condition. The study reveals the pattern followed by the business in the
past. It also bring out the relationship and interdependence of different
industries which helps in interpretation of changes in the right perspective.
The analysis gives an idea about the components of the time series and
their movement in the past. Various indices such as index of production,
prices, bank deposits, money rates, foreign exchange position etc. can
provide information about short and long term variations, the general
trend, the ups downs in a business.
The study of the past data and the comparison of the estimated and
actual values helps in pinpoint ing the areas of shortcoming which can be
overcome. For successful business forescasting co -ordination of all
departments such as production, sales, marketing is sine -qua-nin, which
result in achieving ultimate corporate goals.
There are different theor ies of Business Forecasting such as
i.Time lag or Sequence Theory
ii.Action and Reaction Theory
iii.Cross Cut Analysis Theory
iv.Specific Historical Analogy Theory
Of these, Time lag or Sequence Theory is most important. It is
based on the fact that there is a tim e lag between the effect of changes at
different stages but there is a sequence followed by these effect e.g. In
80’s, the invention of silicon ships brought fourth and fifth generation
computers in use. The computers were introduced in various fields such
as front -line and back house banking, airlines and railways reservation,
new communication technique, home appliances like washing machine
etc. this, in turn, increase the demand for qualified personnel in electronic
filed to manufacture, handle and mai ntain these sophisticated machine. It
has result in mad rush for admission to various branches of electronics and
computer engineering in the recent past.
By applying any one of the these forecasting theories, business
forecasting can be made. It shoul d be noted that while collecting the data
for analysis, utmost care has to be taken so as to increase the reliability of
estimates. The information should be collected by export investigators,
over a long period of time. Otherwise, it may lead to wrong co nclusions.
EXERCISE
1.What is a time series ? Describe the various components of a time
series with suitable example.
2.What are seasonal variation ? Explain briefly with example.
3.Describe the secular trend component of a time series,munotes.in
105
4.What are the method of determining trend in a time series?
5.Compare method of moving average and least squares of estimating
trend component.
6.Find the trend values using the method of semi -averages for the
following data expressing production in thousand unit of a company
for 7 years.
7.Explain the method to calculate 3 yearly and 4 yearly moving
averages.
8.What are the merits and demerits of the method of moving average?
9.Explain the simple average method to find the seasonal indices of a
time series
10.Calculate trend by consider ing three yearly moving average for the
following time series of price indices for the years 2000 -2007. Also
plot on the graph the trend values.
Year 2000 2001 2002 2003 2004 2005 2006 2007
Price Index 111 115 116 118 119 120 122 124
11.Determine the tr end for the following data using 3 yearly moving
averages. Plot the graph of actual time series and the trend values.
Year 1989 1990 1991 1992 1993 1994 1995 1996 1997
Sales
in‘000Rs 24 28 30 33 34 36 35 40 44
12.Determine the trend for th e following data using 3 yearly moving
averages. Plot the graph of actual time series and the trend values.
Year 1977 1978 1979 1980 1981 1982 1983 1984
Sales
in‘000Rs 46 54 52 56 58 62 59 63
13.Determine the trend for the following data us ing 3 yearly moving
averages. Plot the graph of actual time series and the trend values.
14.
Year 1979 1980 1981 1982 1983 1984 1985 1986
Profit in
lakhs of Rs 98 100 97 101 107 110 102 105
15.Determine the tren d for the following data using 5 yearly moving
averages. Plot the graph of actual time series and the trend values.
Year 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000
Values 34 37 35 38 37 40 43 42 48 50 52munotes.in
106
16.Determine the trend for the following data giving the pro duction of
steel in million tons, using 5 yearly moving averages. Plot the graph of
actual time series and the trend values.
Year 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982
Production 28 30.5 32 36.8 38 36 39.4 40.6 42 45 43.5
17.Find five -yearly moving average for the following data which
represents production in thousand unit of a small scale industry. Plot the
given data as well as the moving average on a graph paper.
Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 19891990
Production 110 104 78 105 109 120 115 110 115 122 130
Ans. The trend values are 101.2, 103.2, 105.4, 111.8, 113.8, 116.4 and
118.4 for the years 1982 to 1988.
18.Find the trend component of the following time series of production in
thousand k ilogram during 1971 -1980. Plot the moving average and the
original time on a graph paper.
Year 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980
Production 12 15 18 17 16 20 23 22 24 25
Ans. The trend values are 16, 17.125, 18.375, 19.625, 21.25, 22. 875 for
the years 1973 to 1978.
19.Fit a straight line trend to the following data representing import in
million Rs. of a certain company. Also find an estimate for the year 2008.
Ans. The straight line trend is y = 49 -x. the trend values are 52, 51, 50,
49, 48, 47 and 46 respectively and the estimate trend for the year 2006 is
44 million Rs.
20.The production of a certain brand of television sets in thousand unit is
given below. Fit a straight line trend to the data. Plot the given data and
the trend line on graph find an estimate for the year 2004.
Ans. The straight line trend is y = 947.71 + 33.43 x. the trend values are
846.42, 879.85, 913.28, 946.71, 1013.57 and 1047. The estimate for the
year 2004 is 1080.43 thermal million.
21.The straight line trend by the method of least squares for the following
data which represents the expenditure in lakhs od Rs. on advertisement ofYear 2000 2001 2002 2003 2004 2005 2006
Import 48 50 58 52 45 41 49
Year 1997 1998 1999 2000 2001 2002 2003
production 865 882 910 925 965 1000 1080munotes.in
107
a certain company. Also find an estimate for the year 2005. Plot the given
data and the trend line on a graph paper.
Ans. The trend i s y = 40.13 +2.9x. the trend values are 19.83, 25.62,
31.43, 37.23, 43.03, 48.83, 54.63 and 60.43, 2005 is 66.23.
22.Use the method of least squares to find straight line trend for the
following time series of production in thousand units 1981 –1988. Al so
estimate trend for the year 2003.
Ans. The straight line trend is y = 91.5 + 1.167 x. the trend values are
83.331, 85.665, 87.999, 90.333, 92.667, 95.001, 97.335 and 99 .669. the
estimate of trend, for the year 2003 is 102.003
23.Calculate seasonal indices for the following data:
Year I II III IV
2003 55 53 57 51
2004 56 55 60 53
2005 57 56 61 54
Ans. 100.59, 98.2, 106.57, 94.61
24.Determine the trend for the fo llowing data giving the production of
wheat in thousand tons from the years 1980 to 1990, using the 5 -yearly
moving averages. Plot the graph of actual time series and the trend
values.
Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990
Producti on 13.5 14.7 17 16.2 18.1 20.4 22 21.2 24 25 26.6
25.Determine the trend for the following data giving the income (in
million dollars) from the export of a product from the year 1988 to
1999. Use the 4 -yearly moving average method and plot the graph of
actual time series and trend values.
Year 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999
Income 340 360 385 470 430 444 452 473 490 534 541 576
26.Using the 4 -yearly moving average method find the trend for the
following data.Year 1997 1998 1999 2000 2001 2002 2003 2004
Expenditure 21 24 32 40 38 49 57 60
Year 1995 1996 1997 1998 1999 2000 2001 2002
Production 80 90 92 83 94 99 92 102
munotes.in
108
Year 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977
Value 102 100 103 105 104 109 112 115 113 119 117
27.Determine the trend for the following data giving the sales (in ’00 Rs.)
of a product per week for 20 weeks. Use appropriate moving average
method.
Week
Sales
Week
Sales1
22
11
292
26
12
343
28
13
364
25
14
355
30
15
356
35
16
397
39
17
438
36
18
489
30
19
5210
32
20
49
28.An online marketing company works 5 -days a week. The day -to-day
total sales (in ‘000 Rs) of their product for 4 weeks are given below.
Using a proper moving average method find the trend values.
Days
Sales
Days
Sales1
12
11
352
16
12
323
20
13
324
17
14
385
18
15
366
20
16
357
26
17
348
25
18
389
27
19
4010
30
20
41
29.Fit a straight line trend to the following data. Draw the graph of the
actual time series and the trend line. Estimate the sales for the years
2000.
Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999
Sales in
‘000 Rs 45 47 49 48 54 58 53 59 62 60 64
30.Fit a straight line trend to the following data. Draw the graph of the
actual time series and the trend line. Estimate the sales for the years
2001.
Year 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998
Profit in
‘000 Rs 76 79 82 84 81 84 89 92 88 90
31.Fit a straight line trend to the following data. Draw the graph of the
actual time series and the trend line. Estimate th e sales for the years
2007.
Year 1998 1999 2001 2002 2003 2004 2005 2006
Profit in
‘000 Rs 116 124 143 135 138 146 142 152munotes.in
109
32.Fit a straight line trend to the following data giving the number of
casualties (in hundred) of motorcyclists without hel met. Estimate the
number for the year 1999.
Year 1992 1993 1994 1995 1996 1997 1998
No of
casualties 12 14.2 15.2 16 18.8 19.6 22.1
33.Fit a straight line trend to the following data. Draw the graph of the
actual time series and the trend line. Estimate the import for the years
2002
Year 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000
Import
in’000 Rs. 55 52 50 53 54 56 58 60 57 59
34.Fit a straight line trend to the following data giving the price of crude
oil per barrel in USD. Dr aw the graph of the actual time series and the
trend line. Estimate the sales for the year 2003.
Year 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001
Price per
barrel 98 102 104.5 108 105 109 112 118 115 120
35.Apply the method of le ast squar es to find the number of student
attending the library in the month of May of the academic year 2005 –
2006 from the following data.
Month Jun Jul Aug Sep Oct Nov Dec Jan Feb Mar Apr
Students
105 120 160 225 180 115 124 138 176 230 180
36.Assuming that the trend is absent, find the seasonal indices for the
following data and also find the deseasonalized values.
Quarters I II III IV
1977
1978
1979
198010
12
16
2412
15
18
2614
18
20
2816
22
24
34
37.Calculate seasonal indices for the follow ing data:
Year I II III IV
2003 55 53 57 51
2004 56 55 60 53
2005 57 56 61 54
Ans. 100.59, 98.2, 106.57, 94.61
munotes.in
110
UNIT -IV
Unit -6
INDEX NUMBERS
Unit Structure :
6.0 Objectives
6.1 Introductions
6.2 Importance of Index Numbers
6.3 Price Index Numbers
6.4 Cost of Living Index Number or Consumer Price Index Number
6.5 Useof Cost of Living Index Numbers
6.6 Real Income
6.7 Demerits of Index Numbers
6.0OBJECTIVES
Tounderstand about the importance of Index Numbers.
Tounderstand different types of Index Numbers and their
computations.
Tounderstand about Real Income andCost of Living Index Numbers .
To understand the problems in constructing Index Numbers.
To know the merits and demerits of Index Numbers.
6.1INTRODUCTION
Every variable undergoes some changes over a period of time or in
different regions or due to some factors affecting it. These changes are
needed to be measure d. In the last chapter we have seen how a time series
helps in estimating the value of a variable in future. But the magnitude of
the changes or variations of a variable, if known, are useful for many more
reasons. For example, if the changes in prices of various household
commodities are known, one can plan for a proper budget for them in
advance. If a share broker is aware of the magnitude of fluctuations in the
price of a particular share or about the trend of the market he can plan his
course of action of buying or selling his shares. Thus, we can feel that
there is a need of such a measure to describe the changes in prices, sales,
profits, imports, exports etc, which are useful fr oma common man to a
business organization.
Index number is an important statistical relative tool to measure the
changes in a variable or group of variables with respect to time,
geographical conditions and other characteristics of the variable(s). Index
number is a relative measure, as it is independent of th e units of the
variable(s) taken in to consideration. This is the advantage of index
numbers over normal averages. All the averages which we studied beforemunotes.in
111
are absolute measures, i.e.they are expressed in units, while index
numbers are percentage values which are independent of the units of the
variable(s). In calculating an index number, a base period is considered for
comparison and the changes in a variable are measured using various
methods.
Though index numbers were initially used for measuring th e
changes in prices of certain variables, now it is used in almost every field
of physical sciences, social sciences, government departments, economic
bodies and business organizations. The gross national product (GNP), per
capita income, cost of living in dex, production index, consumption,
profit/loss etc every variable in economics uses this as a tool to measure
the variations. Thus, the fluctuations, small or big, in the economy are
measure dby index numbers. Hence it is called as a barometer of
economic s.
6.2IMPORTANCE OF INDEX NUMBERS
The important characteristics of Index numbers are as follows:
(1)It is a relative measure : As discussed earlier index numbers are
independent of the units of the variable(s), hence it a special kind of
average which can be used to compare different types of data expressed in
different units at different points of time.
(2)Economical Barometer : A barometer is an instrument which measure s
the atmospheric pressure. As the index numbers measure all the ups and
downs in the econ omy they are hence called as the economic barometers.
(3)To generalize the characteristics of a group : Many a time it is
difficult to measure the changes in a variable in complete sense. For
example, it is not possible to directly measure the changes in a bu siness
activity in a country. But instead if we measure the changes in the factors
affecting the business activity, we can generalize it to the complete
activity. Similarly the industrial production or the agricultural output
cannot be measured directly.
(4)To forecast trends : Index numbers prove to be very useful in
identifying trends in a variable over a period of time and hence are used to
forecast the future trends.
(5)To facilitate decision making : Future estimations are always used for
long term and short term planning and formulating a policy for the future
by government and private organizations. Price Index numbers provide the
requisite for such policy decisions in economics.
(6)To measure the purchasing power of money and useful in deflating :
Index numb ers help in deciding the actual purchasing power of money.
We often hear from our elders saying that “ In our times the salary was
just Rs. 100 a month and you are paid Rs. 10,000, still you are not happy!”munotes.in
112
The answer is simple (because of index numbers!) that the money value of
Rs. 100, 30 years before and now is drastically different. Calculation of
real income using index numbers is an important tool to measure the
actual income of an individual. This is called as deflation.
There are different types of index numbers based on their requirement
like, price index, quantity index, value index etc. The price index is again
classified as single price index and composite price index.
6.3PRICE INDEX NUMBERS
The price index numbers are classified as shown in the following diagram:
Notations:
P0: Price in Base Year Q0: Quantity in Base Year
P1: Price in Current Year Q1: Quantity in Current Year
Thesuffix ‘0’stands for the base year and the suffix ‘1’stands for the
current year .
6.3.1 Simple (U nweighted) Price Index N umber By Aggregative
Method
In this method we define the price index number as the ratio of sum of
prices in current year to sum of prices in base year and express it in
percentage. i.e.multiply the quotient by 100.
Symbolical ly, … (1)
Steps for computation:
1.The total of all base year prices is calculated and denoted by0P.
2.The total of all current year pri ces is calculated and denoted by
1P.
3.Using the above form ula, simple price index number is computed.
Example 1
Forthe following data, construct the price index number by simple
aggregative method:
I=1
0P
P
x 100munotes.in
113
Price inCommodity Unit1985 1986
A Kg 10 12
B Kg 4 7
C Litre 6 7
D Litre 8 10
Solution: Follo wing the steps for computing the index number, we find
the totals of the 3rdand 4thcolumns as shown below:
Price inCommodity Unit1985( P0) 1986(P1)
A Kg 10 12
B Kg 4 7
C Litre 6 7
D Litre 8 10
I=1
0P
P
x100 =36
28x 100 = 128.57
Meaning of the value of I:
I= 128.57 means that the prices in 1986, as compared with that in 1985
have increased by 28.57 %.
6.3.2 Simple (Unweighted) Price Index Number by Average of
Price Relatives Meth od
In this method the price index is calculated for every commodity
and its arithmetic mean is taken. i.e.the sum of all price relative is divided
by the total number of commodities.
Symbolically, if there are ncommodities in to consideration, then
the simple price index number of the group is calculated by the formula:
… (2)
Steps for computation
1.The price relatives for each commodity are calculated by the formula:
1
0x 100P
P.
2.The total of these price relatives is calcul ated and denoted as:
1
0x 100P
P
.
3.The arithmetic mean of the price realtives using the above formula no.
(2) gives the required price index number.I=1
01x 100P
n P
munotes.in
114
Example 2
Construct the simple price index number for the following data
using average of price relatives method:
Price inCommodity Unit1997 1998
Rice Kg 10 13
Wheat Kg 6 8
Milk Litre 8 10
Oil Litre 15 18
Solution : In this method we have to find price relatives for every
commodity and then total these price relatives. Following the steps for
computing as mentioned above, we introduce first, the column of price
relatives. The table of computation is as follows:
Price in
Commodity Unit1997( P0)1998( P1)1
0x 100P
P
Rice Kg 10 13 130
Wheat Kg 6 8 133.33
Milk Litre 8 10 125
Oil Litre 15 18 120
Total: 508.33
Now, n= 4 and the total of price relatives is 508.33
1
01x 100PIn P
=508.33
4= 127.08
The prices in 1998 have increased by 27 % as compared with in 1997.
Remar k:
1.The simple aggregative method is calculated without taking into
consideration the units of individual items in the group. This may give
a misleading index number.
2.This problem is overcome in the average of price relatives method, as
the individual price relatives are computed first and then their average
is taken.
3.Both the methods are unreliable as they give equal weightage to all
items in consideration which is not true practically.
6.3.3 Weighted Index Numbers by Aggregative Method
In this method w eights assigned to various items are considered in
the calculations. The products of the prices with the corresponding
weights are computed; their totals are divided and expressed in
percentages.
Symbolically, if Wdenotes the weights assigned and P0,P1have
their usual meaning, then the weighted index number using aggregative
method is given by the formula:munotes.in
115
… (3)
Steps to find weighted index number using aggregative method
1.The columns of P1WandP0Ware introduced.
2.The totals of thes e columns are computed.
3.The formula no. (3) is used for computing the required index number.
Example 3
From the following data, construct the weighted price index number:
Commodity A B C D
Price in 1982 6 10 4 18
Price in 1983 9 18 6 26
Weight 35 30 20 15
Solution : Following the st eps mentioned above, the table of computations
is as follows:
Using the totals from the table, we have
Weighted Index Number I=1
0PW
PW
x100 =1365
860x 100 = 158.72
Remark :
There are different formulae based on what to be taken as the weight while
calculating the weighted index numbers. Based on the choice of the weight
we are going to study here three types of weighted index numbers: (1)
Laspeyre’s Index Number, (2) Paasche’s Index Number and (3) Fisher’s
Index Numb er.
(1)Laspeyre’s Index Number :
In this method, Laspeyre assumed the base quantity ( Q0) as the
weight in constructing the index number. Symbolically, P0,P1and Q0
having their usual meaning, the Laspeyre’s index number denoted by ILisCommodit
yWeight
(W)Price
in 1982
(P0)P0WPrice in
1983
(P1)P1W
A 35 6 210 9 315
B 30 10 300 18 540
C 20 4 80 6 120
D 15 18 270 26 390
Total - -0PW =
860- 1PW = 1365I=1
0PW
PW
x 100
munotes.in
116
given by the formu la:
… (4)
Steps to compute IL:
1. The columns of the products P0Q0andP1Q0are introduced.
2. The totals of these columns are computed.
3. Using the above formula no. (4), ILis computed.
Example 4
From the data given below, construct the Laspeyre’s index number:
1965 1966CommodityPrice Quantity Price
A 5 12 7
B 7 12 9
C 10 15 15
D 18 5 20
Solution : Introducing the columns of the products P0Q0and P1Q0, the
table of computation is completed as shown below:
1965 1966
Commodity Price
(P0)Quantity
(Q0)Price
(P1)P0Q0 P1Q0
A 5 12 7 60 84
B 7 12 9 84 108
C 10 15 15 150 225
D 18 5 20 90 100
Total - - - 0 0P Q = 3841 0PQ = 517
Using the totals from the table and substituting in the formula no. (4), we
have
1 0
0 0517x 100384LPQIP Q x 100 = 134.64
(2)Paasche’s Index Number :
In this method, Paasch assumed the current year quantity ( Q1) as the
weight for constructing the index number. Symbolically, P0,P1andQ1
having their usual meaning, the Paasche’s index number denoted by IPis
given by the formula:
… (5)
The steps for computing IPare similar to that of IL.1 0
0 0LPQIP Qx 100
1 1
0 1PPQIP Qx 100munotes.in
117
Example 5
From the data given below, construct the Paasche’s index number:
1985 1986CommodityPrice Price Quantity
A 5 8 10
B 10 14 20
C 6 9 25
D 8 10 10
Solution : Introducing the columns of the products0 1P Qand1 1PQ, the table
of computations is completed as shown below:
1985 1986
Commodity Price
(P0)Price
(P1)Quantity
(Q1)P0Q1 P1Q1
A 5 8 10 50 80
B 10 14 20 200 280
C 6 9 25 150 225
D 8 10 10 80 100
Total - - -P0Q1
= 480P1Q1=
685
Using the totals from the table and substituting in th e formula no. (5), we
have:1 1
0 1PPQIP Qx 100 =685
480x 100 = 142.71
(3)Fisher’s Index Number :
Fisher developed his own method by using the formulae of Laspeyre and
Paasche. He defined the index number as the geometric m ean of ILandIP.
Symbolically, the Fisher’s Index number denoted as IFis given by the
formula: IF= xL PI I =1 0 1 1
0 0 0 1xPQ PQ
P Q P Q
x 100 .. (6)
Note :
1.The multiple 100 is outside the square root sign .
2.While computing products of t he terms, care should be taken to
multiply corresponding numbers properly.
Example 6
From the following data given below, construct the (i) Laspeyre’s index
number, (ii) Paasche’s index number and hence (iii) Fisher’s index
number.
1975 1976ItemPrice Quantity Price Quantity
A 4 12 6 16
B 2 16 3 20
C 8 9 11 14munotes.in
118
Solution: Introducing four columns of the products of P0Q0,P0Q1,P1Q0
andP1Q1, the table of computations is completes as shown below:
From the table, we have P0Q0= 152, P0Q1= 216, P1Q0= 219 and
P1Q1= 310
IL=1 0
0 0PQ
P Q
x 100 =219
152x 100 = 144.08
IP=1 1
0 1PQ
P Q
x 100 =310
216x 100 = 143.52
IF= xL PI I =144.08 x 143.52 = 143.8
Remark :
1.Laspeyre’s index number though popular has a drawback that it does
not consider the change in consumption over a period. (as it does not
take into account the current quantity).
2.Paasch e’s index number overcomes this by assigning the current year
quantity as weight.
3.Fisher’s index number being the geometric mean of both these index
numbers, it considers both the quantities. Hence it is called as the ideal
index number.
Example 7
From th e following data given below, construct the Kelly’s index number:
Base Year Current YearItemPrice Quantity Price Quantity
A 18 20 24 22
B 9 10 13 16
C 10 15 12 19
D 6 13 8 15
E 32 14 38 18
Solution: Introducing the columns of Q=0 1
2Q Q,P0QandP1Q, the table
of computations is completed as shown blow:Item P0 Q0 P1 Q1 P0Q0P0Q1P1Q0P1Q1
A 4 12 6 16 48 64 72 96
B 2 16 3 20 32 40 48 60
C 8 9 11 14 72 112 99 154
Total 152 216 219 310
munotes.in
119
Item Q0 Q1 Q P0 P0Q P1 P1Q
A 20 22 21 18 378 24 504
B 10 16 13 9 117 13 169
C 15 19 17 10 170 12 204
D 13 15 14 6 84 8 112
E 14 18 16 32 512 38 608
Total 1261 -- 1597
From the table, we have P0Q= 1261 and P1Q= 1597
IK=1
0PQ
P Q
x 100 =1597
1261x 100 = 126.65
6.3.4 Weighted Index Numbers using ave rage of price relatives
method
This is similar to what we have seen in subsection 7.3.2. Here the
individual price relatives are computed first. These are multiplied with the
corresponding weights. The ratio of the sum of the products and the total
value o f the weight is defined to be the weighted index number.
Symbolically, if Wdenotes the weights and Idenote the price relatives
then the weighted index number is given by the formula:IW
W
… (8)
One of the important weighted index num ber is the cost of living index
number , also known as the consumer price index (CPI) number .
6.4 COST OF LIVING INDEX NUMBER OR
CONSUMER PRICE INDEX NUMBER
There are two methods for constructing this index number:
(1) Aggregative expenditure method and (2) Family Budget Method
(1)In aggregative expenditure method we construct the index number by
taking the base year quantity as the weight. In fact this index number is
nothing but the Laspeyre’s index number.
(2)In family budget method, value weights are com puted for each item in
the group and the index number is computed using the formula:
IW
W
, where I=1
0P
Px 100 and W=P0Q0 … (9)munotes.in
120
Example 8
A survey of families in a city revealed the following inf ormation:
Item Food Clothing FuelHouse
RentMisc.
%
Expenditure30 20 15 20 15
Price in 1987 320 140 100 250 300
Price in 1988 400 150 125 250 320
What is the cost of living index number for 1988 as compared to that of
1987?
Solution : Here % expendi ture is taken as the weight ( W). The table of
computations is as shown below:
Item P0 P1I=1
0P
Px 100%
Expenditure
(W)IW
Food 320 400 125 30 3750
Clothing 140 150 107.14 20 2142.8
Fuel 100 125 125 15 1875
House Rent 250 250 100 20 2000
Miscellaneous 300 320 106.67 15 1600.05
Total W=100 11367.85
From the table, we have W= 100 and IW= 11376.85
cost of living index number =IW
W
=11367.85
100= 113.68
6.5USE OF COST OF LIVING INDEX NUMBERS
1.These index numbers reflect the effect of rise and fall in the economy
or change in prices over the standard of living of the people.
2.These index number s help in determining the purchasing power of
money which is the reciprocal of the cost of living index number.
3.It is used in deflation. i.e.determining the actual income of an
individual. Hence it also used by the management of government or
private orga nizations to formulate their policies regarding the wages,
allowance to their employees.
6.6 REAL INCOME
As discussed earlier in this chapter, index numbers are very useful
in finding the real income of an individual or a group of them, which
facilitat es the different managements to decide their wage policies. Themunotes.in
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process of measuring the actual income vis -a-vis the changes in prices is
called as deflation .
The formula for computing the real income is as follows:
Real Income of a year =Money Income for the year
Price Index of that yearx 100
Example 12
Calculate the real income for the following data:
Year 1990 1991 1992 1993 1994 1995
Income in
Rs.800 1050 1200 1600 2500 2800
Price
Index100 105 115 125 130 140
Solution : The real income is calculated by the formula:
real income =Money Income for the year
Price Index of that yearx 100
The table of computation of real income’s is completed as shown below:
Year Income in Rs.Price
IndexReal Income
1990 800 100 800
1991 1050 1051050
105x100 = 1000
1992 1200 1151200
115x 100 = 1043
1993 1600 1251600
125x 100 = 1280
1994 2500 1302500
130x 100 = 1923
1995 2800 1402800
140x 100= 2000
6.7DEMERITS OF INDEX NUMBERS
(1)There are numero us types and methods of constructing index
numbers. If an appropriate method is not applied it may lead to wrong
conclusions.
(2)The sample selection may not be representative of the complete series
of items.
(3)The base period selection also is personalized and hence may be
biased.
(4)Index number is a quantitative measure and does not take into
account the qualitative aspect of the items.
(5)Index numbers are approximations of the changes, they may not
accurate.munotes.in
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Check Your Progress
1.Define Index Numbers.
2.Write a shor t note on the importance of Index Numbers.
3.“Index Numbers are the Economical barometers”. Discuss this
statement with examples.
4.Discuss the steps to construct Index Numbers.
5.What are the problems in constructing an Index Number?
6.Define Cost of Living Index Number and explain its importance.
7.What do you mean by (i) Chain Based Index Number and (ii) Fixed
Base Index Number? Distinguish between the two.
8.Define (i) Laspeyre’s Index Number, (ii) Paasche’s Index Number
and (iii) Fisher’s Index Number. What is the difference between the
three? Which amongst them is called as the ideal Index Number?
Why?
9.What are the demerits of Index Numbers?
10.From the following data, construct the price index number by simple
aggregative method:
Price inCommodity Unit1990 1991
A Kg 14 18
B Kg 6 9
C Litre 5 8
D Litre 12 20
Ans: 148.65
11.From the following data, construct the price index number for 1995,
by simple aggregative method, with 1994 as the base:
Price inCommodity Unit1994 1995
Rice Kg 8 10
Wheat Kg 5 6.5
Oil Litre 10 13
Eggs Dozen 4 6
Ans: 131.48
12.From the following data, construct the price index number for 1986,
by average of price relatives method:
Price inCommodity Unit1985 1986
Banana Dozen 4 5
Rice Kg 5 6
Milk Litre 3 4.5
Slice Bread One Pac ket 3 4
Ans: 132.08munotes.in
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13.From the following data, construct the price index number, by method
of average of price relatives:
Price inCommodity Unit1988 1990
A Kg 6 7.5
B Kg 4 7
C Kg 10 14
D Litre 8 12
E Litre 12 18
Ans: 148
14.From the following data, construct the price index number for 1998,
by (i) simple aggregative method and (ii) simple average of price relatives
method, with 1995 as the base:
Price inCommodity Unit1995 1998
Rice Kg 12 14
Wheat Kg 8 10
Jowar Kg 7 9
Pulses Kg 10 13
Ans: (i) 124.32, (ii) 125.06
15.From the following data, construct the weighted price index number:
Commodity A B C D
Price in 1985 10 18 36 8
Price in 1986 12 24 40 10
Weight 40 25 15 20
Ans: 121.29
16.From the following data, construct the index number using (i) simple
average of price relatives and (ii) weighted average of price relatives:
Ans: (i) 130.13, (ii) 128.82Price inCommodity Weight1988 1990
Rice 4 8 10
Wheat 2 6 8
Pulses 3 8 11
Oil 5 12 15munotes.in
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17.From the data given below, construc t the Laspeyre’s index number:
1975 1976CommodityPrice Quantity Price
A 5 10 8
B 6 15 7.5
C 2 20 3
D 10 14 12
Ans: 131.40
18.From the data given below, construct the Paasche’s index number:
1980 1985CommodityPrice Price Quantity
A 4 7 10
B 14 22 16
C 5 7 30
D 8 10 21
Ans: 144.67
19.From the following data given below, construct the (i) Laspeyre’s
index number, (ii) Paasche’s index number and hence (iii) Fisher’s index
number.
1980 1990CommodityPrice Quantity Price Quantity
A 6 15 9 21
B 4 18 7.5 25
C 2 32 8 45
D 7 20 11 29
Ans: (i) 203.83, (ii) 203.37, (iii) 203.60
20.From the following data given below, construct the (i) Laspeyre’s
index number, (ii) Paasche’s index number and (iii) Fisher’s index
number.
Base Year Current Yea rCommodityPrice Quantity Price Quantity
Cement 140 200 167 254
Steel 60 150 95 200
Coal 74 118 86 110
Limestone 35 50 46 60
Ans: (i) 103.98, (ii) 127.4, (iii) 115.09munotes.in
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21.From the following data given below, construct the Fisher’s index
number:
Base Year Current YearCommodit yPrice Quantity Price Quantity
A 2 8 4 14
B 6 14 7 20
C 8.5 10 12 15
D 14 8 19 12
E 22 60 38 85
Ans: 131.37, 120.15
22.From the following data, construct the aggregative price index
numbers by taking the average price of the t hree years as base.
Commodity Price in 1980 Price in 1981 Price in 1982
A 10 12 16
B 16 19 25
C 5 7 10
Ans: 81.58, 100, 134.21
23.From the following data, construct the price index number by taking
theprice in 1978 as the base price using aggregative m ethod:
CommodityPrice in
1978Price in1979Price in
1980
A 16 18 24
B 4 6 7.5
C 11 15 19
D 20 28 30
Ans: 131.37, 120.15
24.From the following data, construct the price index number by taking
the price in 1998 as the base price:
Commodity WeightPrice in
1998Price in
1999Price in
2000
A 3 12 15 20
B 1 8 9 11
C 4 16 20 25
D 2 15 18 22munotes.in
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25.From the following data, construct (i) IL, (ii) IP, (iii) IF
1969 1970CommodityPrice Quantity Price Quantity
Rice 2 10 3 12
Wheat 1.5 8 1.9 10
Jowa r 1 6 1.2 10
Bajra 1.2 5 1.6 8
Pulses 4 14 6 20
Ans: 144.4, 144.56, 144.28
26.Construct the cost of living index number for 1980 using the Family
Budget Method:
Ans: 192.95
27.Construct the cost of living index number for the following data with
base year as 1989.
Ans: for 199 0: 114.49, for 1991 : 132.20
28.A survey of families in a city revealed the following information:
Item Food Clothing FuelHouse
RentMisc.
%
Expenditure30 20 15 20 15
Price in
1987320 140 100 250 300
Price in
1988400 150 125 250 320
What is the cost of living index number for 1 988 as compared to that of
1987 ? Ans: 113.65Price inItem Quantity1975 1980
A 10 5 7
B 5 8 11
C 7 12 14.5
D 4 6 10
E 1 250 600
Price inItem Weight1989 1990 1991
Food 4 45 50 60
Clothing 2 30 33 38
Fuel 1 10 12 13
House Rent 3 40 42 45
Miscellaneous 1 5 8 10munotes.in
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29.Construct the consumer price index number for the following
industrial data:
Ans: 167.30
30.Calculate the real income for the following data:
Year 1988 1989 1990 1991 1992 1993
Income in
Rs.500 550 700 780 900 1150
Price
Index100 110 115 130 140 155
31.The employees of Australian Steel lt d. have presented the following
data in support of their contention that they are entitled to a wage
adjustment. Dollar amounts shown represent the average weekly take
home pay of the group:
Year 1973 1974 1975 1976
Pay in $ 260.50 263.80 274 282.50
Index 126.8 129.5 136.2 141.1
Compute the real wages based on the take home pay and the price indices
given. Also compute the amount of pay needed in 1976 to provide buying
power equal to that enjoyed in 1973.
32.Calculate the real income for the following d ata:
Year 1977 1978 1979 1980 1981 1982
Income in
Rs.250 300 350 500 750 1000
Price
Index100 105 110 120 125 140Item Weight Price Index
Industrial Production 30 180
Exports 15 145
Imports 10 150
Transpor tation 5 170
Other activity 5 190
munotes.in
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33.The per capita income and the corresponding cost of living index
numbers are given below. Find the per capita real income:
Year 1962 1963 1964 1965 1966 1967
per capita
income220 240 280 315 335 390
cost of
living I.N.100 110 115 135 150 160
34.The following data gives the salaries (in ’00 Rs.) of the employees of
Hindusthan Constructions Ltd with the cost of living index number . Find
the real income and suggest how much allowance should be paid to them
to maintain the same standard of living.
Year 1990 1991 1992 1993 1994 1995
Income 12 14 17 20 24 28
Price
Index100 120 135 155 180 225
35.The income of Mr. Bhushan Damle in 1 999 was Rs. 8,000 per month.
If he gets an increment of Rs. 1,200 in 2000 and the price index being 115
with base as 1999, can you conclude that Mr. Damle has got an increment
which will maintain his standard of living as compared with the previous
year?
munotes.in
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UNIT -V
Unit -7
PROBABILITY DISTRIBUTIONS
Unit Structure :
7.1 Introduction
7.2 Mathematical Expectation and Variance
7.3 Binomial Distribution
7.4 Normal Distribution
7.1INTRODUCTION
In the previous chapter we have seen that outcomes of an
experime nt can be expressed in numbers. In an experiment of throwing a
die, the possible outcomes are expressed as 1, 2, 3, 4, 5, or 6. In
experiments like tossing a coin or picking a card or drawing a ball from a
bag, the outcomes are numbers. But they can be ass igned values like in
tossing a coin, the outcome of heads can be assigned value 0 and that of a
tails can be assigned value 1. Thus, in some way all points in a sample
space can be assigned numerical values.
A relation which assigns every outcome of an e xperiment to a real
number is called as a random variable also called as stochastic variable.
We can also say thus, that a random variable assures the probability for
every outcome of an experiment.
Example 1:
In the above example of throwing a die, the random variable say X
takes values {1, 2, 3, 4, 5, 6}.
In an experiment of tossing two coins, we can assume the random
variable as the number of heads (or tails) in an outcome. The sample space
isS= {HH, HT, TH, TT}. If number of heads denotes the val ue of the
random variable ( X), then the first outcome has 2 heads, second and third
has 1 heads and fourth outcome has no heads. Thus, X= {2, 1, 1, 0}. We
will have a different random variable ( Y) if we take the number of tails as
the counter for the rand om variable. In that case Y= {0, 1, 1, 2}.
Discrete random variable:
A random variable which takes discrete (distinct) values or to say
in mathematical words as the variable which takes finite or countably
infinite values is called as discrete random var iable .munotes.in
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Example 2:
The scores on a die, the number of calls received at a call centre,
number of letters typed by a secretary, number of strikes in a factory etc
are examples of discrete random variables.
Probability Distribution:
We know that with every value of the random variable ( X) there is
a probability ( P(X)) assigned for that particular outcome. The set of all
values of the random variable along with their corresponding probabilities
is called as the probability distribution of the random variable .
Example 3:
In an experiment of tossing three coins, if the random variable X
denotes the number of heads in every outcome then the probability
distribution table of Xis as shown below:
X 0 1 2 3
P(X) 1/8 3/8 3/8 1/8
In an experiment of throwing two dice, if the random variable
represents the sum of the scores on the upper faces of both the dice then its
probability distribution table is as shown below:
X Events P(X)
2 (1, 1) 1/36
3 (1, 2), (2, 1) 2/36 = 1/18
4 (1, 3), (2, 2), (3, 1) 3/36 = 1/12
5 (1, 4), (2, 3), (3, 2), (4, 1) 4/36 = 1/9
6 (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) 5/36
7(1, 6), (2, 5), (3, 4), (4, 3), (5,
2), (6, 1)6/36 = 1/6
8 (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) 5/36
9 (3, 6), (4, 5), (5, 4), (6, 3) 4/36 = 1/9
10 (4, 6), (5, 5), (6, 4) 3/36 = 1/12
11 (5, 6), (6, 5) 2/36 = 1/18
12 (6,6) 1/36
The events are explicitly written to count the probability. It is not
expected to write the events every time. One should also observe that the
total of the column of probabiliti es is 1, which we already know from the
previous chapter that ( )iP A = 1 for all events Ai.
The function P(X) is called the probability function ofX.
The probability distribution of a random variable is called as
discrete probability d istribution and the corresponding probability
function is called as probability mass function .munotes.in
131
7.2MATHEMATICAL EXPECTATION and VARIANCE
Let Xbe a random variable taking values x1,x2,…xnwith
corresponding probabilities1p,2p, …,nprespectively. The Expectation
ofx, denoted as E(x) is given by the formula:
1 1 2 2 ( ) ....n n i i E x p x p x p x p x , whereip= 1
In cases of games of chance if the amount received b y a player if
he wins a game is a, the amount he looses if he does not win the game is b
and the probability of winning the game is pthen the expectation is given
by the formula: ap bp, where 1p p.The negative sign b ecause of the
loss the person suffers.
The expectation of a random variable is called as its mean . The
mean of a random variable is denoted by the Greek letter (pronounce as
‘mu’ ofmusic ).
Laws of Expectation
IfXandYare two r andom variables then
1.E(X)0.
2.The expected value of their sum X + Y is given by : E(XY) =E(X)
E(Y).
3.Similarly, E(aXbY) =aE(X)bE(Y).
4.The expected value of their product XYis given by: E(XY) =E(X).E(Y)
Variance of a random variable
Variance of a random variable Xdenoted by V(X) is the square of
the standard deviation of Xand is calculated by the formula: V(X) =E(X2)
–[E(X)]2
Example 4:
A die is thrown at random. What is the expectation and variance of
the number on it?
Ans: When a die is thrown the possible outcomes are X = {1, 2, 3, 4, 5,
6} with each having a probability of 1/6. This can be tabulated as follows:
xi 1 2 3 4 5 6
pi 1/6 1/6 1/6 1/6 1/6 1/6
E(X) =1 1 2 2 ....n n p x p x p x = (1 + 2 + 3 + 4 + 5 + 6) x 1/6 = 21/6 = 3.5
Now, E(X2) = 1 x 1/6 + 22x 1/6 + 32x 1/6 + 42x 1/6 + 52x 1/6 + 62x 1/6
= (1 + 4 + 9 + 25 + 36) x 1/6 = 75/6 = 12.5
V(X) =E(X2)–[E(X)]2= 12.5 –(3.5)2= 12.5 –12.25 = 0.25
Example 5:
Three coins are tossed. What is the expectation of a heads
occurring?
Ans: When three coins are tossed the sample space is as follows:
S= {HHH ,HHT ,HTH ,THH ,HTT,THT,TTH,TTT}munotes.in
132
The outcomes for a head are: 0 heads, 1 heads, 2 heads and 3 heads.
The corresponding probabilities are shown below:
xi(heads) 0 1 2 3
pi 1/8 3/8 3/8 1/8
E(no.of heads ) =E(x) =i ip x =1 3 3 10 x 1 x 2 x 3 x8 8 8 8 =12
8
E(x) = 1.5
Example 6:
The probability that it will rain on a day is 0.6. Find the
expectation of umbrella seller who earns a profit of Rs. 3050 if it rains and
a loss of Rs. 250 if it does n ot rain on a day.
Ans: Given p= 0.6p= 1–0.6 = 0.4, a= 3050 and b= 250
the mathematical expectation = 3050 x 0.6 –250 x 0.4 = 1830 –
100 = 1730
Exercise
(1)Define ( a) random vari able , ( b) discrete random variable, ( c)
probability mass function
(2)Define ( a) mean and ( b) of a random variable.
(3)Prepare a frequency distribution table of tossing 3 coins.
(4)Prepare a frequency distribution table of throwing two dice.
(5)If it rains, a dealer i n umbrella can earn Rs. 300 per day and if it does
not rain he can lose Rs.80 per day. What is the expectation if the
probability of a rainy day is 0.57?
(6)A person plays a game where he earns Rs. X2if he gets Xon a die.
Find the mean and variance of his earning.
(7)In a game, a person Xthrows a coin three times. He is paid Rs. 400 if
he gets a heads all three times. The entry fee for the game is Rs. 80.
What is the mathematical expectation of X?
(8)A shop owner earns a profit of Rs. 1000 per day. On a holiday he
suffers a loss of Rs. 300 per day. If the probability that a day is a
holiday is 0.14, find his mathematical expectation.
(9)A die is thrown at random. What is the expectation of the number on
it?
(10)If two dice are thrown, what is the expectation of the sum of the
sample points?
(11)Ifndice are thrown, what is the expectation of the sum of the sample
points?munotes.in
133
(Hint: In above two problems use laws of expectation)
(12)Find the mean and variance when a coin is tossed two times.
(13)If three coins are tossed, what is the ex pectation and variance of the
number of tails?
(14)A person draws 2 balls from a bag containing 6 white and 5 black
balls. He is paid Rs. 22, if he draws both balls of white color and Rs.
11 if he draws one of each color. Find his expectation.
(15)A person draws 3 balls from a bag containing 3 white, 4 red and 5
black balls. He is offered Rs. 10, Rs. 5 and Rs. 2 if he draws three
balls of same color, 2 balls of same color and 1 ball of each color
respectively. Find his expectation.
(16)The probability there is at least one error in accounts statement
prepared by A is 0.25, by B is 0.35 and by C is 0.4. If A, B and C
prepared 10, 16 and 20 statements respectively, find the expected
number of error free statements.
(17)Three Mathematics teachers X, Y and Z were given 120, 200 and 150
papers of an examination to assess respectively. The probability that
there is totaling mistake in a paper by A, B and C is 0.2, 0.55 and
0.25 respectively. Find the expected number of ( a) papers in which
error is possible and ( b) error free paper s.
(18)IfXandYare two independent discrete random variables such that
E(X) = 12, E(Y) = 20, then find the expectation of Z= 2X+ 3Y.
(19)In a lottery game there are 10 tickets. 3 tickets have a prize of Rs. 2,
2 tickets have a prize of Rs. 5 and 1 ticket has a prize of Rs. 10. The
remaining tickets are blank. Find the expectation of a player winning
a prize.
(20)A newspaper agent earns Rs.200 a day if there is some breaking news
in it and loses Rs. 20 a day if there is no breaking news. The
probability that there is a breaking news in the paper is 0.45. Find the
expectation of his earnings.
(21)A gambler draws a card from a pack of 52 cards. He earns Rs. 104 if
it is an ace, Rs. 52 if it is a King or a Queen, Rs. 26 if it is a Jack and
loses Rs. 13 if it is any other c ard. Find his expectation.
(22)Find the mean and variance for the following probability distribution:
X -5 0 5 10 15
P(X) 1/3 1/9 1/3 1/9 1/9
(23)Find the mean and variance for the following probability
distribution:munotes.in
134
X 0 2 4 6 8 10
P(X) 0.05 0.15 0.02 0.28 0.3 0.2
(24)The probability distribution of daily demand of cell phones in a
mobile gallery is given below. Find the mean and variance.
Demand 5 10 15 20
Probability 0.4 0.22 0.28 0.10
(25)The probability distribution of number of divorce cases withdrawn
per da y from a court is given below. Find the mean and variance.
No. of
Cases0 1 2 3 4 5
Probability 0.2 0.26 0.12 0.22 0.08 0.12
(26)At the famous ‘Jumbo Vada’ shop near Dadar station in Mumbai,
the probability distribution of arrival of number of customers pe r minute is
given below. Find the expectation of customers per minute.
No of
Customers0 2 4 6 8
Probability 0.24 0.22 0.26 0.18 0.1
(27)Mr. Chinmay has bought a new motorcycle from ‘Swastik
Agency’. The agency offers after sales service contract for Rs. 60 0 for
four years. From a market survey the probability distribution of the
expenses on service in four years for the same brand is known and is given
below. Should Mr. Chinmay pay for the service contract?
Expenses 200 400 600 800 1000 1200
Probability 0.34 0.22 0.1 0.08 0.22 0.04
7.3BINOMIAL DISTRIBUTION
In many situations we see that a same experiment is repeated
number of times. Also the probabilities of the outcomes are fixed
irrespective of the previous trials. Let us consider the example of tossi ng a
coin three times. Here one experiment of tossing a coin is done three times
and the probability of getting a head or getting a tail is always the same
i.e.1/2. Let us call the event of getting a heads as a ‘success’ and getting a
tails as a ‘failure’ and the corresponding probabilities as pandqwhere p +
q= 1 (or q =1–p). Consider an outcome ‘HTT’. With our notations just
introduced, there is one success and 2 failures in this outcome. The
probability of this outcome is obtained by multiplying t he individual
probabilities. So we have the probability of HTT as1
2x1
2x1
2=1
8orpxq
xq=pq2. But if the order is not important then the probability of getting
onesuccess and two failures is 3 x pq2, as there are three such outcomes
vizHTT, THT and TTH. It is easy to do so if the number of trials is less.munotes.in
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When the number of trial is more and the probabilities of outcomes are
fixed either a success or a failure, Be rnoulli characterized such probability
distributions and called them as Binomial Distributions .
Characteristics of Binomial Distributions
1.The experiment consists of n(a finite number) number of trials.
2.The outcome of every trial is either a success or a failure and is
independent of the previous trial(s).
3.The probabilities of success (and hence failures) remains constant for
every trial.
4.Ifpdenotes the probability of success, q= 1 –p denotes the
probability of failure and rdenote the number of succes ses in all n
trials then the probability of rnumber of successes is given by the
formula: P(X = r ) =n r n r
rC p q. This is the probability mass function
ofr. Here r= 0, 1, 2, 3, ……., n.
5.The probabilities given in the formula are nothing but the terms in the
binomial expansion of ( p + q )n. Hence the name given to this
distribution is Binomial Distribution.
6.Mean of a Binomial Distribution is given by =np.
7.Variance of a Binomial Distribution is given by npq. Hence its
standard deviation is npq .
Example 7:
The mean of a binomial distribution is 12 and standard deviation is
3. Calculate n,pandq.
Ans: Given np= 12, npq = 3npq= 9
12 x q= 9q= 9/12 = ¾
p= 1–q= 1–¾ = ¼
nx ¼ = 12 n= 12 x 4 = 48
Thus, n= 48, p= ¼ and q= ¾
Example 8:
The mean and variance of a binomial distribution are 14 and 9.
Comment on this statement.
Ans: From the given information, np= 14 and npq = 9npq=
81
14 x q= 81q= 81/14 = 5.78
We know that pr obability of any outcome is never greater than 1.
Thus, the information given is inconsistent.
Example 9
The probability that a youth exercises every day is 0.6. Find the
probability that out of 5 youths selected ( i) none of them do exercise, ( ii)
atleast one exercises.munotes.in
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Ans: Given p= 0.6q= 1–p =1–0.6 = 0.4 and n= 5
(i) To find probability that none of them exercise means r= 0
We know that: P(X=r) =n r n r
rC p q
P(X= 0) =5 0 5 0
0(0.6) (0.4)C= (0.4)5= 0.01024
(ii) To find probability of atleast one does exercise. This is a
complementary event of no one does exercise, whose probability we have
already found.
P(atleast one does exercise) = 1 –P(X= 0) = 1 –0.01024 = 0.98976
Example 10
The average rainfall in a 30 days’ month is 50%. Find the
probability that ( i) the first four days of a given week will be fine and the
remaining will be wet, ( ii) rain will fall on just three days of a week.
Ans: Given p= 50% = 0.5 q= 0.5
(i) To find probability that the first four days of a given week will be fine
and the remaining will be wet. Here the days of rainfall is fixed.
probability for rainfall in remaining days of a week = (0.5)3(0.5)4
= 0.0078
(ii) To find the probability that rain will fall on just three days of a week.
Here which three days is specified.
P(X =3) =7 3 4
3(0.5) (0.5)C =7 x 6 x 5
3 x 2x 0.0078 = 0.27
Example 12
The food inspect or along with his colleagues comes for an
inspection of a drug. The number of faulty drug tablets found in every
sample pack of 16 tablets is given below. The following table shows the
distribution of 160 tablets.
No of faulty
tablets0 1 2 3 4 5
No of sa mples 26 22 30 35 36 11
(i) Fit a binomial distribution and find the expected frequencies, if the
chance of the machine being defective is ½
(ii) Find the mean and standard deviation of the fitted distribution.
Ans: (i) Given p= ½q= ½ , n= 5 and N= 160
We will now find the probabilities of the number of faulty tablets
by finding the binomial expansion of ( p + q )5.
Now, ( p + q )5=5 4 3 2 2 3 4 51 5 10 10 5 1p p q p q p q pq q
Substituting values of pandq. we have
(p + q )5=1 5 10 10 5 1
32 32 32 32 32 32munotes.in
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Now, the numbers1 5 10 10 5 1, , , , ,32 32 32 32 32 32are the corresponding probabilities
of 0, 1, 2, 3, 4 and 5 no. of faulty tablets.
The expected frequencies are calculated by multiplying each probability
with N= 160
Thus, the observed frequencies w ith the expected frequencies are tabulated
as shown below:
No of faulty
tablets0 1 2 3 4 5
Observed
Frequency26 22 30 35 36 11
Expected
Frequency5 25 50 50 25 5
(ii) Mean = np= 5 x ½ = 2.5
standard deviation = npq =1 15 x x2 2= 1.12
Exercise
(1)Ifn= 10 and p= ¼, find the mean and variance of the binomial
distribution.
(2)The mean of a binomial distribution is 40 and standard deviation is 6.
Calculate n,pandq.
(3)The mean of a binomial distribution is 6 and its st andard deviation is
. Find n,pandq.
(4)The mean of a binomial distribution is 3 and the variance is 1.2, find
nandp.
(5)The mean of a binomial distribution is 5 and variance is 5/2. Find
P(X= 4).
(6)The mean and variance of a binomial distributio n are 3 and 6/5. Find
P(X >3).
(7)Ifn= 8 and p= 2/3, find the mean and standard deviation of the
binomial distribution.
(8)Comment on “The mean and variance of a binomial distribution are 4
and 6”.
(9)Comment on “The mean and standard deviation of a binomial
distribution are 6 and 4”.
(10)A bag contains 10 white and 5 black balls. If 4 balls are selected at
random, find the probability that ( i) 3 are black balls, ( ii) atleast one
black ball is selected.
(11)Out of the total passengers travelling by BEST buses 40% do not
have the exact money for their tickets. If a conductor gives tickets tomunotes.in
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15 passengers, find the probability that less than 5 passengers will
pay exact money.
(12)In a government department an officer is not in his chair 12 out of 30
days in a month. If the se nior officer investigates his office 6 times,
what is the probability that he is not in his chair 3 times.
(13)The probability that internet users will buy a product from an online
marketing advertisement is 0.35. If 100 users login the site, find the
probabil ity that15 or more than that would actually buy the product?
(14)The EGS of central government guarantees employment to every 4
out of 6 unemployed persons. If 100 people are selected at random
from a village, find the probability that 10 of them have received
employment.
(15)The probability that a student from distance education passes is 0.6.
Find the probability that out of 6 students selected ( i) no student
passes, ( ii) atleast one student passes.
(16)The probability that an evening college student will graduate is 0.4.
Determine the probability that out of 5 students ( i) none, ( ii) one, ( iii)
atleast one student will graduate.
(17)The overall passing percentage of students enrolled in a distance
education institute is 35%. Find the probability that out 6 students 3
students pass.
(18)The probability that a micro wave oven is found to be defective is
0.2. If 6 microwave ovens are selected, find the probability that all
are defective.
(19)The probability of a defective bolt is 1/10. Find the mean and
variance of defective bolts o ut of a total of 400 bolts.
(20)5 out of 25 items are found to be defective. If 4 items are selected
find the probability distribution of the defective items.
(21)6 out of 50 items in a lot are found to be defective. Find the
probability of the following if 4 item s are selected: ( i) one defective
item, ( ii) 3 defective items, ( iii) at most 3 defective items.
(22)6% of articles in a given bundle are defective. Find the probability
that in a sample of 5 articles, none is defective.
(23)The probability that a long range missi le hits a target is 0.8. If 4
missiles are shot, find the probability that ( i) exactly two will hit the
target, ( ii) atleast two will hit the target.
(24)The probability of a man hitting a target is ¼. If h fires 7 times, what
is the probability that he hits t he target atleast twice?
(25)The average rainfall in a 30 days’ month is 40%. Find the probability
that ( i) the first four days of a given week will be fine and the
remaining will be wet, ( ii) rain will fall on just three days of a week.
(26)Out of 1000 families w ith 4 children each, how many would you
expect to have ( i) 2 boys and 2 girls, ( ii) atleast one boy, ( iii) no girlmunotes.in
139
(27)If on an average rain falls on 10 days in every 30 days, find the
probability that ( i) the first three days are fine and the remaining are
wet, (ii) the rain will fall on just three days of a week
(28)A box contains 60 Alphanso mangoes of which 8 are normal
mangoes. 8 mangoes are selected by the food inspector. Find the
probability that ( i) 6 mangoes are normal type, ( ii) atleast one is an
Alphanso type mango.
(29)The incidence of a certain disease is such that on an average 20% of
workers suffer from it. If 10 workers are selected at random, find the
probability that ( i) exactly 2 workers suffer from the disease, ( ii) not
more than 2 workers suffer fro m the disease, ( iii) not more than 2
workers suffer from the disease.
(30)An unbiased die is tossed 3 times. Find the probability of obtaining
(i) no six, ( ii) all sixes.
(31)An unbiased die is tossed 4 times. Find the probability of obtaining
(i) atleast one six, (ii) 4 sixes.
(32)Four coins are tossed simultaneously. What is the probability of
getting ( i) 4 heads, ( ii) 2 heads, 2 tails, ( iii) atleast one heads
(33)Five coins are tossed 3200 times. Find the frequency distribution of
heads and tails. Also find the mean and variance.
(34)8 coins are tossed 256 times. Find the expected frequencies of
success. Also find the mean and variance of the fitted values.
(35)The screws produced by a certain machine were checked by
examining the number of defectives in a sample of 12. The foll owing
table shows the distribution of 128 samples.
(36)
No of
defectives0 1 2 3 4 5 6 7
No of samples 7 6 19 35 30 23 7 1
(i) Fit a binomial distribution and find the expected frequencies, if the
chance of the machine being defective is 0.5
(ii) Find the m ean and standard deviation of the fitted distribution.
7.4NORMAL DISTRIBUTION
The Binomial distribution is a discrete probability distribution as
the random variable considered is a discrete. But for many practical
problems related to sales volume, he ight, weight of an individual or a
product, length of an item, strength, resistance, life of an electrical
instrument the random variable is of continuous type
Normal distribution is the most commonly used continuous
probability distribution. The probabi lity density function of a normal
distribution is:munotes.in
140
P(X) =21
2 1
2X
e
for–
distri bution.
Characteristics of a Normal curve
Fig7.1
1.The curve of the normal curve is bell shaped.
2.The curve is symmetrical about the central vertical line corresponding
to the mean of the distribution.
3.The peak of the curve is obtain ed at X=.
4.All the three measures of central tendencies mean, median and mode
coincide for a normal curve.
5.There are two tails of the curve which extend infinitely in both
positive and negative X -axis and never touch the axis.
6.The area under the normal curve is unity i.e.1. This is because the
area represents the probabilities of the variable and the sum of all
probabilities as we know is 1. Due to symmetry, the area to the left of
the mean is exactly 50% i.e.0.5 and similarly the area to the right of
the mean is also 0.5. Thus, for a normal curve, P(X) =P(X).
7.The standard deviation determines the spread of the distribution
around the m ean. As shown in Fig 9.2, if the value of is small then
the curve will be narrow and if the value of is large the curve will
become wider indicating the deviations of data around the mean.
8.The area covered b etween is 68.26% of the total area under the
normal curve.
9.The area covered between 2is 95.44% of the total area under the
normal curve.
10.The area covered between 3is 99.74% of the total area under the
normal curve.
Fig7.2
Area under the Normal Curve
IfXis a random normal variable, then the formula to calculate the
probability function uses methods of integration which are quite
cumbersome. This lead to a procedure of defining a new variable z=
X
, called as the standard normal variate (S.N.V.). It is observed that
the mean of this SNV is zero and its standard deviation is unity. Now the
problem of finding the probabilities for X=X1, is equated with finding the
area under the standard normal curve at z=z1. The area between z =0 andmunotes.in
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z=z1is read from the standard normal tables. This table is provided in the
appendix.
Now let us see some examples regarding how to read the standard
normal tables:
Example 13:
To find the area under the curve when z= 1.35
Ans: We look at the first column of the table for the number
1.3 and then move horizontally till the column of 0.05
which corresponds to our 1.35 (1.3 + 0.05). The value
in this cell is 0.4115. Thus, the area under the curve for
z= 1.35 is 0.4115. This also means that z= 1.35 represents 41.15%
of the total area.
Fig 7.3
Example 14:
To find area under the curve for z> 1.62
Ans: Repeating the previous step we find the value of
1.6 + 0.02 from th e table, this is 0.4474.
Now, we want to find area for z> 1.62. What we
Have got is the area between z= 0 and z= 1.62. The
total area to the right of z= 0 is 0.5. Thus the required
area is 0.5 –0.4474 = 0.0526
Fig 7.4
Example 15:
To find area un der the normal curve between z>–1.25
Ans: We know that the normal curve is symmetric about the
mean. So we repeat the same steps as described above
to find the area between z= 0 and z= 1.25. From the table this area is
0.3944. Now, the area represe nting z>–1.25 is the area between 0 to –
1.25 plus the remaining 50% area i.e0.5
Thus, the required area is 0.3944 + 0.5 = 0.8944
Fig 7.5
Example 16
To find area between –1.2z2
Ans: The required ar ea is split into two areas:
–1.2z0 and 0 z2
The first area as we have already seen how to calculate
is 0.3849 and the second area is 0.4772
Thus, the re quired area is a sum of these two areas
which is 0.8621munotes.in
142
Now let us solve some actual problems using our knowledge of
reading the standard normal table values.
Fig 7.6
Example 17:
The heights of 1000 students in a college are normally distributed
with m ean 160 cm and standard deviation 12 cm. How many students will
be there such that ( i), their heights are greater than 165 cm ( ii) their
heights are between 150 cm and 170 cm, ( iii) heights are less than 145cm.
Ans: Given: N= 1000, = 160,=12
(i)X= 165
z=X
=165 160
12=0.42
P(X> 165) = P(z> 0.42)
The area under the normal curve represented by
z= 0 and z =0.42 is 0.1628
The probability of students with heights greater than 165 cm is 16.28%
and the number of students are 1000 x 0.1628 = 162.8 163
(ii)X1= 150 and X2= 170
1
1Xz
=150 160
12=–0.83
and2
2Xz
=170 160
12= 0.83
The required probability is P(X1XX2)
=P(–0.83z0.83)
=P(–0.83z0) + P(0z0.83)
=P(0z0.83) + P(0z0.83)
= 2P(0z0.83)
From th e table we have the required probability as 2 x 0.2967 = 0.5934
Thus , the number of students whose heights are between 150cm
and 170 cm is
1000 x 0.5934 = 593.4 593.
(iii)X= 145
Xz
=145 160
12=–1.25
The area represented by z–1.25,
Now, the area between z =0 and z= 1.25 is 0.3944
the required area is 0.5 –0.3944 = 0.1056
the number of students with heights less than 145cm are 106.
Fig 7.7munotes.in
143
Example 18:
If 2.28% of teachers in a State have salary less than Rs. 4,500 and
30.85% teachers have salary greater than Rs. 7,000. Find the average
salary and standard deviation.
Ans: (In such kind of problems we do not find the area but we find the
value of z corresponding to the area known. In simple words, we look
inside the table and find the corresponding zvalue from the first column
and first row.)
Since 2.28% of teachers have salary less than 4500, the remaining
47.72% lie to the right of X= 4500. The area to the right of X= 4500 is
0.5–0.028 = 0.4772. Now look in the table, to find z-value for this is –2.
(As it is to the left of the mean)
4500z
=–2 4500 –=–2 … (1)
Since 30.85% teachers have salary more than Rs. 7000, the area to
the right of X= 7000 is 0.3085. So, the remaining area to the left of X=
7000 is 0.5 –0.3085 = 0.1915
Thez-value corresponding to th is area is 0.5 from the table
7000z
= 0.5 7000 –= 0.5 ... (2)
Solving (1) and (2) we have = 1000 and hence = 6500. Thus, the
mean salary is Rs. 6,500 and standard deviation is Rs. 1000
Exercise
(1)Define ( i)standard normal curve, ( ii) Standard normal variate.
(2)State the properties of standard normal curve.
(3)Using the table of areas under a norm al curve, find the probabilities
of: (a)P(0z1.2), ( b)P(0z2.3), ( c)P(–1z0), (d)P(–1z
1), (e)P(–1.5z0.2), ( f)P(z-1.6), ( g)P(z2).
(4)If a random variate X is normally distributed with mean 45 and
standard deviatio n 12, find the probabilities of the following: ( a)X
30, (b)X55, (c)X60, (d) 35X50
(5)IfXis a normal random variable with mean 14 and standard
deviation 6, find the probabilities of: ( a)X10, ( b)X16, ( c)X
18, (d) 10X20.
(6)If X 1and X 2are two random variates with mea ns 30, 25 and standard
deviations 16, 12 respectively, find P(60Y80), where Y= 3X 1+
2X2.
(7)The heights of 600 students in a college are normally distributed with
mean 154 cm and standard deviation 16 cm. How many students will
be there such that ( i) their heights are between 140 cm and 160 cm,munotes.in
144
(ii) their heights are greater than 175 cm, ( iii) heights are less than
135 cm.
(8)The average life of a cartridge is 7 days with standard deviation 1.5
days. If the life e xpectancy of a cartridge shows a normal
distribution, what is the probability that a cartridge functions for
more than 10 days?
(9)The weights of 1000 chocolates of a brand are normally distributed
with mean weight as 15 mg and standard deviation 4mg. Find th e
number of chocolates with weights greater than 20mg.
(10)The marks of 1200 students in a College show a normal distribution
with mean 56 and standard deviation 6. Estimate the number of
students with marks ( i) less than 50, ( ii) greater than 60, ( iii) betwee n
45 and 65.
(11)The mean height of soldiers is 68.22 inches with variance 10.8. If the
heights show a normal distribution, find the number of soldiers out of
a regiment of 1000 whose height is greater than 6 feet.
(12)The marks obtained by students are normally d istributed with mean
65 and variance 25. What is the probability of students getting marks
more than 75?
(13)To pass a physical test for Air force, the height of a cadet should be
at most 162cm. Out of 1400 cadets appeared for the physical test
how many could not clear it, if their average height was 160 cm with
standard deviation 9cm. Assume that the heights of the cadets show a
normal distribution.
(14)A manufacturer knows from his experience that the resistance of
resistors he produces is normally distributed wi th mean 100 ohms
and standard deviation 2 ohms. What is the percentage of resistors
having resistance between 98 and 102 ohms?
(15)The average life of a battery is 150 minutes with standard deviation
14 min. If the average life of battery shows a normal distri bution,
find the probability that a battery works for more than 180 min.
(16)The mean and standard deviation of average monthly salaries of 6000
people are Rs. 20,000 and Rs. 3,500 respectively. Assuming that the
data shows a normal distribution, find the ( i)number of people with
salaries greater than Rs. 25,000, ( ii) salaries between Rs. 15,000 and
Rs. 20, 000 and ( iii) less than Rs. 10,000.
(17)If 1.88% of teachers in a State have salary less than Rs. 3,500 and
4.85% teachers have salary greater than Rs. 8,000. F ind the average
salary and standard deviation.
(18)In an examination marks obtained by students in Mathematics,
Statistics and Economics are normally distributed with average
marks 51, 53 and 46 and standard deviation 15, 12 and 16
respectively. Find the proba bility that the total marks are ( i) 180 and
above, ( ii) less than 90.munotes.in
145
(19)The aggregate of students in a FYBSc class of a College is normally
distributed with mean 425 marks and standard deviation 36 marks. A
student is said to pass if he scores 40% of the tot al of 700 marks.
Find the number of students ( i) who have passed, ( ii) whose
aggregate is less than 325 marks.
(20)The average diameter of a needle is measure as 0.1 mm with a
standard deviation of 0.005mm. If 1600 samples are selected, find
the number of nee dles with diameter ( i) less than 0.08mm, ( ii) greater
than 0.2mm and ( iii) between 0.6mm and 0.3mm
(21)The daily sales of a firm are normally distributed with mean Rs. 8000
and standard deviation Rs. 100. What is the probability that on a
certain day sales wil l be less than Rs. 8200 and what is the
percentage of days with sales between Rs. 8050 and Rs. 8250?
(22)1000 light bulbs are installed in new factory, show a normal
distribution with mean life of 120 days and standard deviation of 20
days. How many bulbs will expire in less than 90 days?
(23)Record kept by goods inwards department of a large factory show
that average number of lorries arriving each week is 248. It is known
that the distribution is normal with standard deviation26. If this
pattern of arrival contin ues, what is the percentage of weeks expected
to have the number of arrivals ( i) less than 230 per week, ( ii) more
than 280 per week.
(24)The income distribution workers in a certain factory were found to be
normal with mean Rs. 500 and standard deviation Rs. 50. There were
228 persons with income above Rs. 600. How many workers were
there in all?
(25)The marks obtained by students in a subject showed a normal
distribution with mean 65 and standard deviation 14. If there were
142 students with marks less than 50, f ind the total number of
students.
(26)For a normal distribution 30% items are below 45, 8% are above 64.
Find the mean and variance.
(27)In a normal distribution 10% item are under 35 and 89% under 63.
Find the mean and standard deviation.
(28)In a class, 30% students have marks less than 40, 33% have marks
between 40 and 50 and the remaining have marks above 50. If the
data is normally distributed, find the average marks and standard
deviation.
(29)A STD booth owner has an average balance of Rs. 230 and standard
deviation of Rs. 30. Assuming that the balance with the owner
behave normally, find the proportion of the balance being ( i) less
than Rs. 180 and ( ii) between Rs. 200 and Rs. 250.
(30)In an examination in a college 46% students secured a pass class and
10% students sec ured first class. If the minimum marks for pass classmunotes.in
146
and first class are 35 and 60 respectively. Find the average marks
obtained by students.
(31)The life of army shoes is normally distributed with mean 8 months
and standard deviation 2 months. If 5000 pairs of shoes are issued ,
how many pairs would be expected to need replacement after 12
months?
(32)The marks obtained by students in an examination are normally
distributed with mean of 70 and standard deviation 6. If the top 5%
students get grade A and the botto m 25% get grade F, what marks is
the lowest A and the highest F?
munotes.in
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QUESTION PAPER PATTERN
MARKS: -100 TIME: -3 HRS
N.B :
1)All questions are compulsory
2)All question carry equal marks
3)Figures to the right indicate marks to a sub -question.
4)Graphs paper will be supplied on request.
5)Use of non -programmable calculator is allowed.
SECTION -I
Q.1 Attempt any four of the following
(a)5 marks (b)5 marks (c)5 marks
(d)5 marks (e)5 marks 20 marks
Q.2 Attempt any four of the following
(a)5 marks (b)5 marks (c)5 marks
(d)5 marks (e)5 marks 20 marks
SECTION -II
Q.3 Attempt any four of the following
(a)5 marks (b)5 marks (c)5 marks
(d)5 marks (e)5 marks 20 marks
Q.4 Attempt any four of the following
(a)5 marks (b)5 marks (c)5 marks
(d)5 marks (e)5 marks 20 marks
Q.5 Attempt any four of the following
(a)5 marks (b)5 marks (c)5 marks
(d)5 marks (e)5 marks 20 marks
munotes.in