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B.B.A. LL.B. (Five Years) Sem 3 Business Statistics 2026 Jan Question Paper with Solutions

KSLU Solved Question Papers

Business Statistics

Previous Year Question Paper with Solution

B.B.A. LL.B. (Five Years) · Sem 3

2026 Jan Examination

munotes.in

Mumbai

First published on munotes.in on 21 October 2026.

Published by munotes.in, Mumbai.

Model answers written and edited by the munotes.in editorial desk.

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The question paper reproduced here is the paper as set by Karnataka State Law University at the 2026 Jan examination.

The questions below are the paper as Karnataka State Law University set it at the 2026 Jan examination, in the order it was set.

MarksPage

The Paper as Set

The questions in this volume are the questions asked at the 2026 Jan examination, reproduced as Karnataka State Law University set them, in the order it set them. Long reading passages are shortened to their first and last lines and our summary; the original paper prints them in full. Only the answers are ours. See the original question paper.

Duration 3 hours · Total marks 80 · 20 questions answered

How to use this volume

Solve the paper first, under exam conditions and against the clock. Then read the answers here and mark your own. Reading a solution before attempting the question feels productive and teaches very little, because recognising an answer is not the same as being able to write one.

Read the original question paper

Q1

Unit I

(a)Draw 'Less than ogive and more than ogive curves from the following data and also locate median value to verify the actual calculations.[10]

Answer

The figures given in the question

Class Interval	Frequency
10-50	10
50-100	30
100-150	50
150-200	40
200-2500	20

For full marks, cover: the less than and more than cumulative frequencies, the plotting points of both ogives, how the curves are drawn, the median read where they cross, and the median by formula to verify it.

The last class is printed as 200-2500. The class limits run 50, 100, 150, 200, so the last class is 200-250, and the working reads it that way. The answer does not depend on it, because the median falls in an earlier class.

Step 1: the cumulative frequencies

An ogive is the graph of a cumulative frequency distribution. A less than ogive plots each upper class limit against the number of items below it. A more than ogive plots each lower class limit against the number of items above it.

Class	f	less than cf	more than cf
10-50	10	10	150
50-100	30	40	140
100-150	50	90	110
150-200	40	130	60
200-250	20	150	20
Total	150

The less than column adds the frequencies from the top. The more than column adds them from the bottom. Both end at N = 150.

Step 2: the plotting points

Less than ogive: upper limit	Number of items below it	More than ogive: lower limit	Number of items above it
10	0	10	150
50	10	50	140

Less than ogive: upper limit	Number of items below it	More than ogive: lower limit	Number of items above it
100	40	100	110
150	90	150	60
200	130	200	20
250	150	250	0

Step 3: drawing the curves

Take the class limits on the X axis, with 1 cm for 25 units, from 0 to 250.
Take the cumulative frequencies on the Y axis, with 1 cm for 10 items, from 0 to 150.
Plot the less than points and join them with a smooth freehand curve. It rises from (10, 0) to (250, 150).
Plot the more than points on the same graph and join them. It falls from (10, 150) to (250, 0).
The two curves cross at one point. Drop a perpendicular from that point to the X axis.
The foot of the perpendicular is the median. The height of the crossing point is N/2.

On this data the curves cross at the point (135, 75). Between 100 and 150 the less than curve rises from 40 to 90, and the more than curve falls from 110 to 60. Both pass through 75 at 135.

Step 4: the median by formula, to verify the graph

N/2 = 150/2 = 75. The 75th item lies in the class 100-150, because the less than cf passes 75 there.

Median = L + ((N/2 - cf)/f) × h. Here L is the lower limit of the median class and cf is the cumulative frequency before it. The letter f is its frequency and h is its width.

Here L = 100, cf = 40, f = 50 and h = 50.

Median = 100 + ((75 - 40)/50) × 50 = 100 + 35 = 135.

The value read from the graph and the value worked by formula agree.

Conclusion. The less than and more than ogives cross at (135, 75), so the median read from the graph is 135. The formula gives the same median, 100 + ((75 - 40)/50) × 50 = 135, which verifies the graph.

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