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Question

The mean deviation about the median for the following data
Marks01010202030304040505060Number of students68141642
is

A
9.35
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B
10.34
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C
11.35
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D
12.33
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Solution

The correct option is B 10.34
ClassMid points(xi)fic.f.|xiM|fi|xiM|01056622.85137.110201581412.85102.820302514282.8539.930403516447.15114.440504544817.1568.650605525027.1554.3

The class interval containing N2th or 25th term is 2030. Therefore 2030 is the median class.

Median =l+N2f1f×h
=20+251414×10
=20+1114×10
=20+7.85=27.85

Mean deviation about median is given by 1N61fi|xiM|
=150×517.1=10.34

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