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Question

Find mean deviation from median for the given data.

x0101020203030404050
f5815166

A
8.28
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B
9.56
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C
10.19
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D
13.65
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Solution

The correct option is B 9.56
Class fi Class frequency(cf) Mid-Value(xi) |xiMedian| fi|xiMedian|
010 5 5 5 23 115
1020 8 13 15 13 104
2030 15 28 25 3 45
3040 16 44 35 7 112
4050 6 50 45 17 102
N=fi=50
Here, N=50, so 502=25 and the cummulative frequency is just greater than N2 is 28.
Thus, 2030 is median class.
l = lower limit of median class =20
F = Cummulative frequency before median class =13
Difference between the class = h=10
Frequency of median class = f=15
Median =l+N2Ff×h=20+251315×10=28
fi×|xiMedian|=478
Mean deviation about the mean =1NfI×|xiMedian|=150×478=9.56

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