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Question

# Find mean deviation from median for the given data.x0âˆ’1010âˆ’2020âˆ’3030âˆ’4040âˆ’50f5815166

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) |xi−Median| fi|xi−Median| 0−10 5 5 5 23 115 10−20 8 13 15 13 104 20−30 15 28 25 3 45 30−40 16 44 35 7 112 40−50 6 50 45 17 102N=∑fi=50Here, N=50, so 502=25 and the cummulative frequency is just greater than N2 is 28.Thus, 20−30 is median class. l = lower limit of median class =20F = Cummulative frequency before median class =13Difference between the class = h=10 Frequency of median class = f=15Median =l+N2−Ff×h=20+25−1315×10=28∑fi×|xi−Median|=478Mean deviation about the mean =1N∑fI×|xi−Median|=150×478=9.56

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