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Question

The age distribution of 100 life insurance policy holders is as follows :

Age (on nearest birth day)1719.52025.52635.53640.54150.55155.55660.56170.5No. of persons5161226141265

Calculate the mean deviation from the median age.


Solution

To make this function continuos, we need to substract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

Agenumbers of peoplefiCumulative FrequencyMid points xi|di|=|xi38.31|fidi16.7519.55518.2520.38101.919.7525.75162122.7515.88254.0825.7535.75123330.757.8894.5635.7540.75265938.250.389.38407550.75147345.757.1299.6850.7555.75128553.2514.62175.4460.7570.7559665.2527.12135.68i=1fi=968i=1fi|di|=988

N=96  N2=48

The cumulative frequency just greater than N2=38 is 59 and the corresponding class is 35.75-40.75

Thus the median class is 35.75 = 40.75

t = 35.75,  f = 26,  f = 33,  h = 5 

Median =l+N2Ff×h

=35.75+(48.3320)×5=35.75+2.88=38.63

Mean deviation from Median =8i=1fi|di|N=988.3696=10.26


Mathematics
RD Sharma
Standard XI

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