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# The age distribution of 100 life insurance policy holders is as follows : Age (on nearest birth day)17−19.520−25.526−35.536−40.541−50.551−55.556−60.561−70.5No. of persons5161226141265 Calculate the mean deviation from the median age.

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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|=|xi−38.31|fidi16.75−19.55518.2520.38101.919.75−25.75162122.7515.88254.0825.75−35.75123330.757.8894.5635.75−40.75265938.250.389.3840−75−50.75147345.757.1299.6850.75−55.75128553.2514.62175.4460.75−70.7559665.2527.12135.6∑8i=1fi=96∑8i=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+N2−Ff×h =35.75+(48.3320)×5=35.75+2.88=38.63 Mean deviation from Median =∑8i=1fi|di|N=988.3696=10.26  Suggest Corrections  0      Similar questions  Explore more