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Question

The age distribution of 100 life-insurance policy holders is as follows:
Age (on nearest birth day) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of persons 5 16 12 26 14 12 6 5
Calculate the mean deviation from the median age.


Solution

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

 
Age  Number of People
fi
Cumulative Frequency   Midpoints
xi
di=xi-38.63  fidi
16.7519.75
 
5 5 18.25 20.38 101.9
19.7525.75 16 21 22.75 15.88 254.08
 
25.7535.75 12 33 30.75 7.88 94.56
 
35.7540.75 26 59 38.25 0.38 9.38
 
40.7550.75 14 73 45.75 7.12 99.68
 
50.7555.75 12 85 53.25 14.62 175.44
 
55.7560.75 6 91 58.25 19.62 117.72
 
60.7570.75 5 96 65.75 27.12 135.6
 
   N=fii=18=96       i=18fidi=988.36
 

 

                                                                                                                                                                               

N=96N2=48The cumulative frequency just greater than  N2=38 is 59 and the corresponding class is  35.75-40.75.Thus, the median class is 35.75-40.75l=35.75, f= 26, F=33, h=5Median =l+N2-Ff×h             = 35.75+48-3326×5                =35.75+2.88             = 38.63Mean deviation from the median =i=18fidiN                                                   =988.3696                                                  =10.29

Mathematics
RD Sharma XI (2015)
Standard XI

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