The age distribution of 100 life insurance policy holders is as follows - begin array - - WhlineWhlineitext No- of persons - 5 - 16 - 12 - 26 - 14 - 12 - 6 - 5 Whlinelend array Calculate the mean deviation from the median age-

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Standard XII

Mathematics

Mean Deviation about Median

The age distr...

Question

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

$\begin{matrix} Age (on nearest birth day) & 17 - 19.5 & 20 - 25.5 & 26 - 35.5 & 36 - 40.5 & 41 - 50.5 & 51 - 55.5 & 56 - 60.5 & 61 - 70.5 No. of persons & 5 & 16 & 12 & 26 & 14 & 12 & 6 & 5 \end{matrix}$

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.

$\begin{matrix} Age & numbers of people f_{i} & Cumulative Frequency & Mid points x_{i} & | d_{i} | = | x_{i} - 38.31 | & f_{i} d_{i} 16.75 - 19.5 & 5 & 5 & 18.25 & 20.38 & 101.9 19.75 - 25.75 & 16 & 21 & 22.75 & 15.88 & 254.08 25.75 - 35.75 & 12 & 33 & 30.75 & 7.88 & 94.56 35.75 - 40.75 & 26 & 59 & 38.25 & 0.38 & 9.38 40 - 75 - 50.75 & 14 & 73 & 45.75 & 7.12 & 99.68 50.75 - 55.75 & 12 & 85 & 53.25 & 14.62 & 175.44 60.75 - 70.75 & 5 & 96 & 65.25 & 27.12 & 135.6 \sum_{i = 1}^{8} f_{i} = 96 & \sum_{i = 1}^{8} f_{i} | d_{i} | = 988 \end{matrix}$

$N = 96$ $∴ \frac{N}{2} = 48$

The cumulative frequency just greater than $\frac{N}{2} = 38$ 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 $= l + \frac{\frac{N}{2} - F}{f} \times h$

$= 35.75 + (\frac{48.33}{20}) \times 5 = 35.75 + 2.88 = 38.63$

Mean deviation from Median $= \frac{\sum_{i = 1}^{8} f_{i} | d_{i} |}{N} = \frac{988.36}{96} = 10.26$

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Similar questions

Q. 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.

A life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

$\begin{matrix} A g e (i n y e a r s) & N u m b e r o f p o l i c y h o l d e r s B e l o w 20 & 2 B e l o w 25 & 6 B e l o w 30 & 24 B e l o w 35 & 45 B e l o w 40 & 78 B e l o w 45 & 89 B e l o w 50 & 92 B e l o w 55 & 98 B e l o w 60 & 100 \end{matrix}$

Calculate mean deviation about median age for the age distribution of 100 persons given below :

$\begin{matrix} Age & 16 - 20 & 21 - 25 & 26 - 30 & 31 - 35 & 36 - 40 & 41 - 45 & 46 - 50 & 51 - 55 No. of persons & 5 & 6 & 12 & 14 & 26 & 12 & 16 & 9 \end{matrix}$

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
$\begin{matrix} Age in years & Number of policy holders Below 20 & 2 Below 25 & 6 Below 30 & 24 Below 35 & 45 Below 40 & 78 Below 45 & 89 Below 50 & 92 Below 55 & 98 Below 60 & 100 \end{matrix}$

Question 3
A life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

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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.