Question

A life insurance agent found the following data for 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 year.

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

Answer 1

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below. 

The cumulative frequencies with their respective class intervals are as follows.

Age (in years)	Frequency (f\(_i\))	Number of policy holders (
18 - 20	2	2
20 - 25	6 - 2 = 4	6
25 - 30	24 - 6 = 18	24
30 - 35	45 - 24 = 21	45
35 - 40	78 - 45 = 33	78
40 - 45	89 - 78 = 11	89
45 - 50	92 - 89 = 3	92
50 - 55	98 - 92 = 6	98
55 - 56	100 - 98 = 2	100

From the table, it can be observed that n = 100. 
Cumulative frequency just greater \(\frac{n}2 ( i.e., \frac{100}2 = 50)\) than is 78, belonging to class interval 35 - 40.
Median class = 35 - 40
Lower limit (\(l\)) of median class = 35
Frequency (\(f\)) of median class = 33
Cumulative frequency (\(cf\)) of median class = 45
Class size (\(h\)) = 5

 Median = \(l + (\frac{\frac{n}2 - cf}f \times h)\)

Median =  \(35  + (\frac{50 - 45}{33} \times 5)\)

Median = 35 + \(\frac{25}{33}\)
Median = 35.76

Therefore, median age is 35.76 years. 

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To find the class mark (xi) for each interval, the following relation is used.  

Class mark  \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)

Taking 11.5 as assumed mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) are calculated according to step deviation method as follows.

Number of letters	Frequency (f\(_i\))	\(\bf{x_i}\)	\(\bf{d_i = x_i -11.5}\)	\(\bf{u_i = \frac{d_i}{3}}\)	\(\bf{f_iu_i}\)
1 - 4	6	2.5	-9	-3	-18
4 - 7	30	5.5	-6	-2	-60
7 - 10	40	8.5	-3	-1	-40
10 - 13	16	11.5	0	0	0
13 - 16	4	14.5	3	1	4
16 - 19	4	17.5	6	2	8
Total	100				-106

From the table, it can be observed that  

\(\sum f_i = 100\)
\(\sum f_iu_i = -106\)

Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})\times h\) 

\(\overset{-}{x}\) = \(11.5 + (\frac{-106 }{100})\times 3\)

\(\overset{-}{x}\) = 11.5 - 3.18
Mean, \(\overset{-}{x}\) = 8.32

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The data in the given table can be written as 
 

Number of letters	Frequency (f\(_i\))
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4
Total	100

From the data given above, it can be observed that the maximum class frequency is 40, belonging to class interval 7 - 10.

Therefore, modal class = 7 - 10
Lower limit (\(l\)) of modal class = 7
Frequency (\(f_1\)) of modal class = 40
Frequency (\(f_0\)) of class preceding the modal class = 30
Frequency (\(f_2\)) of class succeeding the modal class = 16
Class size (\(h\)) = 3

Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)

Mode = \(7 + (\frac{40 - 30 }{ 2(40) - 30 - 16}) \times3\)

Mode =\(7+ [\frac{10}{34}] \times 3\)

Mode = \(7 +( \frac{ 30}{ 34})\)
Mode = 7 + 0.88
Mode = 7.88

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.