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

The class widths in this dataset are not uniform, eliminating the need to adjust frequencies based on class intervals. The provided frequency table is of the "less than" type, using upper class limits. Policies were issued to individuals aged 18 years and older, but younger than 60 years. Consequently, the class intervals and their corresponding cumulative frequencies are presented as follows.

The cumulative frequencies and their respective class intervals are detailed below.

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

The total number of observations, n, is 100. The cumulative frequency just greater than \( \frac{n}{2} \) (which is \( \frac{100}{2} = 50 \)) is 78, corresponding to the class interval 35 - 40. This is identified as the median class.
The lower limit (\(l\)) of the median class is 35.
The frequency (\(f\)) of the median class is 33.
The cumulative frequency (\(cf\)) of the median class is 45.
The class size (\(h\)) is 5.

The median is calculated using the formula: 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, the median age is 35.76 years.

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To determine the class mark (\(x_i\)) for each interval, the following formula is used:
Class mark (\(x_i\)) = \( \frac{\text{Upper limit + Lower limit}}{2} \)

Using 11.5 as the assumed mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) are calculated using the step deviation method as presented below.

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, the sum of frequencies is \( \sum f_i = 100 \) and the sum of \( f_iu_i \) is \( \sum f_iu_i = -106 \).

The mean (\( \overset{-}{x} \)) is calculated using the formula: \( \overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i}) \times h \)

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

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

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The data from the table can be represented 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

The highest class frequency is 40, which falls within the class interval 7 - 10. This is identified as the modal class.
The lower limit (\(l\)) of the modal class is 7.
The frequency (\(f_1\)) of the modal class is 40.
The frequency (\(f_0\)) of the class preceding the modal class is 30.
The frequency (\(f_2\)) of the class succeeding the modal class is 16.
The class size (\(h\)) is 3.

The mode is calculated using the formula: 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

The median number of letters in surnames is 8.05, the mean is 8.32, and the modal size is 7.88.