Statistics | Exercise 13.3

Question 3

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.

Solution

Understand the Question

The given data is presented in a cumulative "less than" format, representing policyholders aged $18$ years onwards up to $60$ years.
First, convert the cumulative frequencies into continuous class intervals and calculate the individual frequencies ( $f_i$ ) by taking the difference between consecutive cumulative frequencies.
Find the median class using $\dfrac{n}{2}$ , where $n = 100$ is the total frequency.
Compute the median using the formula: $\text{Median} = l + \left( \dfrac{\dfrac{n}{2} - cf}{f} \right) \times h$

Step 1 · Convert to Class Intervals

Given that policies are given to persons aged $18$ years onwards up to $60$ years, we convert the cumulative data into class intervals and find the frequencies:

\begin{array}{|c|c|c|} \hline \text{Age (in years)} & \text{Number of policy holders } (f_i) & \text{Cumulative frequency } (cf) \\ \hline 18 - 20 & 2 & 2 \\ \hline 20 - 25 & 6 - 2 = 4 & 6 \\ \hline 25 - 30 & 24 - 6 = 18 & 24 \\ \hline 30 - 35 & 45 - 24 = 21 & 45 \\ \hline 35 - 40 & 78 - 45 = 33 & 78 \\ \hline 40 - 45 & 89 - 78 = 11 & 89 \\ \hline 45 - 50 & 92 - 89 = 3 & 92 \\ \hline 50 - 55 & 98 - 92 = 6 & 98 \\ \hline 55 - 60 & 100 - 98 = 2 & 100 \\ \hline \text{Total } (n) & 100 & \\ \hline \end{array}

Step 2 · Identify the Median Class

Total frequency $n = 100$

$\dfrac{n}{2} = \dfrac{100}{2} = 50$

The cumulative frequency just greater than $50$ is $78$ , which corresponds to the class interval $35 - 40$ .

Therefore, the median class is $35 - 40$ .

From the median class:

Lower limit ( $l$ ) $= 35$
Frequency ( $f$ ) $= 33$
Class size ( $h$ ) $= 40 - 35 = 5$
Cumulative frequency of preceding class ( $cf$ ) $= 45$

Step 3 · Calculate Median

Using the median formula:

$\text{Median} = l + \left[ \dfrac{\dfrac{n}{2} - cf}{f} \right] \times h$

Substitute the values:

\begin{aligned} \text{Median} &= 35 + \left[ \dfrac{50 - 45}{33} \right] \times 5 \\[1.1em] &= 35 + \left[ \dfrac{5}{33} \right] \times 5 \\[1.1em] &= 35 + \dfrac{25}{33} \\[0.6em] &= 35 + 0.7575\dots \\[0.6em] &\approx 35.76 \end{aligned}

Answer

$35.76\text{ years}$

Common Mistakes

Treating Cumulative Frequencies as Class Frequencies: The values given in "less than" form are cumulative frequencies ( $cf$ ), not simple frequencies ( $f_i$ ). You must subtract consecutive terms to find the actual frequencies.
Using the Wrong $cf$ in the Formula: $cf$ in the median formula represents the cumulative frequency of the class preceding the median class ( $45$ ), not of the median class itself ( $78$ ).
Unequal First Interval: Note that the first interval is $18 - 20$ (width $2$ ), while the remaining intervals have width $5$ . Since the median class is $35 - 40$ , we use $h = 5$ .

Question 3

More questions in Exercise 13.3