Question

An SBI health insurance agent found the following data for distribution of ages of 100 policy holders. Find the modal age and median age of the policy holders.
Age (in yrs): 15-20, 20-25, 25-30, 30-35, 35-40, 40-45, 45-50, 50-55, 55-60
Number of policy holders: 2, 4, 18, 21, 33, 11, 3, 6, 2

• A. Modal Age = 36.76 yrs, Median Age = 35.76 yrs
• B. Modal Age = 35.76 yrs, Median Age = 36.76 yrs
• C. Modal Age = 38.50 yrs, Median Age = 35.50 yrs
• D. Modal Age = 36.76 yrs, Median Age = 38.25 yrs

Hint:

Notice that both calculations are centered around the class interval 35--40.
Always verify that your final calculated values for median and mode lie strictly within this class interval.
Since both 35.76 and 36.76 lie between 35 and 40, this acts as an excellent check of your arithmetic accuracy!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a grouped frequency distribution table representing the ages of 100 policy holders. We need to calculate:
1. The Median age of the policy holders.
2. The Modal age of the policy holders.

Step 2: Key Formula or Approach:
1.

Median formula:
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]
where \(l\) is the lower limit of the median class, \(N = \sum f_i = 100\), \(cf\) is the cumulative frequency of the preceding class, \(f\) is the frequency of the median class, and \(h\) is the class width.
2.

Mode formula:
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
where \(l\) is the lower limit of the modal class, \(f_1\) is the frequency of the modal class, \(f_0\) is the frequency of the preceding class, \(f_2\) is the frequency of the succeeding class, and \(h\) is the class width.

Step 3: Detailed Explanation:
1.

Calculate the Median age:
Construct the cumulative frequency (\(cf\)) table:
- 15--20: \(f = 2 \implies cf = 2\)
- 20--25: \(f = 4 \implies cf = 6\)
- 25--30: \(f = 18 \implies cf = 24\)
- 30--35: \(f = 21 \implies cf = 45\)
- 35--40: \(f = 33 \implies cf = 78\)
- 40--45: \(f = 11 \implies cf = 89\)
- 45--50: \(f = 3 \implies cf = 92\)
- 50--55: \(f = 6 \implies cf = 98\)
- 55--60: \(f = 2 \implies cf = 100\)
Here, \(N = 100 \implies \frac{N}{2} = 50\).
The cumulative frequency just greater than 50 is 78, which corresponds to the class interval 35--40.
Therefore, the median class is 35--40.
- Parameters: \(l = 35\), \(cf = 45\), \(f = 33\), \(h = 5\).
\[ \text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5 = 35 + \frac{25}{33} \approx 35 + 0.76 = 35.76\ \text{years} \]

2.

Calculate the Modal age:
- Identify the modal class:
The class with the highest frequency (33) is 35--40.
- Parameters: \(l = 35\), \(f_1 = 33\), \(f_0 = 21\), \(f_2 = 11\), \(h = 5\).
\[ \text{Mode} = 35 + \left( \frac{33 - 21}{2(33) - 21 - 11} \right) \times 5 \]
\[ \text{Mode} = 35 + \left( \frac{12}{66 - 32} \right) \times 5 \]
\[ \text{Mode} = 35 + \left( \frac{12}{34} \right) \times 5 = 35 + \frac{60}{34} \approx 35 + 1.76 = 36.76\ \text{years} \]

Step 4: Final Answer:
The modal age of the policy holders is 36.76 years and the median age is 35.76 years, which corresponds to option (A).