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: Find the modal class using an equivalent, regrouped version of the mode formula.
The class with the highest frequency, 33, is $35$-$40$, so this is the modal class. Take $l = 35$, $f_1 = 33$ (modal class frequency), $f_0 = 21$ (frequency just before it), $f_2 = 11$ (frequency just after it), and $h = 5$ (class width).
Instead of writing the denominator as $2f_1 - f_0 - f_2$, split it into two separate differences added together:
\[ \text{Mode} = l + \frac{f_1 - f_0}{(f_1 - f_0) + (f_1 - f_2)} \times h \]

Step 2: Compute each piece separately.
\[ f_1 - f_0 = 33 - 21 = 12, \qquad f_1 - f_2 = 33 - 11 = 22 \]
\[ \text{Mode} = 35 + \frac{12}{12 + 22} \times 5 = 35 + \frac{12}{34} \times 5 = 35 + \frac{60}{34} \approx 35 + 1.76 = 36.76 \text{ years} \]

Step 3: Find the median class using a "more than" cumulative frequency table instead of the usual "less than" table.
Build cumulative totals counting from the top of the table downward (how many policy holders are older than each lower boundary):
More than 15: 100, more than 20: 98, more than 25: 94, more than 30: 76, more than 35: 55, more than 40: 22, more than 45: 11, more than 50: 8, more than 55: 2.
Since $\frac{N}{2} = 50$, we look for the class where the "more than" cumulative frequency first drops to 50 or below. "More than 35" gives 55 (still above 50), and "more than 40" gives 22 (below 50), so the median lies within the class $35$-$40$.

Step 4: Apply the median formula using this class.
The cumulative frequency of the class before $35$-$40$ (that is, up to 35) taken from the usual ascending count is $cf = 45$, the frequency of the median class is $f = 33$, the lower limit is $l = 35$, and $h = 5$:
\[ \text{Median} = 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{25}{33} \approx 35 + 0.76 = 35.76 \text{ years} \]

Step 5: Final Answer.
The modal age is 36.76 years and the median age is 35.76 years, so option (A) is correct. \[ \boxed{\text{Mode} = 36.76 \text{ yrs},\ \text{Median} = 35.76 \text{ yrs}} \]