The following is a cumulative frequency distribution (less than type) of 900 persons? each of age 30 years and above.

What is the modal age (in years) (correct to one decimal place)?

Age below (in years)	40	50	60	70	80
Number of persons	120	250	650	850	900

Correct option is C

Concept Used:

Mode = $$L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$​​

Where:

L = Lower boundary of the modal class.

$$f_1$$​​ = Frequency of the modal class.

$$f_0​$$​= Frequency of the class before the modal class.

$$f_2$$​​ = Frequency of the class after the modal class.

h = Width of the class interval.

Solution:

The highest frequency is 400,

so the modal class is 50 – 60

​$$L = 50, \ \ f_1 = 400, \ \ f_0 = 130, \ \ f_2 = 200, \ \ h = 10$$​​

Now,

Mode = $$L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$$​​

​$$= 50 + \left( \frac{400 - 130}{2(400) - 130 - 200} \right) \times 10$$ 

​$$= 50 + \left( \frac{270}{800 - 330} \right) \times 10$$​​

​$$= 50 + \left( \frac{270}{470} \right) \times 10$$​​

​$$= 50 + (0.5745 \times 10)$$​​

= 50 + 5.7 = 55.7

The modal age is 55.7 years.