Question

The Age groups of workers are given in the table :

$\begin{array}{cc}\text{Age}& \text{No. of workers}\\ 25-30& 30\\ 30-35& 28\\ 35-40& 22\\ 40-45& 30\\ 45-50& 35\\ 50-55& 24\end{array}$

Find the Median from the given data.

• A. 42.33
• B. 38.776
• C. 40.75
• D. 47.25

Answer 1

The correct option is C

40.75

The cumulative frequency table can be drawn as :

$\begin{array}{cc}\text{Age}& \text{No. of workers(Cumulative)}\\ <30& 30\\ <35& 58\\ <40& 80\\ <45& 110\\ <50& 145\\ <55& 169\end{array}$

N = 169 (An odd number)

Median = $\frac{(169+1)}{2}th$ observation =85th observation.

So, the Median class is 40 - 45

We know that there are 30 workers from the 80th worker to the 110th worker whose age is between 40 and 45 years.

However, we do not know their individual age

We will divide 5 from 40 to 45 into 30 equal parts and assume that one observation is in each subdivision.

We will assume that the value of each observation in a subdivision is the mid - value of the subdivision.

So, the age of the 81st worker is the mid value of 40 and $40\frac{5}{30}$ft, that is $40\frac{5}{60}$

So, from the 81st worker to the 85th worker, there are 4 workers.

If the 81st term of the AP is $40\frac{5}{60}$ and the common difference is $\frac{5}{30}$,

Then the 85th term = $40\frac{5}{60}$ + $4\times \frac{5}{30}$

= $40+\frac{5+40}{60}$

= 40 + 0.75

= 40.75