The frequency distribution of the age of students in a class of 40 students is given below.

If the mean deviation about the median is 1.25, then $$4x + 5y$$ is equal to :

Total students = 40

Given frequencies:
5 + 8 + 5 + 12 + x + y = 40
$$\Rightarrow30+x+y=40$$
$$\Rightarrow x+y=10\quad...(1)$$
Cumulative frequencies:

• 15 → 5
• 16 → 13
• 17 → 18
• 18 → 30

So 20th and 21st observations lie in age 18 ⇒ median = 18

$$\text{MD}=\frac{\sum_{ }^{ }f|x-18|}{40}=1.25$$

$$\sum_{ }^{ }f|x-18|=1.25\times40=50$$

Now compute:
$$=5|15-18|+8|16-18|+5|17-18|+12|18-18|+x|19-18|+y|20-18|$$

= 5(3) + 8(2) + 5(1) + 12(0) + x(1) + y(2)
= 15 + 16 + 5 + x + 2y
= 36 + x + 2y
36 + x + 2y = 50
$$\Rightarrow x+2y=14\quad...(2)$$

Solve (1) & (2)
x + y = 10
x + 2y = 14

Subtract:
$$y=4\Rightarrow x=6$$

4x + 5y = 4(6) + 5(4) = 24 + 20 = 44