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:

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Solution

Q.
The frequency distribution of the age of students in a class of 40 students is given below. [2024]

Age 15 16 17 18 19 20

No. of students 5 8 5 12 $x$ $y$

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

1 43

2 44

3 46

4 47

Ans.
(2)

Age No. of Students C.F.

15 5 5

16 8 13

17 5 18

18 12 30

19 $x$ $30 + x$

20 $y$ $30 + x + y$

$30 + x + y = 40$ (Given)

$\Rightarrow x + y = 10$ ...(i)

Median of the given data = 18

$Mean deviation about median = \frac{\sum f_{i} | x_{i} - 18 |}{\sum f_{i}}$

$\sum f_{i} | x_{i} - 18 | = 5 \times 3 + 8 \times 2 + 5 \times 1 + 12 \times 0 + x \times 1 + y \times 2$

$= 15 + 16 + 5 + x + 2 y = 36 + x + 2 y$

$\Rightarrow \frac{36 + x + 2 y}{40} = 1.25$ (Given)

$\Rightarrow x + 2 y = 14$ ...(ii)

On solving (i) and (ii), we get

$x = 6, y = 4$

Now,

$4 x + 5 y = 4 \times 6 + 5 \times 4 = 24 + 20 = 44$

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