Find the missing frequencies and the median for the following distribution if the mean is 1.46.

Solution:

(1) Let the missing frequencies be x and y.

Given:

$N=200$

 

$86+x+y=200$

$x=114-y \quad \ldots . .(1)$

We know that mean, $\bar{X}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$1.46=\frac{140+x+2 y}{200}$

$x+2 y+140=292$

$x+2 y=152$......(2)

Solving (1) and (2), we get

$114-y+2 y=152$

$y=38$

Therefore,

$x=114-38$

$=76$

Hence, the missing frequencies are 38 and 76.

(2) Calculation of median.

Now, we have $N=200$.

So, $\frac{N}{2}=100$

Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.

Hence, the median is 1.