Find the missing frequency (p) for the following distribution whose mean is 7.68.

Solution:

Given:

Mean $=7.68$

First of all prepare the frequency table in such a way that its first column consist of the values of the variate $\left(x_{i}\right)$ and the second column the corresponding frequencies $\left(f_{i}\right)$.

Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing $\left(f_{i} x_{i}\right)$.

Then, sum of all entries in the column second and denoted by $\sum f_{i}$ and in the third column to obtain $\sum f_{i} x_{i}$.

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

$7.68=\frac{303+9 p}{41+p}$

By using cross multiplication method,

$303+9 p=314.88+7.68 p$

$9 p-7.68 p=314.88-303$

$1.32 p=11.88$

$p=\frac{11.88}{1.32}$

$=9$

Hence, $p=9$