Find the missing frequency (p) for the following distribution whose mean is 7.68.
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, $p=9$