# The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50.

Question:

The mean of the following frequency distribution is $62.8$ and the sum of all the frequencies is 50 . Compute the missing frequency $f_{1}$ and $f_{2}$.

Solution:

It is given that mean $=62.8$ and $N=50$.

Let the assumed mean A = 50 and h = 20.

$\sum f_{i}=50$

$30+f_{1}+f_{2}=50$

$f_{1}=20-f_{2}$......(1)

We know that mean, $\bar{X}=A+h\left(\frac{1}{N} \sum f_{i} u_{i}\right)$

Now, we have $\sum f_{i}=30+f_{1}+f_{2}, \sum f_{t} u_{t}=28-f_{1}+f_{2}, h=20$ and $A=50$.

Putting the values in the above formula, we have

$62.8=50+20\left(\frac{1}{30+f_{1}+f_{2}} \times\left(28-f_{1}+f_{2}\right)\right)$

$62.8-50=20\left(\frac{1}{30+f_{1}+f_{2}} \times\left(28-f_{1}+f_{2}\right)\right)$

$12.8\left(30+f_{1}+f_{2}\right)=20\left(28-f_{1}+f_{2}\right)$

$32.8 f_{1}-7.2 f_{2}=176$ .......(2)

Putting the value of $f_{1}$ in $(2)$, we get

$32.8\left(20-f_{2}\right)-7.2 f_{2}=176$

$32.8 \times 20-32.8 f_{2}-7.2 f_{2}=176$

$656-176=40 f_{2}$

$f_{2}=\frac{480}{40}$

$=12$

Putting the value of $f_{2}$ in $(1)$, we get

$f_{1}=20-12$

$=8$

Hence, the missing frequency $f_{1}=8$ and $f_{2}=12$.