Question

The mean of the following frequency distribution is $56$, but the frequencies $f_1$ and $f_2$ in classes $20-40$ and $80-100$ respectively are missing. Find the missing frequencies.

Class-interval	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$	Total
Frequency	$16$	$f_1$	$25$	$16$	$f_2$	$10$	$90$

Answer 1

Step 1: Finding the mean:

Completing the table, we get:

Class-interval	$x_i$	Frequency $f_i$	$f_i{x}_i$
$0-20$	$10$	$16$	$160$
$20-40$	$30$	$f_1$	$30f_1$
$40-60$	$50$	$25$	$1250$
$60-80$	$70$	$16$	$1120$
$80-100$	$90$	$f_2$	$90f_2$
$100-120$	$110$	$10$	$1100$
		$\sum _{}{f}_i=90$	$\sum _{}{f}_i{x}_i=3630+30f_1+90f_2$

The formula of the mean is given by:

$\overline{x}=\left(\frac{\sum _{}{f}_i{x}_i}{\sum _{}{f}_i}\right)$

$f_i$ is the frequency of observations, and

$x_i$ is the mid-value of the class interval.

The mean of the given distribution is given as $56$.

Also, from the sum of frequencies we have

$$\begin{gathered} 67+f_1+f_2=90 \\ \Rightarrow f_1+f_2=23 \end{gathered}$$ …(1)

Step 2: Finding the values of ${\mathit{f}}_{\mathbf{1}}$ and ${\mathit{f}}_{\mathbf{2}}$:

Substituting the values we get,

$$\begin{gathered} 56=\left(\frac{3630+30f_1+90f_2}{90}\right) \\ \Rightarrow 56\times 90=3630+30f_1+90f_2 \\ \Rightarrow 5040=3630+30f_1+90f_2 \\ \Rightarrow 30f_1+90f_2=5040-3630 \end{gathered}$$

$\Rightarrow f_1+3f_2=47$ …(2)

From equation (1),

$f_2=23-f_1$

Substituting it in equation (2) we get,

$$\begin{gathered} f_1+3\left(23-f_1\right)=47 \\ \Rightarrow f_1+69-3f_1=47 \\ \Rightarrow 2f_1=22 \\ \therefore f_1=11 \end{gathered}$$

Now, substituting it in equation (1), we get

$$\begin{gathered} 11+f_2=23 \\ \Rightarrow f_2=12 \end{gathered}$$

Hence, the required values are $f_1=11$ and $f_2=12$.