Question

The arithmetic mean of the following frequency distribution is $62.8$ and the sum of all frequencies is $50$. Compute the missing frequencies $f_1$ and $f_2$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$	Total
Frequency	$5$	$f_1$	$10$	$f_2$	$7$	$8$	$50$

Answer 1

Step 1 Finding mean:

Completing the table,

Class-interval	Mid-value $x_i$	Frequency $f_i$	$f_i{x}_i$
$0-20$	$10$	$5$	$50$
$20-40$	$30$	$f_1$	$30f_1$
$40-60$	$50$	$10$	$500$
$60-80$	$70$	$f_2$	$70f_2$
$80-100$	$90$	$7$	$630$
$100-120$	$110$	$8$	$880$
		$\Sigma f_i=30+f_1+f_2$	$\Sigma f_i{x}_i=2060+30f_1+70f_2$

The formula for mean is $\overline{x}=\frac{\Sigma f_i{x}_i}{\Sigma f_i}$.

where $\Sigma f_i$ is the total frequency and $x_i$ is the mid-value of class-intervals.

From the question, we have $\Sigma f_i=50$.

$$\begin{gathered} \Rightarrow 50=30+f_1+f_2 \\ \Rightarrow f_1+f_2=20 \\ \Rightarrow f_1=20-f_2 \end{gathered}$$

Step 2 Finding the values of $f_1$ and $f_2$:

Substituting the values in the formula for mean,

$\Rightarrow 62.8=\frac{2060+30f_1+70f_2}{50}$

Substituting $f_1=20-f_2$,

$$\begin{gathered} \Rightarrow 3140=2060+30\left(20-f_2\right)+70f_2 \\ \Rightarrow 3140=2060+600-30f_2+70f_2 \\ \Rightarrow 40f_2=480 \\ \therefore f_2=12 \end{gathered}$$

Now the value of $f_1$ would be

$$\begin{gathered} \Rightarrow f_1=20-12 \\ \therefore f_1=8 \end{gathered}$$

Hence, the values of missing frequencies are obtained as $f_1=8$ and $f_2=12$.