Question

The mean of the following frequency table is 50. But the frequencies $f_1$ and $f_2$ in classes $20-40$ and $60-80$ are missing. Find the missing frequencies.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$Total$
Frequency	$17$	$f_1$	$32$	$f_2$	$19$	$100$

Answer 1

Step 1: The formula used to solve this problem:

Arithmetic Mean$=A+h\left[\frac{1}{N}\sum f_i{u}_i\right]$

where,

$N=$ total number of frequencies

$h=$ class size(units between each of the class interval)

$A=$ Assumed mean(Total mid value of the class interval)

$u_i=\frac{x_i-A}{h}$

$f_i=$Sum of the frequency

Step 2: Make the required frequency distribution table:

First, we have to prepare a table to find the value of mid values and $f_i{u}_i$.

Class	Frequency	Mid values	$u_i=\frac{x_i-A}{h}$	$f_i{u}_i$
$0-20$	$17$	$10$	$-2$	$-34$
$20-40$	$f_1$	$30$	$-1$	$-f_1$
$40-60$	$32$	$50$	$0$	$0$
$60-80$	$f_2$	$70$	$1$	$-f_2$
$80-100$	$19$	$90$	$2$	$38$
	$N=\sum f_i=63+f_1+f_2$			$\sum f_i{u}_i=4-f_1+f_2$

From the above table we get the values of,

$N=\sum f_i=120$

$A=120$

$h=120$

$68+f_1+f_2=120$

$f_1+f_2=52\to \left(1\right)$

Step 3: Find the missing frequencies

The given value of mean$=50$.

We will get, Arithmetic Mean $=A+h\left[\frac{1}{N}\sum f_i{u}_i\right]$

$50+20\left[\frac{4-f_1+f_2}{120}\right]=50$

$\frac{4-f_1+f_2}{6}=0$

$4-f_1+f_2=0$

$f_1-f_2=4\to \left(2\right)$

While solving equation (1) and (2), we get

$f_1+f_2=52$

$f_1-f_2=4$

$------$

$2f_1=56$

$f_1=28$

Substitute $f_1=28$ in equation (2)

$f_1-f_2=4$

$28-f_2=4$

$-f_2=4-28$

$f_2=24$

Therefore, $f_1=28$ and $f_2=24.$

Final Answer:

Hence , the missing frequencies are $f_1=28$ and $f_2=24.$