Question

Compute the missing frequencies $f_1$ and $f_2$ in the following data if the mean is $166\frac{9}{26}$ and the sum of the frequency is $52$ .

Class	$140-150$	$150-160$	$160-170$	$170-180$	$180-190$	$190-200$	Total
Frequency	$5$	$f_1$	$20$	$f_2$	$6$	$2$	$52$

Answer 1

Step 1 Finding the mean:

The given table is completed as below

Class	Frequency $\left(f_i\right)$	Mid values $\left(x_i\right)$	$\left(f_i{x}_i\right)$
$140-150$	$5$	$145$	$725$
$150-160$	$f_1$	$155$	$155f_1$
$160-170$	$20$	$165$	$3300$
$170-180$	$f_2$	$175$	$175f_2$
$180-190$	$6$	$185$	$1110$
$190-200$	$2$	$195$	$390$
Total	$\sum _{}{f}_i=33+f_1+f_2$		$\sum _{}\left(f_i{x}_i\right)=5525+155f_1+175f_2$

The formula for mean is $Mean\left(\overline{x}\right)=\frac{\sum _{}\left(f_i{x}_i\right)}{\sum _{}{f}_i}$.

where $\sum _{}{f}_i$ is the total frequency and $x_i$ is the mid-values of the class-intervals.

It is given that $\overline{x}=166\frac{9}{26}$ and $\sum _{}{f}_i=52$.

Adding the frequencies,

$$\begin{gathered} \Rightarrow 33+f_1+f_2=52 \\ \Rightarrow f_1+f_2=19 \\ \Rightarrow f_1=19-f_2 \end{gathered}$$

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

Substituting values in the formula for mean,

$\Rightarrow 166\frac{9}{26}=\frac{5525+155f_1+175f_2}{52}$

Converting mixed fraction and substituting value of $f_1$,

$$\begin{gathered} \Rightarrow \frac{4325}{26}\times 52=5525+155(19-f_2)+175f_2 \\ \Rightarrow 180=20f_2 \\ \therefore f_2=9 \end{gathered}$$

Find the value of $f_1$ by putting the value of $f_2$ in $f_1=19-f_2$ .

$$\begin{gathered} \Rightarrow f_1=19-f_2 \\ \therefore f_1=10 \end{gathered}$$

Hence, the missing frequencies are $f_1=10$ and $f_2=9$.