Question

If the variance of the following frequency distribution:

Class	$10-20$	$20-30$	$30-40$
Frequency	$2$	$x$	$2$

is $50$, then $x$ is equal to

Answer 1

Step 1. Make a probability distribution table :

Class	$X_i$	Frequency $\left(f_i\right)$	$X_i{f}_i$	$\overline{X}=\frac{\sum _{}{X}_i{f}_i}{\sum _{}{f}_i}$	${(\overline{X}-X_i)}^2$	${(\overline{X}-X_i)}^2{f}_i$
$10-20$	$15$	$2$	$30$	$\frac{100+25x}{4+x}=25$	${\left(25-15\right)}^2=100$	$100·2=200$
$20-30$	$25$	$x$	$25x$	$\frac{100+25x}{4+x}=25$	${\left(25-25\right)}^2=0$	$0·x=0$
$30-40$	$35$	$2$	$70$	$\frac{100+25x}{4+x}=25$	${\left(25-35\right)}^2=100$	$100·2=200$
		$\sum _{}{f}_i=4+x$	$\sum _{}{X}_i{f}_i=100+25x$			$\sum _{}{(\overline{X}-X_i)}^2{f}_i=400$

Step 2. Find the value of $x$.

The variance is given as:

$\mathrm{Variance}=\frac{\sum _{}{(\overline{X}-X_i)}^2{f}_i}{\sum _{}{f}_i}$

Now, substitute the value of variance as $50$,$\sum _{}{(\overline{X}-X_i)}^2{f}_i=400$, and $\sum _{}{f}_i=4+x$ in the above formula:

$\begin{array}{rcl}50& =& \frac{400}{4+x}\\ & \Rightarrow & 4+x=\frac{400}{50}\\ & \Rightarrow & 4+x=8\\ & \Rightarrow & x=8-4\\ & \Rightarrow & x=4\end{array}$

So, the value of $x$ for which the variance of the given frequency distribution is $50$ is $4$.

Hence, the value of $x$ is $4$.