Q. The variance of the numbers $2, 3, 11 $ and $x$ is $\frac{49}{4}$. Find the value of $x$.
Statistics
Solution:
From given data, we make the following table.
$x$
$x^2$
$2$
$4$
$3$
$9$
$11$
$121$
$x$
$x^2$
$\Sigma x = 16+x $
$\Sigma x^2 = 134+x^2$
But we know that, variance =$\frac{\Sigma x^{2}}{n} -\left(\frac{\Sigma x}{n}\right)^{2} $
$ \Rightarrow \frac{134+x^{2}}{4} -\left(\frac{16+x}{4}\right)^{2} = \frac{49}{4} $ (given)
$ \Rightarrow \frac{134+x^{2}}{4} - \frac{\left(256+x^{2}+32x\right)}{16} = \frac{49}{4}$
$ \Rightarrow \frac{3x^{2}-32x+280}{16} = \frac{49}{4}$
$\Rightarrow 3x^{2}-32x+280 = 196$
$ \Rightarrow 3x^{2}-32x+84 = 0$
$ \Rightarrow \left(x-6\right)\left(3x-14\right)=0$
$ \Rightarrow x=6, x= \frac{14}{3}$
Therefore, the values of $x$ are $6$ and $\frac{14}{3}$.
| $x$ | $x^2$ |
| $2$ | $4$ |
| $3$ | $9$ |
| $11$ | $121$ |
| $x$ | $x^2$ |
| $\Sigma x = 16+x $ | $\Sigma x^2 = 134+x^2$ |