Question

If the standard deviation of the numbers $2,3, x$ and $11$ is $3.5$, then find the value of $3 x^{2}-32 x+85$.

Answer 1

Mean $\bar{x}=\frac{\sum x}{n}$
$\Rightarrow \bar{x}=\frac{2+3+x+11}{4}=\frac{16+x}{4}\,\,\, ..(i)$
$\sigma^{2}=\frac{1}{n} \sum x^{2}-(\bar{x})^{2}$
$\Rightarrow(3.5)^{2}=\frac{1}{4}\left(4+9+x^{2}+121\right)-\left(\frac{16+x}{4}\right)^{2}\ldots[$ From (i) $]$
$\Leftrightarrow \frac{49}{4}=\frac{1}{4}\left(134+x^{2}\right)-\frac{\left(x^{2}+32 x+256\right)}{16}$
$\Leftrightarrow 196=4 x^{2}+536-x^{2}-32 x-256$
$\Leftrightarrow 3 x^{2}-32 x+280=196$
$\Leftrightarrow 3 x^{2}-32 x+84=0$
$\Leftrightarrow 3 x^{2}-32 x+85=1$