Question

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

Answer 1

Answer: 1

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