Let $X=\{x \in N: 1 \leq x \leq 17\}$ and $\mathrm{Y}=\{\mathrm{ax}+\mathrm{b}: \mathrm{x} \in \mathrm{X}$ and $\mathrm{a}, \mathrm{b} \in \mathrm{R}, \mathrm{a}>0\}$. If mean and variance of elements of $\mathrm{Y}$ are 17 and 216 respectively then $a+b$ is equal to :
Correct Option: 1
$\sigma^{2}=$ variance
$\mu=$ mean
$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$
$\mu=17$
$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$
$\Rightarrow 9 \mathrm{a}+\mathrm{b}=17$ .......(1)
$\sigma^{2}=216$
$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$
$\Rightarrow \quad \frac{\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$
$\Rightarrow \quad a^{2} 81-18 \times 9 a^{2}+a^{2} 3 \times(35)=216$
$\Rightarrow a^{2}=\frac{216}{24}=9 \quad \Rightarrow \quad a=3(a>0)$
$\Rightarrow$ From (1), $\mathrm{b}=-10$
So, $a+b=-7$