Let X={mathrm{x} in mathrm{N}: 1 leq mathrm{x | vedclass

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Question

$\sigma^{2}=$ variance

$\mu=$ mean

$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$

$\mu=17$

$\Rightarrow \frac{\sum_{x=1}^

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

$\sigma^{2}=$ variance

$\mu=$ mean

$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\muight)^{2}}{n}$

$\mu=17$

$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$

$\Rightarrow \quad 9 a+b=17$

$\sigma^{2}=216$

$\Rightarrow \quad \frac{\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$

$\Rightarrow \frac{\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$

$\Rightarrow \quad a^{2} 81-18 	imes 9 a^{2}+a^{2} 3 	imes(35)=216$

$\Rightarrow \quad$ From $(1), b=-10$

So, $a+b=-7$

Let $X=\{\mathrm{x} \in \mathrm{N}: 1 \leq \mathrm{x} \leq 17\}$ 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 $Y$ are $17$ and $216$ respectively then $a + b$ is equal to

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