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13.Statistics
hard
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
A
$-7$
B
$7$
C
$9$
D
$-27$
(JEE MAIN-2020)
Solution
$\sigma^{2}=$ variance
$\mu=$ mean
$\sigma^{2}=\frac{\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$
$\mu=17$
$\Rightarrow \frac{\sum_{x=1}^{17}(a x+b)}{17}=17$
$\Rightarrow \quad 9 a+b=17$
$\sigma^{2}=216$
$\Rightarrow \quad \frac{\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$
$\Rightarrow \frac{\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$
$\Rightarrow \quad a^{2} 81-18 \times 9 a^{2}+a^{2} 3 \times(35)=216$
$\Rightarrow \quad$ From $(1), b=-10$
So, $a+b=-7$
Standard 11
Mathematics