# Prove the following

Question:

Let $X=\{x \in \mathbf{N}: 1 \leq x \leq 17\}$ and $Y=\{a x+b: x \in X$ and $a, b \in \mathbf{R}, a>0\} .$ If mean and variance of elements of $Y$ are 17 and 216 respectively then $a+b$ is equal to :

1. (1) 7

2. (2) $-7$

3. (3) $-27$

4. (4) 9

Correct Option: , 2

Solution:

$\because \bar{x}=\frac{1+2+3+\ldots .+17}{17}=\frac{17 \times 18}{17 \times 2}=9$

$\bar{y}=a \bar{x}+b=\frac{a(1+2+3+\ldots . .+17)}{17}+b=17$

$\Rightarrow \frac{a \cdot(17 \cdot 18)}{17 \cdot 2}+b=17 \Rightarrow 9 a+b=17 \ldots(\mathrm{i})$

$\operatorname{Var}(x)=\sigma A^{2}=\frac{\sum x^{2}}{n}-(\bar{x})^{2}$

$=\frac{1^{2}+2^{2}+\ldots .+17^{2}}{17}-(9)^{2}$

$=\frac{17 \cdot 18 \cdot 35}{6 \cdot 17}-(9)^{2}=105-81=24$

$\operatorname{Var}(y)=a^{2} \operatorname{Var}(x)=a^{2} \cdot 24=216$

$a^{2}=\frac{216}{24}=9 \Rightarrow a=3$

$\therefore$ From (i), $b=17-9 a=17-27=-10$

$\therefore a+b=3+(-10)=-7$