Question

Let $X=\{x \in N: 1 \leq x \leq 17\}$ and $Y=\{a x+b: x \in X$ and $a, b \in R, a>0\} .$ 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

Answer 1

Answer: (A)

$\sigma^{2}=$ variance
$\mu=$ mean
$\sigma^{2}=\frac{\displaystyle\sum_{i=1}^{n}\left(x_{i}-\mu\right)^{2}}{n}$
$\mu=17$
$\Rightarrow \frac{\displaystyle\sum_{x=1}^{17}(a x+b)}{17}=17$
$\Rightarrow 9 a+b=17 \dots$(i)
$\sigma^{2}=216$
$\Rightarrow \frac{\displaystyle\sum_{x=1}^{17}(a x+b-17)^{2}}{17}=216$
$\Rightarrow \frac{\displaystyle\sum_{x=1}^{17} a^{2}(x-9)^{2}}{17}=216$
$\Rightarrow a^{2} 81-18 \times 9 a^{2}+a^{2} 3 \times(35)=216$
$\Rightarrow a^{2}=\frac{216}{24}=9$
$ \Rightarrow a=3(a>0)$
$\Rightarrow $ From $(1), b=-10$
So, $a+b=-7$