Question

Let $X=\{x\in N:1\le x\le 17\}$ and $Y=\{ax+b:x\in X$ 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
${\sigma }^2=\frac{\sum _{i=1}^n{\left(x_i-\mu \right)}^2}{n}$
$\mu =17$
$\Rightarrow \frac{\sum _{x=1}^{17}(ax+b)}{17}=17$
$\Rightarrow 9a+b=17\dots$(i)
${\sigma }^2=216$
$\Rightarrow \frac{\sum _{x=1}^{17}(ax+b-17{)}^2}{17}=216$
$\Rightarrow \frac{\sum _{x=1}^{17}{a}^2(x-9{)}^2}{17}=216$
$\Rightarrow a^{2}81-18\times 9a^2+a^{2}3\times (35)=216$
$\Rightarrow a^2=\frac{216}{24}=9$
$\Rightarrow a=3(a>0)$
$\Rightarrow$ From $(1),b=-10$
So, $a+b=-7$