Answer:
Option B
Explanation:
We know that , if x1 , x2..........xn are n observations , then their standard deviation is given by
$\sqrt{\frac{1}{n}\sum x_i^2}-(\frac{\sum x_{i}}{2})^{2}$
We have , $(3.5)^{2}=\frac{(2^{2}+3^{2}+11^{2})}{4}-(\frac{2+3+a+11}{4})$
$\Rightarrow \frac{49}{4}=\frac{(4+9+a^{2}+121)}{4}-(\frac{16+a}{4})^{2}$
$\Rightarrow \frac{49}{4}=\frac{(134+a^{2})}{4}-(\frac{256+a^{2}+32a}{16})$
$\Rightarrow \frac{49}{4}=\frac{4a^{2}+536-256-a^{2}-32a}{16}$
$\Rightarrow 49\times4=3a^{2}-32a+280$
$\Rightarrow 3a^{2}-32a+84=0$
