Question

Let the mean and the variance of $5$ observations $x_1,x_2,x_3,x_4,x_5$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively.
If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $\left(4a+x_5\right)$ is equal to:

• A. 13
• B. 15
• C. 17
• D. 18

Answer 1

Answer: (B)

$\overline{x}=\frac{\sum x_i}{5}=\frac{24}{5}$
$\Rightarrow \sum x_i=24$
${\sigma }^2=\frac{\sum x_i^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25}$
$\Rightarrow \sum x_i^2=154$
$x_1+x_2+x_3+x_4=14$
$\Rightarrow x_5=10$
${\sigma }^2=\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\frac{49}{4}=a$
$x_1^2+x_2^2+x_3^2+x_4^2=4a+49$
$x_5^2=154-4a-49$
$\Rightarrow 100=105-4a\Rightarrow 4a=5$
$4a+x_5=15$