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 $\left(4 a+x_{5}\right)$ is equal to:

• A. 13
• B. 15
• C. 17
• D. 18

Answer 1

Answer: (B)

$\overline{ x }=\frac{\displaystyle\sum x _{ i }}{5}=\frac{24}{5} $
$\Rightarrow \displaystyle\sum x _{ i }=24$
$\sigma^{2}=\frac{\displaystyle\sum x _{ i }^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}{25}$
$\Rightarrow \displaystyle\sum x _{ i }^{2}=154$
$x _{1}+ x _{2}+ x _{3}+ x _{4}=14$
$\Rightarrow x _{5}=10$
$\sigma^{2}=\frac{ x _{1}^{2}+ x _{2}^{2}+ x _{3}^{2}+ x _{4}^{2}}{4}-\frac{49}{4}= a$
$x _{1}^{2}+ x _{2}^{2}+ x _{3}^{2}+ x _{4}^{2}=4 a +49$
$x _{5}^{2}=154-4 a -49$
$\Rightarrow 100=105-4 a \Rightarrow 4 a =5$
$4 a + x _{5}=15$