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13.Statistics
hard
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
$18$
(JEE MAIN-2022)
Solution
$\bar{x}=\frac{\sum x_{i}}{5}=\frac{24}{5} \Rightarrow \sum x_{i}=24$
$\sigma^{2}=\frac{\sum x_{i}^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}{25}$
$\Rightarrow \sum x_{i}^{2}=154$
$x_{1}+x_{2}+x_{3}+x_{4}=14$
$\Rightarrow x_{5}=10$
$\sigma^{2}=\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}{4}-\frac{49}{4}=a$
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4 a+49$
$x_{5}^{2}=154-4 a-49$
$\Rightarrow 100=105-4 a \Rightarrow 4 a=5$
$4 a+x_{5}=15$
Standard 11
Mathematics