Let the mean and the variance of 5 observatio | vedclass

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Question

$\bar{x}=\frac{\sum x_{i}}{5}=\frac{24}{5} \Rightarrow \sum x_{i}=24$

$\sigma^{2}=\frac{\sum x_{i}^{2}}{5}-\left(\frac{24}{5}\right)^{2}=\frac{194}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

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

Diameters	$33-36$	$37-40$	$41-44$	$45-48$	$49-52$
No. of circles	$15$	$17$	$21$	$22$	$25$

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

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