Let mean and variance of 5 observations x_{1}, x_{2}, x_{3}, x_{4}, x

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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

$13$

$15$

$17$

$18$

(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

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$\mathrm{x}$	$\mathrm{x}_{1}=2$	$\mathrm{x}_{2}=6$	$\mathrm{x}_{3}=8$	$\mathrm{x}_{4}=9$
$\mathrm{f}$	$4$	$4$	$\alpha$	$\beta$

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

Solution

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Let the mean and variance of the frequency distribution

$\mathrm{x}$ $\mathrm{x}_{1}=2$ $\mathrm{x}_{2}=6$ $\mathrm{x}_{3}=8$ $\mathrm{x}_{4}=9$

$\mathrm{f}$ $4$ $4$ $\alpha$ $\beta$

be $6$ and $6.8$ respectively. If $x_{3}$ is changed from $8$ to $7 ,$ then the mean for the new data will be: