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Let the mean and the variance of 5 observations $\mathrm{x_{1}, x_{2}, x_{3}, x_{4}, x_{5} be \frac{24}{5} and \frac{194}{25}}$ 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:
Option: 1
$13$

Option: 2
$15$

Option: 3
$17$

Option: 4
$18$

Answers (1)

$\mathrm{\text { Mean }=\frac{24}{5} }$

$\mathrm{\Rightarrow \frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}=\frac{24}{5}}$

$\mathrm{\Rightarrow \quad x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=24}$ ..........(i)

$\mathrm{Mean\: of\: x_{1}, x_{2}, x_{3}, x_{4}=\frac{7}{2}}$

$\mathrm{\Rightarrow \quad \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}=\frac{7}{2}} \\$

$\mathrm{\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}=14 } \\$ ............(ii)

$\mathrm{\Rightarrow x_{5}=10 }$ [from (i) and (ii)]

$\mathrm{\text { Variance }=\frac{194}{25}} \\$

$\mathrm{\Rightarrow \frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}}{5}-\frac{576}{25}=\frac{194}{25} } \\$

$\mathrm{\Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=154} \\$

$\mathrm{\left.\Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=54 \text { (as } x_{5}=10\right)}$

$\mathrm{\therefore \text { Variance of } x_{1}, x_{2}, x_{3}, x_{4}} \\$

$\mathrm{\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}{4}-\frac{49}{4}=a }\\$

$\mathrm{\Rightarrow a=\frac{54}{4}-\frac{49}{4} }\\$

$\mathrm{\Rightarrow a=\frac{5}{4}} \\$

$\mathrm{\therefore 4 a+x_{5}=5+10=15}$

Hence the correct answer is option 2.

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Kshitij

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