13.Statistics
hard

The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:

A

$5$

B

$8$

C

$4$

D

$6$

(JEE MAIN-2021)

Solution

$n_{1}=100 \quad n=250$

$\therefore n_{2}=250-100 \Rightarrow n_{2}=150$

$\bar{x}=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}}$

$15.6=\frac{100(15)+(150)\left(\bar{x}_{2}\right)}{250}$

$\Rightarrow \overline{\mathrm{x}}_{2}=16$

$\overline{\mathrm{X}}_{1}=15 \quad\quad\quad\quad \Rightarrow$

$\sigma_{1}^{2}=V_{1}(x)=9 \quad \sigma^{2}=\operatorname{Var}(x)=13.44$

$\sigma^{2}=\frac{\mathrm{n}_{1} \sigma_{1}^{2}+\mathrm{n}_{2} \sigma_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}$

$\mathrm{n}_{2}=150, \overline{\mathrm{x}}_{2}=16, \mathrm{~V}_{2}(\mathrm{x})=\sigma_{2}$

$13.44=\frac{100 \times 9+150 \times \sigma_{2}^{2}}{250}+\frac{100 \times 150}{(250)^{2}} \times 1$

$\Rightarrow \sigma_{2}^{2}=16 \Rightarrow \sigma_{2}=4$

Standard 11
Mathematics

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