Question

The average marks of $10$ students in a class was $60$ with standard deviation $4$. While the average marks of other $10$ students was $40$ with a standard deviation $6$. If all the $20$ students are taken together, their standard deviation will be.

• A. $5.0$
• B. $7.5$
• C. $9.8$
• D. $11.5$

Answer 1

Answer: (D)

$11.5$

Here,

$n_1=10,{\overline{x}}_1=60$ and ${\sigma }_1=4$

$\therefore$ $\sum x_i=60\times 10=600$

Now, $n_2=10,{\overline{x}}_2=40$ and ${\sigma }_2=6$

$\therefore$ $\sum x_i=10\times 40=400$

$\Rightarrow$ Total marks of $20$ students $\left({\sum }_{i=1}^{20}x\right)=400+600=1000$

$\Rightarrow$ $\overline{x}=\frac{1000}{20}=50$

Now, ${\sigma }_1^2=(4{)}^2$

$\Rightarrow$ $\frac{1}{n_1}{\sum }_{i=1}^{10}{x}_i^2-(\overline{x_1}{)}^2=16$

$\Rightarrow$ ${\sum }_{i=1}^{10}{x}_i^2=n_1(16+(\overline{x_1}{)}^2)$

$=10(16+3600)$

$\Rightarrow$ ${\sum }_{i=1}^{10}{x}_i^2=36160$

Now, ${\sigma }_2^2=(6{)}^2$

$\Rightarrow$ $\frac{1}{n_2}{\sum }_{i=11}^{20}{x}_i^2-(\overline{x_2}{)}^2=36$

$\Rightarrow$ ${\sum }_{i=11}^{20}{x}_i^2=(36+(\overline{x_2}{)}^2)n_2$

$=(36+1600)10$

$\Rightarrow$ ${\sum }_{i=11}^{20}{x}_i^2=16360$

Now, required standard deviation $\left(\sigma \right)=\sqrt{\frac{1}{n}\sum x_i^2-{\overline{x}}^2}$

$=\sqrt{\frac{1}{20}(36160+16360)-(50{)}^2}$

$=\sqrt{2626-2500}$

$=\sqrt{126}$

$=11.2$