Question

The mean of two samples of sizes $200$ and $300$ were found to be $25$ and $10$ respectively. Their standard deviations were $3$ and $4$ respectively. The varience of combined sample size of $500$ is

• A. $64$
• B. $65.2$
• C. $67.2$
• D. $64.2$

Answer 1

The correct option is C $67.2$

Given : First sample size$\left(n_1\right)=200$, Second sample size$\left(n_2\right)=300$

First mean $\left({\overline{x}}_1\right)=25$, Second mean $\left({\overline{x}}_2\right)=10$

S.D.$\left({\sigma }_1\right)=3$ and S.D.$\left({\sigma }_2\right)=4$

Combined mean, $\overline{x}=\frac{200\times 25+300\times 10}{500}=\frac{80}{5}=16$

Let $d_1=\overline{x_1}-\overline{x}=25-16=9$

and $d_2=\overline{x_2}-\overline{x}=10-16=-6$

$\therefore {\sigma }^2=\frac{n_1({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2)}{n_1+n_2}$

$=\frac{200(9+81)+300(16+36)}{500}=67.2$