Mathematics - Statistics Question with Solution | TestHub

Let the mean is denoted by $\overline{x_i}$, variance by ${{\sigma }_i}^2$, so standard deviation will be ${\sigma }_i$, now as per given data we get,

$\begin{array}{ccccc}A& & B& & A+B\\ {\overline{x}}_1=40& & {\overline{x}}_2=55& & \overline{x}=50\\ {\sigma }_1=\alpha & & {\sigma }_2=30-\alpha & & {\sigma }_2=350\\ n_1=100& & n_2=n& & 100+n\end{array}$

So, using the formula of mean in $A+B$ we get,

$\overline{x}=\frac{100\times 40+55n}{100+n}=50$

$\Rightarrow n=200$

Now using the formula of variance we get,

${\sigma }_1^2=\frac{\sum x_i^2}{100}-{40}^2={\alpha }^2$

And ${\sigma }_2^2=\frac{\sum y_i^2}{200}-{55}^2={\left(30-\alpha \right)}^2$

So, variance of $A+B$ will be,

${\sigma }^2=\frac{\sum x_i^2+\sum y_i^2}{300}-{50}^2={350}^2$

$\Rightarrow {350}^2=\frac{\left(1600+{\alpha }^2\right)\times 100+\left[3025+{\left(30-\alpha \right)}^2\right]\times 200}{300}-{50}^2$

$\Rightarrow {\alpha }^2-40\alpha +300=0$

$\Rightarrow \alpha =10$ or $30$

And $\alpha =30$ is not possible

So, ${\sigma }_1{}^2+{\sigma }_2{}^2={10}^2+{20}^2=500$