Question

Let the mean and standard deviation of marks of class A of $100$ students be respectively $40$ and $\alpha$ (> 0 ), and the mean and standard deviation of marks of class B of $n$ students be respectively $55$ and 30 $-\alpha$. If the mean and variance of the marks of the combined class of $100+ n$ students are respectively $50$ and $350$ , then the sum of variances of classes $A$ and $B$ is :

• A. 500
• B. 450
• C. 650
• D. 900

Answer 1

The Correct Option is A

A	B	A+B
\(\overline{x_1}=40\)	\(\overline{x_2}=55\)	\(\overline{x}=50\)
\(\sigma_2=\alpha\)	\(\sigma_2=30-\alpha\)	\(\sigma^2=350\)
\(n_1=100\)	\(n_2=n\)	\(100+n\)

\(\overline{x}=\frac{100\times40+55n}{100+n}\)
5000 + 50n = 4000 + 55n
1000 = 5n
n = 200

${\sigma }_1^2=\frac{\sum x_i^2}{100}-40^2$
${\sigma }_2^2=\frac{\sum x_j^2}{100}-55^2$
$350={\sigma }^2=\frac{\sum x_i^2+\sum x_j^2}{300}-(\overline{x}{)}^2$
$350=\frac{\left(1600+{\alpha }^2\right)\times 100+\left[(30-\alpha {)}^2+3025\right]\times 200}{300}-(50{)}^2$
$2850\times 3={\alpha }^2+2(30-\alpha {)}^2+1600+6050$
$8550={\alpha }^2+2(30-\alpha {)}^2+7650$
${\alpha }^2+2(30-\alpha {)}^2=900$
${\alpha }^2-40\alpha +300=0$
$\alpha =10,30$
${\sigma }_1^2+{\sigma }_2^2=10^2+20^2=500$
So, the correct option is (A) : 500