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

Answer: (A)

500

Answer 2

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

σ12​=100∑xi2​​−402
σ22​=100∑xj2​​−552
350=σ2=300∑xi2​+∑xj2​​−(x)2
350=300(1600+α2)×100+[(30−α)2+3025]×200​−(50)2
2850×3=α2+2(30−α)2+1600+6050
8550=α2+2(30−α)2+7650
α2+2(30−α)2=900
α2−40α+300=0
α=10,30
σ12​+σ22​=102+202=500
So, the correct option is (A) : 500