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 :

Question

If the variance of the data $ 2,3,5,8,12 $ is $ \sigma^2 $ and the mean deviation from the median for this data is $ M $, then $ \sigma^2 - M $ is:

Answer 1

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

σ_{1}^{2} = \frac{\sum x _{i}^{2}}{100} - 4 0^{2}

σ_{2}^{2} = \frac{\sum x _{j}^{2}}{100} - 5 5^{2}

350 = σ^{2} = \frac{\sum x _{i}^{2} + \sum x _{j}^{2}}{300} - (\overline{x})^{2}

350 = \frac{( 1600 + α ^{2} ) \times 100 + [ ( 30 - α ) ^{2} + 3025 ] \times 200}{300} - (50)^{2}

2850 \times 3 = α^{2} + 2 (30 - α)^{2} + 1600 + 6050

8550 = α^{2} + 2 (30 - α)^{2} + 7650

α^{2} + 2 (30 - α)^{2} = 900

α^{2} - 40 α + 300 = 0

α = 10, 30

σ_{1}^{2} + σ_{2}^{2} = 1 0^{2} + 2 0^{2} = 500

So, the correct option is (A) : 500

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

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The Correct Option is A

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