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Statistics Part 1

Section: Mathematics

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Question : 4 of 100

Marks: +1, -0

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:

[31-Jan-2023 Shift 2]

500
650
450
900

Solution:

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

\; 5000 + 50 n = 4000 + 55 n

\; 1000 = 5 n

\; 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

\; 2850 = \frac{(1600+\alpha^2) \times 100 + [(30-\alpha)^2+3025] \times 200}{300} - (50)^2

\; 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

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