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

Solution

Maths sir · 5 steps

Use the combined mean to find n first, then the combined variance gives a quadratic in α.

1. Find n. The combined mean is (100 × 40 + 55n)/(100 + n) = 50. This gives 4000 + 55n = 5000 + 50n, so 5n = 1000 and n = 200.
2. Combined variance. Each class adds its own variance plus the square of its mean's distance from the combined mean. Here d_A = 40 - 50 = -10 and d_B = 55 - 50 = 5.

   σ² = (n_A(σ_A² + d_A²) + n_B(σ_B² + d_B²))/(n_A + n_B)

3. Put in numbers. 350 × 300 = 100(α² + 100) + 200((30-α)² + 25). So 105000 = 100α² + 200(30-α)² + 15000, and dividing by 100 gives 900 = α² + 2(30-α)².
4. Solve for α. Expanding gives 3α² - 120α + 900 = 0, that is α² - 40α + 300 = 0, so α = 10 or 30. With α = 30, class B's standard deviation 30 - α becomes 0, which would mean all 200 students scored exactly 55. Reject it, so α = 10.
5. So. σ_A² = 10² = 100 and σ_B² = 20² = 400. Their sum is 500, option (d).

The idea to keep. Combining groups: variance is not just the weighted average of variances. Each group also adds nᵢ dᵢ², where dᵢ is its mean minus the combined mean.

Answer from the NTA final answer key. Solution by PrepWiser.