An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If μ is the average marks of girls and σ² is the variance of marks of 50 candidates, then μ + σ² is equal to ________________.

Solution

Maths sir · 4 steps

Get the girls' mean from the total first, then rebuild each group's sum of squares.

1. Girls' mean. Total marks of all 50 is 50 × 15 = 750. Boys contribute 20 × 12 = 240, so 30μ = 510 and μ = 17.
2. Sum of squares. Variance = (Σ x²)/n - x̄², so Σ x² = n(σ² + x̄²). Boys give 20(2 + 144) = 2920 and girls give 30(2 + 289) = 8730, a total of 11650.

   σ² = 11650/50 - 15² = 233 - 225 = 8

3. Quick check. Combined variance = average of the group variances + spread of the group means: (20 × 2 + 30 × 2)/50 + (20 × 30 × (17-12)²)/50² = 2 + 6 = 8.
4. So. μ + σ² = 17 + 8 = 25.

The common slip: 19. Taking the combined variance as just 2, because both groups have variance 2, gives 17 + 2 = 19. That ignores the gap between the two group means.

The idea to keep. To combine two groups, never average the variances alone. Convert each group to Σ x² = n(σ² + x̄²), add, then subtract the square of the overall mean.

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