Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is ___________.

Solution

Maths sir · 4 steps

Shifting does not change variance, but the two groups now sit further apart, and that gap adds spread.

1. Shift each set. Subtracting 3 from every element of A moves its mean to 5 - 3 = 2 and leaves its variance at 12. Adding 2 to every element of B moves its mean to 8 + 2 = 10 and leaves its variance at 20.
2. Mean of C. Both parts have 5 elements, so the mean of C is (5 × 2 + 5 × 10)/10 = 6.
3. Sum of squares. For each part, Σ x² = n(σ² + x̄²). The shifted A gives 5(12 + 4) = 80 and the shifted B gives 5(20 + 100) = 600, so C has Σ x² = 680.

   σ_C² = (Σ x²)/10 - x̄_C² = 680/10 - 6² = 68 - 36 = 32

4. So. Mean plus variance of C is 6 + 32 = 38, option (c).

The idea to keep. Adding a constant moves the mean and leaves the variance alone. To combine groups, go back to Σ x² = n(σ² + x̄²) for each group, add them up, then subtract the square of the new mean.

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