Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to __________.

Solution

Maths sir · 4 steps

The shifts are not the same for every number, so the spread changes. Rebuild the sum of squares and recompute.

1. Old mean and squares. The old mean is (2n × 6 + n × 3)/3n = 15n/3n = 5. From the variance formula below, (Σ x²)/3n = 4 + 5² = 29, so Σ x² = 87n.

   σ² = (Σ xᵢ²)/N - x̄²

2. New mean. The first group's sum goes from 12n to 12n + 2n = 14n. The second group's sum goes from 3n to 3n - n = 2n. The new mean is 16n/3n = 16/3.
3. New sum of squares. Each (x+1)² adds 2x + 1, so the first group adds 2(12n) + 2n = 26n. Each (y-1)² adds -2y + 1, so the second group adds -2(3n) + n = -5n. The new sum of squares is 87n + 26n - 5n = 108n.
4. So. k = 108n/3n - (16/3)² = 36 - 256/9 = 68/9, so 9k = 68.

The common slip: 36. Treating the change as a plain shift and keeping the variance at 4 gives 9k = 36. The two groups move in opposite directions, so the spread grows.

The idea to keep. Variance stays the same only when every number moves by the same amount. When groups shift differently, find the new Σ x and Σ x² from the group sums and use σ² = (Σ x²)/N - x̄².

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