Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be 24/5 and 194/25 respectively. If the mean and variance of the first 4 observation are 7/2 and a respectively, then (4a + x₅) is equal to:

Solution

Maths sir · 4 steps

Turn mean and variance into a sum and a sum of squares, then subtract the first four to get x5.

1. Find x5 from sums. Sum of all 5 is 5 × 24/5 = 24. Sum of the first 4 is 4 × 7/2 = 14. So x₅ = 24 - 14 = 10.

   σ² = (Σ xᵢ²)/n - x̄²

2. Sum of squares. For all 5: (Σ xᵢ²)/5 = 194/25 + 576/25 = 770/25, so Σ xᵢ² = 154. Remove x₅² = 100, and the first 4 give x₁² + x₂² + x₃² + x₄² = 54.
3. Variance of four. a = 54/4 - (7/2)² = 27/2 - 49/4 = 5/4, so 4a = 5.
4. So. 4a + x₅ = 5 + 10 = 15, which is option (b).

The idea to keep. Mean gives you Σ xᵢ and variance gives you Σ xᵢ². Work with these two totals: adding or removing an observation is then a simple subtraction.

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