Let x₁, x₂, ..., x₁₀ be ten observations such that Σᵢ₌₁¹⁰ (xᵢ - 2) = 30, Σᵢ₌₁¹⁰ (xᵢ - β)² = 98, β > 2, and their variance is 4/5. If μ and σ² are respectively the mean and the variance of 2(x₁ - 1) + 4β, 2(x₂ - 1) + 4β, ..., 2(x₁₀ - 1) + 4β, then βμ/σ² is equal to :

Solution

Maths sir · 5 steps

Find the mean first, then split the sum of squares about β into the spread plus a shift term.

1. Mean of x. Σ (xᵢ - 2) = 30 gives Σ xᵢ = 30 + 20 = 50, so x̄ = 5. The variance 4/5 then gives Σ (xᵢ - 5)² = 10 × 4/5 = 8.
2. Split about beta. Write xᵢ - β = (xᵢ - 5) + (5 - β) and square. The cross term is 2(5-β)Σ (xᵢ - 5), which is zero because deviations from the mean add to zero.

   Σ (xᵢ - β)² = Σ (xᵢ - 5)² + 10(5 - β)² ⇒ 98 = 8 + 10(5-β)²

3. Find beta. (5 - β)² = 9, so β = 8 or β = 2. Since β > 2, β = 8.
4. New data. Each new value is 2xᵢ - 2 + 32 = 2xᵢ + 30. The mean scales and shifts: μ = 2 × 5 + 30 = 40. The variance ignores the shift and scales by 2²: σ² = 4 × 4/5 = 16/5.
5. So. βμ/σ² = (8 × 40)/(16/5) = (320 × 5)/16 = 100, which is option (a).

The idea to keep. The sum of squares about any point a equals nσ² + n(x̄ - a)². For ax + b, the mean becomes ax̄ + b and the variance becomes a²σ², because adding b does not change the spread.

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