Consider 10 observations x₁, x₂, …, x₁₀ such that Σᵢ₌₁¹⁰(xᵢ-α)=2 and Σᵢ₌₁¹⁰(xᵢ-β)²=40, where α, β are positive integers. Let the mean and the variance of the observations be 6/5 and 84/25 respectively. Then β/α is equal to :

Solution

Maths sir · 4 steps

Turn the mean and variance into Σ xᵢ and Σ xᵢ², then expand both given sums.

1. Find alpha. The mean is 6/5, so Σ xᵢ = 10 × 6/5 = 12. Then Σ (xᵢ - α) = 12 - 10α = 2, which gives α = 1.
2. Find sum of squares. Variance = (Σ xᵢ²)/10 - (mean)², so (Σ xᵢ²)/10 = 84/25 + 36/25 = 24/5. Hence Σ xᵢ² = 48.

   Σ (xᵢ-β)² = Σ xᵢ² - 2βΣ xᵢ + 10β² = 48 - 24β + 10β² = 40

3. Solve for beta. This simplifies to 5β² - 12β + 4 = 0, that is (β - 2)(5β - 2) = 0. So β = 2 or β = 2/5, and only β = 2 is a positive integer.
4. So. β/α = 2/1 = 2, which is option (a).

The idea to keep. Whenever sums like Σ (xᵢ - a) or Σ (xᵢ - b)² appear, expand them. Everything reduces to Σ xᵢ and Σ xᵢ², which the mean and variance hand you.

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