Let a₁, a₂, … a₁₀ be 10 observations such that Σₖ₌₁¹⁰ aₖ=50 and Σ_(∀ k < j) aₖ · aⱼ=1100. Then the standard deviation of a₁, a₂, …, a₁₀ is equal to :

Solution

Maths sir · 4 steps

Square the sum: it gives you every aₖ² plus twice the pair-products, so Σ aₖ² falls out.

1. Square the sum. When you expand (a₁ + a₂ + … + a₁₀)², each aₖ² appears once and each product aₖ aⱼ with k < j appears twice.

   (Σ aₖ)² = Σ aₖ² + 2Σ_(k<j) aₖ aⱼ

2. Find the squares. 50² = Σ aₖ² + 2 × 1100, so 2500 = Σ aₖ² + 2200. That gives Σ aₖ² = 300.
3. Variance. The mean is 50/10 = 5. Variance = (Σ aₖ²)/n - ā² = 300/10 - 25 = 30 - 25 = 5.
4. So. Standard deviation = √5, which is option (d).

The idea to keep. When a question hands you the sum and the sum of pairwise products, use (Σ a)² = Σ a² + 2Σ_(k<j) aₖ aⱼ to get Σ a². The variance comes straight after that.

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