If Σᵢ₌₁ⁿ ( xᵢ - a ) = n and Σᵢ₌₁ⁿ ( xᵢ - a )² = na
(n, a > 1) then the standard deviation of n
observations x₁ , x₂ , ..., xₙ is :

Solution

Maths sir · 3 steps

Shift every xᵢ by a: the spread does not change, and the given sums become easy to use.

1. Shift the data. Let yᵢ = xᵢ - a. Subtracting the same number from every observation moves the data but does not change its spread, so xᵢ and yᵢ have the same standard deviation.
2. Read off sums. Now the data tell us Σ yᵢ = n and Σ yᵢ² = na. So the mean of the yᵢ is ȳ = n/n = 1.

   σ² = (Σ yᵢ²)/n - ȳ² = na/n - 1² = a - 1

3. So. The variance is a - 1, so the standard deviation is σ = √(a - 1), option (d).

The idea to keep. Variance and standard deviation do not change when you add or subtract a constant from every value. Use σ² = (Σ y²)/n - ȳ² on the shifted data.

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