Let the mean and the variance of 20 observations x₁, x₂, …, x₂₀ be 15 and 9 , respectively. For α ∈ R, if the mean of (x₁+α)²,(x₂+α)², …,(x₂₀+α)² is 178 , then the square of the maximum value of α is equal to ________.

Solution

Maths sir · 4 steps

Expand the square, then turn the mean of xᵢ² into variance plus mean squared.

1. Mean of squares. Variance is the mean of the squares minus the square of the mean. So the mean of xᵢ² is 9 + 15² = 234.
2. Expand and average. Each term is xᵢ² + 2α xᵢ + α². Averaging over all 20 terms gives the mean of xᵢ², plus 2α times the mean of xᵢ, plus α².

   234 + 30α + α² = 178 ⇒ α² + 30α + 56 = 0

3. Solve. The quadratic factorises as (α + 2)(α + 28) = 0, so α = -2 or α = -28.
4. So. The maximum value is α = -2, since -2 > -28. Its square is (-2)² = 4.

The common slip: 784. Choosing α = -28 because it is larger in size gives 784. The maximum of two negative numbers is the one closer to zero, -2.

The idea to keep. Whenever a question asks for the mean of squares, use mean of x² = variance + (mean)². Then expand the bracket and average term by term.

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