If the mean and variance of the data 65,68,58,44,48,45,60, α, β, 60 where α> β, are 56 and 66.2 respectively, then α²+β² is equal to _________.

Solution

Maths sir · 4 steps

The mean fixes α+β, the variance fixes Σ x²; subtract the eight known squares.

1. Total from mean. There are 10 values with mean 56, so their total is 560. The eight known values add to 448, and 60 appears twice. So α + β = 112.

   σ² = (Σ xᵢ²)/n - x̄² ⇒ 66.2 = (Σ xᵢ²)/10 - 3136

2. Sum of squares. This gives (Σ xᵢ²)/10 = 3202.2, so Σ xᵢ² = 32022. The known squares are 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600 = 25678.
3. Quick check. α² + β² = 32022 - 25678 = 6344. With α + β = 112, this gives α = 62 and β = 50. These satisfy α > β, so the data is consistent.
4. So. α² + β² = 6344.

The common slip: 9944. Counting 60 only once in the sum of squares gives 32022 - 22078 = 9944, not 6344. The value 60 appears twice in the data.

The idea to keep. Unknown data values in a mean and variance question: the mean gives their sum, and the variance gives their sum of squares. You do not need the individual values unless they are asked.

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