Related Formula
Mean x = (Σ xᵢ)/(n)$$\text{Mean } \bar{x} = \frac{\sum x_i}{n}$$
Variance σ² = (Σ xᵢ²)/(n) - x²$$\text{Variance } \sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$
Core Logic
Let the 9 unchanged numbers be x₁, x₂, , x₉$x_1, x_2, \dots, x_9$.
Case 1 (with α$\alpha$):
Σi=1⁹ xᵢ + α = 10 × 10 = 100 ⇒ Σi=1⁹ xᵢ = 100 - α$$\sum_{i=1}^9 x_i + \alpha = 10 \times 10 = 100 \Rightarrow \sum_{i=1}^9 x_i = 100 - \alpha$$
Σi=1⁹ xᵢ² + α²10 - (10)² = 2 ⇒ Σi=1⁹ xᵢ² + α² = 1020 ⇒ Σi=1⁹ xᵢ² = 1020 - α²$$\frac{\sum_{i=1}^9 x_i^2 + \alpha^2}{10} - (10)^2 = 2 \Rightarrow \sum_{i=1}^9 x_i^2 + \alpha^2 = 1020 \Rightarrow \sum_{i=1}^9 x_i^2 = 1020 - \alpha^2$$
Step 1: Applying Replacement Conditions
Case 2 (with β$\beta$):
Mean = 10.1
Σi=1⁹ xᵢ + β10 = 10.1 ⇒ 100 - α + β = 101 ⇒ β - α = 1$$\frac{\sum_{i=1}^9 x_i + \beta}{10} = 10.1 \Rightarrow 100 - \alpha + \beta = 101 \Rightarrow \beta - \alpha = 1$$
Variance = 1.99
Σi=1⁹ xᵢ² + β²10 - (10.1)² = 1.99$$\frac{\sum_{i=1}^9 x_i^2 + \beta^2}{10} - (10.1)^2 = 1.99$$
Σi=1⁹ xᵢ² + β² = 10(1.99 + 102.01) = 10(104) = 1040$$\sum_{i=1}^9 x_i^2 + \beta^2 = 10(1.99 + 102.01) = 10(104) = 1040$$
Step 2: Solving for Alpha and Beta
Substitute Σi=1⁹ xᵢ² = 1020 - α²$\sum_{i=1}^9 x_i^2 = 1020 - \alpha^2$:
1020 - α² + β² = 1040 ⇒ β² - α² = 20$$1020 - \alpha^2 + \beta^2 = 1040 \Rightarrow \beta^2 - \alpha^2 = 20$$
(β - α)(β + α) = 20$$(\beta - \alpha)(\beta + \alpha) = 20$$
Since β - α = 1$\beta - \alpha = 1$:
(1)(α + β) = 20 ⇒ α + β = 20$$(1)(\alpha + \beta) = 20 \Rightarrow \alpha + \beta = 20$$
(Solving gives α = 19/2, β = 21/2$\alpha = 19/2, \beta = 21/2$).
Pattern Recognition
When a single term is replaced, immediately establish the difference of sums and difference of sum-of-squares. The identity β² - α²$\beta^2 - \alpha^2$ gracefully factors to use the established β - α$\beta - \alpha$ value.
Chapter Mix
Class 11 Maths: Statistics