Question

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively, then $\alpha^2 + \beta^2$ is equal to

Answer 1

Step 1: Calculate the Mean $\bar{x}$

Given that the mean $\bar{x} = 56$.

Step 2: Calculate the Variance $\sigma^2$

Given that the variance $\sigma^2 = 66.2$.

Step 3: Set up the Equations for $\alpha$ and $\beta$

Using the formula for variance with mean and variance values:

$\frac{\alpha^2 + \beta^2 + 25678}{10} - (56)^2 = 66.2$

Step 4: Solve for $\alpha^2 + \beta^2$

Rearranging, we find:

$\alpha^2 + \beta^2 = 6344$

So, the correct answer is: 6344