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
Correct Answer: 6344
Solution and Explanation
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
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