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

Question

For a statistical data \( x_1, x_2, \dots, x_{10} \) of 10 values, a student obtained the mean as 5.5 and \[ \sum_{i=1}^{10} x_i^2 = 371. \] He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively.
The variance of the corrected data is:

Answer 1

Step 1: Mean Calculation

The mean (\(\bar{x}\)) is provided as 56.

Step 2: Variance Calculation

The variance (\(\sigma^2\)) is provided as 66.2.

Step 3: Equation Setup for \(\alpha\) and \(\beta\)

Using the variance formula with the given mean and variance:

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

Step 4: Solve for \(\alpha^2 + \beta^2\)

Rearranging the equation yields:

\[ \alpha^2 + \beta^2 = 6344 \]

The final result is: 6344

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: Mean Calculation

Step 2: Variance Calculation

Step 3: Equation Setup for \(\alpha\) and \(\beta\)

Step 4: Solve for \(\alpha^2 + \beta^2\)

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