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
Approach Solution - 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
Approach Solution -2
Step 1: Use the formula for the mean.
Mean \( \bar{x} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
There are 10 observations: \[ \frac{65 + 68 + 58 + 44 + 48 + 45 + 60 + \alpha + \beta + 60}{10} = 56 \]
Simplify: \[ (65 + 68 + 58 + 44 + 48 + 45 + 60 + 60) + \alpha + \beta = 560 \] \[ 448 + \alpha + \beta = 560 \] \[ \boxed{\alpha + \beta = 112} \]
Step 2: Use the variance formula.
Variance \( \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 \)
Given: \[ \sigma^2 = 66.2, \quad \bar{x} = 56, \quad n = 10 \] Substitute: \[ 66.2 = \frac{\sum x_i^2}{10} - 56^2 \] \[ \frac{\sum x_i^2}{10} = 66.2 + 3136 = 3202.2 \] \[ \sum x_i^2 = 32022 \]
Step 3: Compute the sum of squares of known values.
Known data: \( 65, 68, 58, 44, 48, 45, 60, 60 \) \[ \sum x_i^2 = 65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2 + \alpha^2 + \beta^2 \]
Compute the known squares: \[ 65^2 = 4225, \quad 68^2 = 4624, \quad 58^2 = 3364, \quad 44^2 = 1936, \] \[ 48^2 = 2304, \quad 45^2 = 2025, \quad 60^2 = 3600 \] There are two 60’s → \( 2 \times 3600 = 7200 \)
Sum of known squares: \[ 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 7200 = 25678 \]
Now: \[ 25678 + \alpha^2 + \beta^2 = 32022 \] \[ \boxed{\alpha^2 + \beta^2 = 32022 - 25678 = 6344} \]
Final Answer:
\[ \boxed{\alpha^2 + \beta^2 = 6344} \]
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