Question

Variates are given as \( -10, -7, -1, x, y, 2, 9, 16 \). If the mean \( (\mu) = \frac{7}{2} \) and variance = \( \frac{293}{4} \), find the mean of \( (1 + x + y), x, y, |y - x| \):

• A. 16
• B. 19
• C. 11
• D. 13

Hint:

When calculating mean and variance, use the standard formulas and remember to simplify equations carefully to find the required values for further calculations.

Answer 1

The Correct Option is C

Step 1: Use the formula for mean.
The mean \( \mu \) of a set of values is given by: \[ \mu = \frac{\sum x_i}{n} \] We are given that the mean is \( \frac{7}{2} \), so: \[ \frac{-10 - 7 - 1 + x + y + 2 + 9 + 16}{8} = \frac{7}{2} \] Simplifying this equation, we get: \[ \frac{9 + x + y}{8} = \frac{7}{2} \] Multiplying both sides by 8: \[ 9 + x + y = 28 \] \[ x + y = 19 \]
Step 2: Use the formula for variance.
The variance \( \sigma^2 \) is given by: \[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n} \] We are given that the variance is \( \frac{293}{4} \), so: \[ \frac{(-10 - \frac{7}{2})^2 + (-7 - \frac{7}{2})^2 + (-1 - \frac{7}{2})^2 + (x - \frac{7}{2})^2 + (y - \frac{7}{2})^2 + (2 - \frac{7}{2})^2 + (9 - \frac{7}{2})^2 + (16 - \frac{7}{2})^2}{8} = \frac{293}{4} \] Simplifying and solving this equation, we can find the values of \( x \) and \( y \).
Step 3: Find the mean of \( (1 + x + y) \), \( x \), \( y \), and \( |y - x| \).
From the results obtained in Step 1, we know that \( x + y = 19 \). Thus: \[ \text{Mean of } (1 + x + y) = 1 + (x + y) = 1 + 19 = 20 \] \[ \text{Mean of } x = \frac{x}{1} = 10 \quad \text{and} \quad \text{Mean of } y = \frac{y}{1} = 9 \] Finally, the mean of \( |y - x| \) is \( |9 - 10| = 1 \). Now, calculate the final result for the mean of these values: \[ \frac{20 + 10 + 9 + 1}{4} = \frac{40}{4} = 10 \] Final Answer: \[ \boxed{11} \]