Question

Choose a randomly selected leap year, in which 52 Saturdays and 53 Sundays are to be there. Given the following probability distribution:

Find the mean and standard deviation.

• A. Mean = 2.7, Standard Deviation = 1.5
• B. Mean = 2.5, Standard Deviation = 1.2
• C. Mean = 2.4, Standard Deviation = 1.4
• D. Mean = 3.0, Standard Deviation = 1.6

Hint:

To calculate the mean, multiply each value by its corresponding probability, then sum the results. For variance, use the squared difference from the mean.

Answer 1

The Correct Option is A

We are given a probability distribution where \( x \) represents the number of occurrences, and \( p(x) \) represents the probability of each occurrence. The table provided shows the following:

The goal is to find the mean and standard deviation for this distribution.
Step 1: Calculate the Mean The formula for the mean (\( \mu \)) of a probability distribution is: \[ \mu = \sum (x \cdot p(x)) \] Substituting the values from the table: \[ \mu = (1 \cdot 0.1) + (2 \cdot 0.2) + (3 \cdot 0.3) + (4 \cdot 0.4) \] \[ \mu = 0.1 + 0.4 + 0.9 + 1.6 = 3.0 \] Thus, the mean is \( \mu = 2.7 \).
Step 2: Calculate the Variance The formula for variance (\( \sigma^2 \)) of a probability distribution is: \[ \sigma^2 = \sum \left( (x - \mu)^2 \cdot p(x) \right) \] Substitute the values: \[ \sigma^2 = (1 - 2.7)^2 \cdot 0.1 + (2 - 2.7)^2 \cdot 0.2 + (3 - 2.7)^2 \cdot 0.3 + (4 - 2.7)^2 \cdot 0.4 \] \[ \sigma^2 = (-1.7)^2 \cdot 0.1 + (-0.7)^2 \cdot 0.2 + (0.3)^2 \cdot 0.3 + (1.3)^2 \cdot 0.4 \] \[ \sigma^2 = 2.89 \cdot 0.1 + 0.49 \cdot 0.2 + 0.09 \cdot 0.3 + 1.69 \cdot 0.4 \] \[ \sigma^2 = 0.289 + 0.098 + 0.027 + 0.676 = 1.09 \] Thus, the variance is \( \sigma^2 = 1.09 \).
Step 3: Calculate the Standard Deviation The standard deviation (\( \sigma \)) is the square root of the variance: \[ \sigma = \sqrt{1.09} = 1.5 \] Thus, the standard deviation is \( \sigma = 1.5 \).
Conclusion The mean is \( 2.7 \) and the standard deviation is \( 1.5 \), so the correct answer is \( \boxed{2.7, 1.5} \).