Questions: The accompanying table shows the probability distribution for x, the number that shows up when a loaded die is rolled. Find the mean. x P(x) 1 0.14 2 0.16 3 0.12 4 0.14 5 0.13 6 0.31 A. μ=0.17 B. μ=3.89 C. μ=3.50 D. μ=3.76

Solution Steps

To find the mean \( \mu \) of the probability distribution, we use the formula:

\[ \mu = \sum_{i=1}^{n} x_i \cdot P(x_i) \]

Substituting the values from the table:

\[ \mu = 1 \times 0.14 + 2 \times 0.16 + 3 \times 0.12 + 4 \times 0.14 + 5 \times 0.13 + 6 \times 0.31 \]

Calculating each term:

\[ \mu = 0.14 + 0.32 + 0.36 + 0.56 + 0.65 + 1.86 = 3.89 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \sum_{i=1}^{n} (x_i - \mu)^2 \cdot P(x_i) \]

Substituting the values:

\[ \sigma^2 = (1 - 3.89)^2 \times 0.14 + (2 - 3.89)^2 \times 0.16 + (3 - 3.89)^2 \times 0.12 + (4 - 3.89)^2 \times 0.14 + (5 - 3.89)^2 \times 0.13 + (6 - 3.89)^2 \times 0.31 \]

Calculating each term:

\[ \sigma^2 = (2.89)^2 \times 0.14 + (1.89)^2 \times 0.16 + (0.89)^2 \times 0.12 + (0.11)^2 \times 0.14 + (1.11)^2 \times 0.13 + (2.11)^2 \times 0.31 \]

\[ \sigma^2 = 8.3521 \times 0.14 + 3.5721 \times 0.16 + 0.7921 \times 0.12 + 0.0121 \times 0.14 + 1.2321 \times 0.13 + 4.4521 \times 0.31 \]

Calculating the total:

\[ \sigma^2 = 1.1693 + 0.5715 + 0.0950 + 0.0017 + 0.1602 + 1.3802 = 3.3779 \approx 3.38 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{3.38} \approx 1.84 \]

Final Answer