Questions: What is P(X>0)?

Solution Steps

The mean \( \mu \) of the discrete random variable \( X \) is calculated as follows:

\[ \mu = E(X) = \sum_{x} x \cdot P(X = x) = 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{3} \]

Calculating this gives:

\[ \mu = 0 + \frac{1}{6} + \frac{2}{6} + 1 = 1.5 \]

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

\[ \sigma^2 = E(X^2) - (E(X))^2 \]

First, we compute \( E(X^2) \):

\[ E(X^2) = \sum_{x} x^2 \cdot P(X = x) = 0^2 \times \frac{1}{3} + 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + 3^2 \times \frac{1}{3} \]

Calculating this gives:

\[ E(X^2) = 0 + \frac{1}{6} + \frac{4}{6} + 3 = 4.833 \]

Now, substituting back into the variance formula:

\[ \sigma^2 = 4.833 - (1.5)^2 = 4.833 - 2.25 = 2.583 \]

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

\[ \sigma = \sqrt{\sigma^2} = \sqrt{2.583} \approx 1.607 \]

To find \( P(X > 0) \):

\[ P(X > 0) = P(X = 1) + P(X = 2) + P(X = 3) = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} \]

Calculating this gives:

\[ P(X > 0) = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.667 \]

Final Answer