Questions: Let X represent a continuous random variable with a uniform distribution over the interval from 0 to 3. What is P(X>0.9) ? 0.3 0.7 2.1 0.9

Solution Steps

For a uniform distribution \( X \) over the interval \([a, b]\), the mean \( E(X) \) is calculated as:

\[ E(X) = \frac{a + b}{2} = \frac{0 + 3}{2} = 1.5 \]

The variance \( \text{Var}(X) \) of a uniform distribution is given by:

\[ \text{Var}(X) = \frac{(b - a)^2}{12} = \frac{(3 - 0)^2}{12} = \frac{9}{12} = 0.75 \]

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

\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{0.75} \approx 0.866 \]

The cumulative distribution function \( F(x; a, b) \) for a uniform distribution is defined as:

\[ F(x; a, b) = \frac{x - a}{b - a}, \quad a \leq x \leq b \]

To find \( P(X > 0.9) \), we can use the CDF:

\[ P(X > 0.9) = 1 - P(X \leq 0.9) = 1 - F(0.9; 0, 3) \]

Calculating \( F(0.9; 0, 3) \):

\[ F(0.9; 0, 3) = \frac{0.9 - 0}{3 - 0} = \frac{0.9}{3} = 0.3 \]

Thus,

\[ P(X > 0.9) = 1 - 0.3 = 0.7 \]

Final Answer