Questions: Question If X ~ U(4.5,18.5) is a continuous uniform random variable, what is P(X<11) ? Select the correct answer below: 1/28 3/28 13/28 9/14 19/28

Solution Steps

The mean \( E(X) \) of a uniform distribution \( U(a, b) \) is given by the formula:

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

Substituting the values \( a = 4.5 \) and \( b = 18.5 \):

\[ E(X) = \frac{4.5 + 18.5}{2} = \frac{23}{2} = 11.5 \]

The variance \( \text{Var}(X) \) of a uniform distribution is calculated using the formula:

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

Substituting the values:

\[ \text{Var}(X) = \frac{(18.5 - 4.5)^2}{12} = \frac{(14)^2}{12} = \frac{196}{12} = 16.3333 \]

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

\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{16.3333} \approx 4.0415 \]

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

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

To find \( P(X < 11) \), we calculate:

\[ P(4.5 \leq X \leq 11) = F(11) - F(4.5) \]

Calculating \( F(11) \):

\[ F(11) = \frac{11 - 4.5}{18.5 - 4.5} = \frac{6.5}{14} \approx 0.4643 \]

And since \( F(4.5) = 0 \):

\[ P(4.5 \leq X \leq 11) = 0.4643 - 0 = 0.4643 \]

Final Answer