Questions: The time to fly between New York City and Chicago is uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes. What is the probability that a flight is between 125 and 140 minutes? Multiple Choice O 0.33 O 0.67 O 1.00 O 0.50 < Prev 1 of 10 Next >

Solution Steps

The mean \( E(X) \) of a uniform distribution is calculated using the formula:

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

Substituting the values \( a = 120 \) and \( b = 150 \):

\[ E(X) = \frac{120 + 150}{2} = 135.0 \]

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

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

Substituting the values:

\[ \text{Var}(X) = \frac{(150 - 120)^2}{12} = \frac{900}{12} = 75.0 \]

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

\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{75.0} \approx 8.6603 \]

To find the probability \( P(125 \leq X \leq 140) \), we use the cumulative distribution function \( F(x; a, b) \):

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

Calculating \( F(140) \) and \( F(125) \):

\[ F(140) = \frac{140 - 120}{150 - 120} = \frac{20}{30} = \frac{2}{3} \approx 0.6667 \]

\[ F(125) = \frac{125 - 120}{150 - 120} = \frac{5}{30} = \frac{1}{6} \approx 0.1667 \]

Thus, the probability is:

\[ P(125 \leq X \leq 140) = F(140) - F(125) = 0.6667 - 0.1667 = 0.5 \]

Final Answer