Questions: The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 8 minutes. Find the probability that a randomly selected passenger has a waiting time less than 0.75 minutes.

Solution Steps

The mean \( E(X) \) of a uniform distribution defined on the interval \([a, b]\) is given by the formula:

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

Substituting \( a = 0 \) and \( b = 8 \):

\[ E(X) = \frac{0 + 8}{2} = 4.0 \]

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

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

Substituting \( a = 0 \) and \( b = 8 \):

\[ \text{Var}(X) = \frac{(8 - 0)^2}{12} = \frac{64}{12} = 5.333 \]

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

\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{5.333} \approx 2.309 \]

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 the probability that a randomly selected passenger has a waiting time less than \( 0.75 \) minutes, we calculate:

\[ P(0 \leq X \leq 0.75) = F(0.75) - F(0) \]

Calculating \( F(0.75) \):

\[ F(0.75) = \frac{0.75 - 0}{8 - 0} = \frac{0.75}{8} = 0.09375 \]

Calculating \( F(0) \):

\[ F(0) = \frac{0 - 0}{8 - 0} = 0 \]

Thus, the probability is:

\[ P(0 \leq X \leq 0.75) = 0.09375 - 0 = 0.09375 \]

Final Answer