Questions: Lexi is in a hurry to get to work and is rushing to catch the bus. She knows that the bus arrives every six minutes during rush hour, but does not know the exact times the bus is due. She realizes that from the time she arrives at the stop, the amount of time that she will have to wait follows a uniform distribution with a lower bound of 0 minutes and an upper bound of six minutes. What is the probability that Lexi will have to wait less than two minutes? Multiple Choice 0.3333 1.0000 0.6667

Solution Steps

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

\[ E(X) = \frac{a + b}{2} = \frac{0 + 6}{2} = 3.0 \]

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

\[ \text{Var}(X) = \frac{(b - a)^2}{12} = \frac{(6 - 0)^2}{12} = \frac{36}{12} = 3.0 \]

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

\[ \sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{3.0} \approx 1.7321 \]

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 Lexi will have to wait less than two minutes, we calculate:

\[ P(0 \leq X \leq 2) = F(2) - F(0) \]

Calculating \(F(2)\) and \(F(0)\):

\[ F(2) = \frac{2 - 0}{6 - 0} = \frac{2}{6} = 0.3333 \] \[ F(0) = \frac{0 - 0}{6 - 0} = 0.0 \]

Thus, the probability is:

\[ P(0 \leq X \leq 2) = 0.3333 - 0.0 = 0.3333 \]

Final Answer