The time spent waiting for a bus is uniformly distributed between 0 and 5 minutes. State answers using decimals. What is the probability that a person will wait at most 3 minutes for the bus? type your answer... What is the probability that a person will wait more than 3 minutes for the bus? type your answer... What the probility that a person will wait between 3 and 4 minutes for the bus? type your answer... Find the 75 th percentile of wait times. type your answer... minutes Find the 80th percentile of wait times. type your answer... minutes What is the mean of these wait times? type yopr answer... minutes What is the standard deviation of these wait times? State the answer to three decimal places. type your answer... \( 24 \% \) of people wait a maximum of how many minutes? type your answer... minutes. \( 24 \% \) of people wait a minimum of how many minutes? type your answer... minutes

To solve the problem, we will use the properties of a uniform distribution. The time spent waiting for a bus is uniformly distributed between 0 and 5 minutes. ### Known Conditions: - The minimum wait time \( a = 0 \) minutes. - The maximum wait time \( b = 5 \) minutes. ### 1. Probability that a person will wait at most 3 minutes: The probability \( P(X \leq x) \) for a uniform distribution is given by: \[ P(X \leq x) = \frac{x - a}{b - a} \] For \( x = 3 \): \[ P(X \leq 3) = \frac{3 - 0}{5 - 0} = \frac{3}{5} = 0.6 \] ### 2. Probability that a person will wait more than 3 minutes: This can be calculated as: \[ P(X > 3) = 1 - P(X \leq 3) = 1 - 0.6 = 0.4 \] ### 3. Probability that a person will wait between 3 and 4 minutes: \[ P(3 < X < 4) = P(X < 4) - P(X \leq 3) \] Calculating \( P(X < 4) \): \[ P(X < 4) = \frac{4 - 0}{5 - 0} = \frac{4}{5} = 0.8 \] Thus, \[ P(3 < X < 4) = 0.8 - 0.6 = 0.2 \] ### 4. Find the 75th percentile of wait times: The \( p \)-th percentile for a uniform distribution is given by: \[ P = a + p(b - a) \] For the 75th percentile (\( p = 0.75 \)): \[ P = 0 + 0.75(5 - 0) = 3.75 \text{ minutes} \] ### 5. Find the 80th percentile of wait times: For the 80th percentile (\( p = 0.80 \)): \[ P = 0 + 0.80(5 - 0) = 4.0 \text{ minutes} \] ### 6. Mean of these wait times: The mean \( \mu \) of a uniform distribution is given by: \[ \mu = \frac{a + b}{2} = \frac{0 + 5}{2} = 2.5 \text{ minutes} \] ### 7. Standard deviation of these wait times: The standard deviation \( \sigma \) of a uniform distribution is given by: \[ \sigma = \frac{b - a}{\sqrt{12}} = \frac{5 - 0}{\sqrt{12}} = \frac{5}{\sqrt{12}} \approx 1.443 \] Rounding to three decimal places, \( \sigma \approx 1.443 \). ### 8. 24% of people wait a maximum of how many minutes? To find the maximum wait time for 24%: \[ P = a + 0.24(b - a) = 0 + 0.24(5 - 0) = 1.2 \text{ minutes} \] ### 9. 24% of people wait a minimum of how many minutes? To find the minimum wait time for 24%, we can use the same percentile concept: \[ P = a + 0.24(b - a) = 0 + 0.24(5 - 0) = 1.2 \text{ minutes} \] ### Summary of Answers: 1. Probability of waiting at most 3 minutes: **0.6** 2. Probability of waiting more than 3 minutes: **0.4** 3. Probability of waiting between 3 and 4 minutes: **0.2** 4. 75th percentile of wait times: **3.75** minutes 5. 80th percentile of wait times: **4.0** minutes 6. Mean of wait times: **2.5** minutes 7. Standard deviation of wait times: **1.443** minutes 8. 24% of people wait a maximum of: **1.2** minutes 9. 24% of people wait a minimum of: **1.2** minutes