10:21PM Wed Feb 26 Uniform Distributions (Statistics 3) 4. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Find \( P(x>19)(1 \) Point) \( -0.2 \) 0.3 1 0.2 More 5. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Find \( P(x>19 \mid x>15)(1 \) Point) 33 0.3 1 0.17

To solve the problems related to the uniform distribution, we need to understand the properties of a uniform distribution. The uniform distribution is defined by two parameters: the minimum value \( a \) and the maximum value \( b \). In this case, the time to change the oil is uniformly distributed between 11 and 21 minutes, so we have: - \( a = 11 \) - \( b = 21 \) ### Problem 4: Find \( P(x > 19) \) 1. **Calculate the total range of the distribution**: \[ \text{Range} = b - a = 21 - 11 = 10 \] 2. **Calculate the range for \( x > 19 \)**: - The values greater than 19 are between 19 and 21. - The range for \( x > 19 \) is: \[ \text{Range for } x > 19 = 21 - 19 = 2 \] 3. **Calculate the probability**: \[ P(x > 19) = \frac{\text{Range for } x > 19}{\text{Total Range}} = \frac{2}{10} = 0.2 \] Thus, the answer for Problem 4 is \( 0.2 \). ### Problem 5: Find \( P(x > 19 \mid x > 15) \) 1. **Calculate the range for \( x > 15 \)**: - The values greater than 15 are between 15 and 21. - The range for \( x > 15 \) is: \[ \text{Range for } x > 15 = 21 - 15 = 6 \] 2. **Calculate the range for \( x > 19 \) within \( x > 15 \)**: - As calculated earlier, the range for \( x > 19 \) is 2. 3. **Calculate the conditional probability**: \[ P(x > 19 \mid x > 15) = \frac{P(x > 19)}{P(x > 15)} = \frac{2/10}{6/10} = \frac{2}{6} = \frac{1}{3} \approx 0.33 \] Thus, the answer for Problem 5 is approximately \( 0.33 \), which corresponds to \( 0.3 \) when rounded. ### Summary of Answers: - Problem 4: \( 0.2 \) - Problem 5: \( 0.3 \)