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

In a uniform distribution, every outcome between the minimum and maximum values is equally likely. In this case, the time to change the oil ranges from 11 to 21 minutes. To calculate \( P(x > 19) \), you would look at the range of values greater than 19, which is 19 to 21 (a span of 2 minutes) over the total span of 10 minutes (11 to 21). Therefore, \( P(x > 19) = \frac{2}{10} = 0.2 \). For \( P(x > 19 \mid x > 15) \), we evaluate the probability of changing the oil in more than 19 minutes given that it takes more than 15 minutes. The new range starts at 15 and ends at 21. Of this range, the portion that satisfies \( x > 19 \) is from 19 to 21 (2 minutes). The total length of the new range (15 to 21) is 6 minutes. Thus, \( P(x > 19 \mid x > 15) = \frac{2}{6} = \frac{1}{3} \approx 0.33 \).