Normal text Problem 18 Question: Fast-food restaurants spend quite a bit of time studying the amount of time cars spend in their drive-thrus. Certainly, the faster the cars get through the drive-through, the more the opportunity for making money. In 2007 , QSR Magazine studied drive-thru times for fast-food restaurants and Wendy's had the best time, with a mean time spent in the drive-thru of 138.5 seconds. Assume drive-thru times are normally distributed with a standard deviation of 29 seconds. a. Find the probability that a randomly selected car will get through Wendy's drive-thru in less than 100 seconds. o Answer: b. Find the probability that a randomly selected car will spend more than 160 seconds in Wendy's drive-thru. o Answer:

To find the probabilities for both scenarios, we can use the z-score formula, which helps in determining how many standard deviations an element is from the mean. a. For the first part, we calculate the z-score for 100 seconds: \[ z = \frac{(X - \mu)}{\sigma} \] \[ z = \frac{(100 - 138.5)}{29} \approx -1.32 \] Using the z-table, the probability of z being less than -1.32 is approximately 0.0918. Thus, the probability that a randomly selected car will get through Wendy's drive-thru in less than 100 seconds is about **9.18%**. b. For the second part, we calculate the z-score for 160 seconds: \[ z = \frac{(160 - 138.5)}{29} \approx 7.34 \] Since 7.34 is highly positive, the probability of a randomly selected car spending more than 160 seconds in the drive-thru is extremely small. In fact, the probability that a randomly selected car will spend more than 160 seconds is approximately **0%** (often interpreted as just about 0.0000).