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: c. Find the proportion of cars that spend between 2 and 3 minutes in Wendy's drive-thru. [Hint: Think about the units for the mean and standard deviation.] a Answer:

To find the probability that a randomly selected car will get through Wendy's drive-thru in less than 100 seconds, we can use the Z-score formula: \( Z = \frac{(X - \mu)}{\sigma} \) Where: - \( X = 100 \) seconds (the value we're interested in) - \( \mu = 138.5 \) seconds (the mean) - \( \sigma = 29 \) seconds (the standard deviation) Calculating the Z-score: \( Z = \frac{(100 - 138.5)}{29} = \frac{-38.5}{29} \approx -1.327 \) Now, we look up this Z-score in the standard normal distribution table or use a calculator to find the probability. The probability of a Z-score of -1.327 is approximately 0.091. So, the probability that a randomly selected car will get through Wendy's drive-thru in less than 100 seconds is about **9.1%**. --- For part b, we're finding the probability that a randomly selected car will spend more than 160 seconds in the drive-thru. Using the Z-score formula again: \( Z = \frac{(X - \mu)}{\sigma} \) Where: - \( X = 160 \) seconds (the value of interest) Calculating the Z-score: \( Z = \frac{(160 - 138.5)}{29} = \frac{21.5}{29} \approx 0.741 \) Now, we can find the probability associated with a Z-score of 0.741. The probability of a Z-score of 0.741 is approximately 0.770 (using a standard normal distribution table). Since we want the probability of spending more than 160 seconds, we subtract this from 1: \( P(X > 160) = 1 - P(Z < 0.741) \approx 1 - 0.770 = 0.230 \). So, the probability that a randomly selected car will spend more than 160 seconds in Wendy's drive-thru is about **23%**. --- For part c, we want to find the proportion of cars that spend between 2 and 3 minutes in the drive-thru. Since 2 and 3 minutes equal 120 and 180 seconds, respectively, we will also calculate Z-scores for these values. For 120 seconds: \( Z = \frac{(120 - 138.5)}{29} \approx -0.636 \) For 180 seconds: \( Z = \frac{(180 - 138.5)}{29} \approx 1.422 \) Now we can find the probabilities for both Z-scores. Using a standard normal distribution table: - The probability for \( Z = -0.636 \) is approximately 0.263. - The probability for \( Z = 1.422 \) is approximately 0.922. To find the proportion between these two values, we subtract the lower probability from the upper probability: \( P(120 < X < 180) = P(Z < 1.422) - P(Z < -0.636) \approx 0.922 - 0.263 = 0.659 \). Thus, the proportion of cars that spend between 2 and 3 minutes in Wendy's drive-thru is about **65.9%**.