A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 151.3 seconds. Assuming drive-through times are normally distributed with a standard deviation of 27 seconds, complete parts (a) through (d) below
(b) What is the probability that a randomly selected car will spend more than 198 seconds in the restaurant's drive-through?
The probability that a randomly selected car will spend more than 198 seconds in the restaurant's drive-through is 0418
(Round to four decimal places as needed.)
(c) What proportion of cars spend between 2 and 3 minutes in the restaurant's dirvethrough?
The proportion of cars that spend between 2 and 3 minutes in the restaurant's drivethrough is 7329
(Round to four decimal places as needed)
(d) Would it be unusual for a càr to spend more than 3 minutes in the restaurant's drivethrough? Why?
The probability that a car spends more than 3 minutes in the restaurant's drive-through is so it be unusual, since the probability is than 0.05 (Round to four decimal places as needed)
Answer
The probability that a car spends more than 3 minutes in the restaurant's drive-through is approximately 0.1439. Since this probability is greater than 0.05, it would not be unusual for a car to spend more than 3 minutes in the restaurant's drive-through. This is represented as \(\boxed{0.1439}\).
Step 1 ：The problem is asking for probabilities related to a normal distribution. The mean (mu) is 151.3 seconds and the standard deviation (sigma) is 27 seconds.
Step 2 ：For part (b), we need to find the probability that a randomly selected car will spend more than 198 seconds in the restaurant's drive-through. This is equivalent to finding the area under the normal distribution curve to the right of 198 seconds.
Step 3 ：For part (c), we need to find the proportion of cars that spend between 2 and 3 minutes (or 120 and 180 seconds) in the restaurant's drive-through. This is equivalent to finding the area under the normal distribution curve between 120 and 180 seconds.
Step 4 ：For part (d), we need to find the probability that a car spends more than 3 minutes (or 180 seconds) in the restaurant's drive-through. This is equivalent to finding the area under the normal distribution curve to the right of 180 seconds.
Step 5 ：The probability that a randomly selected car will spend more than 198 seconds in the restaurant's drive-through is approximately 0.0418. This is represented as \(\boxed{0.0418}\).
Step 6 ：The proportion of cars that spend between 2 and 3 minutes in the restaurant's drive-through is approximately 0.7329. This is represented as \(\boxed{0.7329}\).
Step 7 ：The probability that a car spends more than 3 minutes in the restaurant's drive-through is approximately 0.1439. Since this probability is greater than 0.05, it would not be unusual for a car to spend more than 3 minutes in the restaurant's drive-through. This is represented as \(\boxed{0.1439}\).
