Question

A study found that the mean amount of times cars spent in drive through a of a certain fast-food restaurant was 136.1 seconds. Assuming drive-through times are normally distributed with a standard deviation of 33 seconds, complete parts a through d below

A. What is the probability that a randomly selected car will get through the restaurant s drive thru in less than 82 seconds?

Answer 1

The probability that a randomly selected car will get through the drive-thru in less than 82 seconds is approximately 5.05%, as determined by calculating the Z-score and using the standard normal distribution.

Step-by-step explanation:

The question asks for the probability that a randomly selected car will get through a restaurant's drive-thru in less than 82 seconds when the mean time is 136.1 seconds and the standard deviation is 33 seconds. Since the drive-through times are normally distributed, we can use the Z-score formula to find this probability. The Z-score is calculated by subtracting the mean from the observed value and dividing by the standard deviation:

• Z = (X - μ) / σ

Substituting into the formula gives:

• Z = (82 - 136.1) / 33 = -54.1 / 33 ≈ -1.64

Using a standard normal distribution table or a calculator with a normal distribution function, we can find the probability that corresponds with a Z-score of -1.64. This will give us the probability that a car will get through the drive-thru in less than 82 seconds.

To find the exact probability, we look up the cumulative area to the left of the Z-score -1.64, which yields approximately 0.0505. Therefore, there is a 5.05% chance that a randomly selected car will get through the drive-thru in less than 82 seconds.