Answer:
a) The probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is 5.59%.
b) The probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is 12.71%
Step-by-step explanation:
Normally distributed problems can be solved by the z-score formula:
On a normaly distributed set with mean
and standard deviation
, the z-score of a value X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.
A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 157.4 seconds. This means that ![\mu = 157.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%20157.4)
Assuming drive-through times are normally distributed with a standard deviation of 33 seconds. This means that
.
(a) What is the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds?
This probability is the pvalue of the zscore of ![X = 105](https://tex.z-dn.net/?f=X%20%3D%20105)
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{105 - 157.4}{33}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B105%20-%20157.4%7D%7B33%7D)
![Z = -1.59](https://tex.z-dn.net/?f=Z%20%3D%20-1.59)
has a pvalue of .0559. This means that there is a 5.59% probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds,
(b) What is the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through?
This is 1 subtracted by the pvalue of the zscore of
.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{195 - 157.4}{33}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B195%20-%20157.4%7D%7B33%7D)
![Z = 1.14](https://tex.z-dn.net/?f=Z%20%3D%201.14)
has a pvalue of .87286. This means that there is a 87.286% probability that a randomly selected car will get through the restaurant's drive-through in less than 195 seconds.
So the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive through is
![P = 1 - .87286 = .1271](https://tex.z-dn.net/?f=P%20%3D%201%20-%20.87286%20%3D%20.1271)