Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 157.4 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of 33 ​seconds, complete parts​ below:
(a) What is the probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds? The probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is (Round to four decimal places as needed.)
(b) What is the probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through? The probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is (Round to four decimal places as needed)

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 157.4$

Assuming​ drive-through times are normally distributed with a standard deviation of 33 ​seconds. This means that $\sigma = 33$.

(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 $X = 105$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{105 - 157.4}{33}$

$Z = -1.59$

$Z = -1.59$ 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 $X = 195$.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{195 - 157.4}{33}$

$Z = 1.14$

$Z = 1.14$ 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

$P = 1 - .87286 = .1271$