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

$P = 1 - .87286 = .1271$

Answer 2

The question is an illustration of a normal distribution.

• The probability that a randomly selected car will get through the restaurant's drive-through in less than 105 seconds is 0.0561
• The probability that a randomly selected car will spend more than 195 seconds in the restaurant's drive-through is 0.1274

The given parameters are:

$\mu = 157.4$

$\sigma= 33$

(a) Probability that a selected car gets through in less than 105 seconds

This means that: x = 105

First, we calculate the z-score

$z = \frac{x - \mu}{\sigma}$

This gives

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

$z = \frac{-52.4}{33}$

$z = -1.588$

So, the required probability is:

$P(x < 105) = P(z < -1.588)$

Using a z probability table, we have:

$P(x < 105) = 0.0561$

(b) Probability that a selected car spends more than 195 seconds

This means that: x = 195

First, we calculate the z-score

$z = \frac{x - \mu}{\sigma}$

This gives

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

$z = \frac{37.6}{33}$

$z = 1.139$

So, the required probability is:

$P(x >195) = P(z > 1.139)$

Using a z probability table, we have:

$P(x > 195) =0.1274$

Read more about probabilities of normal distributions at:

brainly.com/question/6476990