Mathematics, 27.09.2019 02:00 milkshakegrande101

Astudy 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)

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Answer from: paige1616

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 0.0561The 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:

link

Answer from: Metlife

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 from: Quest

108

Answer from: Quest

She will need 5/8 bags for each square yard.

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