a study found that the mean amount of time cars spent in drive 2 of 12 A study f

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a study found that the mean amount of time cars spent in drive

2 of 12 A study found that the mean amount of time cars spent in drnve-through Click here to view the Click here to view the standard normal distribution table (page 2) (a) What is the d normal distribution table (page 1) of a certain fast-food restaurant was 137.7 seconds Ass probability that a randomly selected car will get through the restaurant's drive-through in less than 94 seconds? The probability that a randomly selected car will get through the restaurants drive through in less than 94 seconds is (Round to four decimal places as needed) (b) What is the probability that a randomly selected car will spend more than 174 seconds in the restaurant's drve-through? The probability that a randomly selected car will spend more than 174 seconds in the restaurants drive-through is Round to four decimal places as needed) (c) What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through? The proportion of cars that spend between 2 and 3 minutes in the restaurants drive-through is Round to four decimal places as needed) (d) Would it be unusual for a car to spend more than 3 minutes in the restaurant's drive-through? Why? The probability that a car spends more than 3 minutes in the restaurant's drive-through is Round to four decimal places as needed) so it be unusual, since the p Enter your answer in each of the answer boxes.

Explanation / Answer

Let X be the random variable that the amount of time cars spent in drive throughs of a certain fast food restaurant.

Here X ~ N(mu = 137.7, sigma = 27)

a) Here we have to find P(X < 94).

Convert x = 94 into z-score.

z-score is defined as,

z = (x - mu) / sigma

z = (94 - 137.7) / 27 = -1.62

Now we have to find P(Z < -1.62)

This probability we can find in excel.

syntax :

=NORMSDIST(z)

where z is z-score.

P(Z < -1.62) = 0.0528

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b) Here we have to find P(X > 174).

z-score for x = 174 is,

z = (174 - 137.7) / 27 = 1.34

Now we have to find P(Z > 1.34)

This probability we can find in excel.

syntax :

=1 - NORMSDIST(z)

where z is z-score.

P(Z > 1.34) = 0.0894

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c) Here we have to find P(2 minute < X < 3 minute).

That is we have to find P(120 < X < 180)

Z-score for x = 120 and x = 180 are,

z = (120 - 137.7) / 27 = -0.66

z = (180 - 137.7) / 27 = 1.57

Now we have to find P(-0.66 < Z < 1.57)

P(-0.66 < Z < 1.57) = P(Z < 1.57) - P(Z < -0.66)

= 0.9414 - 0.2561

= 0.6853

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d) Now in this part we have to find P(X > 3 minute) or P(X > 180)

P(X > 1.57) = 0.0586

Here Probability > 0.05

The observation 1.57 is not unusual it is usual.

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