Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was

140.9140.9

seconds. Assuming​ drive-through times are normally distributed with a standard deviation of

2424

​seconds, complete parts​ (a) through​ (d) below.Click here to view the standard normal distribution table (page 1).

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Click here to view the standard normal distribution table (page 2).

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​(a) What is the probability that a randomly selected car will get through the​ restaurant's drive-through in less than

105105

​seconds?The probability that a randomly selected car will get through the​ restaurant's drive-through in less than

105105

seconds is

nothing.

​(Round to four decimal places as​ needed.)

​(b) What is the probability that a randomly selected car will spend more than

184184

seconds in the​ restaurant's drive-through?The probability that a randomly selected car will spend more than

184184

seconds in the​ restaurant's drive-through is

nothing.

​(Round to four decimal places as​ needed.)

​(c) What proportion of cars spend between

22

and

33

minutes in the​ restaurant's drive-through?The proportion of cars that spend between

22

and

33

minutes in the​ restaurant's drive-through is

nothing.

​(Round to four decimal places as​ needed.)

Answer 1

Answer A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 140.9 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of 24​seconds.

Solution:

μ = 140.9

σ = 24

​(a) The probability that a randomly selected car will get through the​ restaurant's drive-through in less than 105 ​ seconds:

P(X < 105) = P(X - μ)/σ < (105-140.9)/24

= P(Z < -1.50)

= 0.0668 (from standard normal distribution table)

The probability that a randomly selected car will get through the​ restaurant's drive-through in less than 105 seconds is 0.0668

​(b) What is the probability that a randomly selected car will spend more than 184 seconds in the​ restaurant's drive-through:

P(X > 184) = P(Z > (184 - 140.9)/24)

= P(Z > 1.80)

= 0.0359 (from standard normal distribution table)

The probability that a randomly selected car will spend more than 184 seconds in the​ restaurant's drive-through is 0.0359

​(c) What proportion of cars spend between 2 and 3 minutes in the​ restaurant's drive-through:

2 minutes = 120 seconds

3 minutes = 180 seconds

P(2 < X <3) = P((120-140.9)/24 < Z <

(180-140.9)/24)

= P(- 0.87 < Z < 1.63)

= P(Z>1.63) - P(Z< -0.87)

(from standard normal distribution table)

= 0.9484 - 0.1922

= 0.7562

The proportion of cars that spend between 2 and 3 minutes in the​ restaurant's drive-through is 0.7562.