40. Wendy's Drive-Through Fast-food restaurants spend quite abit of time studying the amount of time cars spend in their drivethroughs. Certainly, the faster the cars get through the drive-through, the more the opportunity for making money. QSRMagazine studied drive-through times for fast-food restaurantsand found Wendy's had the best time, with a mean time spentin the drive-through of 138.5 seconds. Assuming drive-throughtimes are normally distributed with a standard deviation of29 seconds, answer the following.(a) What is the probability that a randomly sclected car will getthrough Wendy's drive-through in less than 100 scconds?(b) What is the probability that a randomly sclected car will spendmore than 160 seconds in Wendy's drive-through?(e) What proportion of cars spend between 2 and 3 minutes inWendy's drive-through?(d) Would it be unusual for a car to spend more than 3 minutesin Wendy's drive-through? Why?
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40. Wendy's Drive-Through Fast-food restaurants spend quite a bit of time studying the amount of time cars spend in their drive throughs. Certainly, the faster the cars get through the drive- through, the more the opportunity for making money. QSR Magazine studied drive-through times for fast-food restaurants and found Wendy's had the best time, with a mean time spent in the drive-through of 138.5 seconds. Assuming drive-through times are normally distributed with a standard deviation of 29 seconds, answer the following. (a) What is the probability that a randomly sclected car will get through Wendy's drive-through in less than 100 scconds? (b) What is the probability that a randomly sclected car will spend more than 160 seconds in Wendy's drive-through? (e) What proportion of cars spend between 2 and 3 minutes in Wendy's drive-through? (d) Would it be unusual for a car to spend more than 3 minutes in Wendy's drive-through? Why?
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It is given that the mean and standard deviation are 138.5 and 29, respectively.
a)
The probability that a randomly selected car will get through Wendy’s drive-through in less than 100 seconds is obtained as follows:
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x -138.5 100-138.5 P(x<100) P 29 29 =P(z-1.33) Using standard normal table -0.0918
b)
The probability that a randomly selected car will spend more than 160 ...
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x-138.5 160 -138.5 P(x160) P 29 29 -P(z>0.74) 1-P(z0.74) =1-0.7704 Using standard normal table) 0.2296
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