Question

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturday nights. Wendy's begins an advertising blitz that touts its late-night drive-through service and convenience. After one week of advertising, the Wendy's marketing department conducts a survey at 200 Wendy's to count the number of cars, X, that arrive at the drive-through between the hours of 12:00 midnight and 12:30 AM on Saturday night.

Question 1. Use the Poisson distribution to calculate the probability that exactly 8 cars will use the drive-through between 12:00 midnight and 12:30 AM on a Saturday night at Wendy's. Do the same for exactly 9 cars.

P(X = 8) (use 4 decimal places)
P(X = 9) (use 4 decimal places)

Question 2. At how many of the 200 restaurants in the survey would you expect exactly 8 customers to use the drive-through? exactly 9 customers?

expected number of 200 restaurants in which exactly 8 customers use the drive-through.
expected number of 200 restaurants in which exactly 9 customers use the drive-through.

Answer 1

1)
`P(X = 8) = 0.1033`

`P(X = 9) = 0.0688`

2) Expected number of 200 restaurants in which exactly 8 customers use the drive-through: 20.66

Expected number of 200 restaurants in which exactly 9 customers use the drive-through: 13.76

Explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

`e = 2.71828` is the Euler number

`\mu` is the mean in the given time interval.

Question 1. Use the Poisson distribution to calculate the probability that exactly 8 cars will use the drive-through between 12:00 midnight and 12:30 AM on a Saturday night at Wendy's. Do the same for exactly 9 cars.

Cars arrive at the Wendy's drive-through at a rate of 1 car every 5 minutes between the hours of 11:00 PM and 1:00 AM. on Saturday nights. This means that during 30 minutes, 6 cars expected to arrive. So
`\mu = 6`.

P(X = 8)

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 8) = (e^(-6)*(6)^(8))/((8)!) = 0.1033`

P(X = 9)

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 9) = (e^(-6)*(6)^(9))/((9)!) = 0.0688`

Question 2. At how many of the 200 restaurants in the survey would you expect exactly 8 customers to use the drive-through? exactly 9 customers?

There is a 10.33 probability that 8 customers would use the drive through for each restaurant.

So of 200, the expected number is

`E(X) = 200*0.1033 = 20.66`

There is a 6.88 probability that 9 customers would use the drive through for each restaurant.

So of 200, the expected number is

`E(X) = 200*0.0688 = 13.76`