Question

Prior to placing an order, the amount of time (in minutes) that a driver waits in line at a Starbucks drive-thru follows an exponential distribution with a probability density function of f(x) = 0.2e−0.2x.

a. What is the mean waiting time that a driver faces prior to placing an order?
b. What is the probability that a driver spends more than the average time before placing an order? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
c. What is the probability that a driver spends more than 10 minutes before placing an order? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
d. What is the probability that a driver spends between 4 and 6 minutes before placing an order? (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

Answer 1

a. The mean waiting time is 5 minutes. b. The probability that a driver spends more than the average time is approximately 0.6321. c. The probability that a driver spends more than 10 minutes before placing an order is approximately 0.1813. d. The probability that a driver spends between 4 and 6 minutes before placing an order is approximately 0.1748.

Step-by-step explanation:

a. To find the mean waiting time, we need to calculate the expected value of the exponential distribution. The mean is given by the formula μ = 1/λ, where λ is the rate parameter of the exponential distribution. In this case, λ = 0.2, so the mean waiting time is μ = 1/0.2 = 5 minutes.

b. To find the probability that a driver spends more than the average time before placing an order, we need to calculate the cumulative distribution function (CDF) of the exponential distribution. The CDF is given by the formula F(x) = 1 - e^(-λx). Substituting λ = 0.2 and x = 5, we get F(5) = 1 - e^(-0.2*5) ≈ 0.6321.

c. To find the probability that a driver spends more than 10 minutes before placing an order, we can calculate 1 minus the CDF at 10 minutes. Using the formula F(x) = 1 - e^(-λx) and substituting λ = 0.2 and x = 10, we get F(10) = 1 - e^(-0.2*10) ≈ 0.1813.

d. To find the probability that a driver spends between 4 and 6 minutes before placing an order, we can subtract the CDF at 4 minutes from the CDF at 6 minutes. Using the formula F(x) = 1 - e^(-λx), we get F(6) - F(4) = (1 - e^(-0.2*6)) - (1 - e^(-0.2*4)) ≈ 0.1748.

Answer 2

The mean waiting time in a Starbucks drive-thru, following an exponential distribution, is 5 minutes. The probability of waiting more than this average is approximately 0.3679, and the probability of waiting more than 10 minutes is approximately 0.1353.

Step-by-step explanation:

The amount of time (in minutes) that a driver waits in line at a Starbucks drive-thru follows an exponential distribution with a probability density function of f(x) = 0.2e−0.2x.

• Mean waiting time = λ^{-1} = (0.2)^{-1} = 5 minutes.
• Probability that a driver spends more than the average time = 1 - P(X ≤ 5) = 1 - (1 - e^{-0.2 × 5}) = e^{-1} ≈ 0.3679.
• Probability that a driver spends more than 10 minutes = 1 - P(X ≤ 10) = 1 - (1 - e^{-0.2 × 10}) = e^{-2} ≈ 0.1353.
• Probability that a driver spends between 4 and 6 minutes = P(X ≤ 6) - P(X ≤ 4) = (1 - e^{-0.2 × 6}) - (1 - e^{-0.2 × 4}) = e^{-0.8} - e^{-1.2} ≈ 0.0996.

Exponential Distribution for Starbucks Drive-Thru

• Probability that it takes less than one minute for the next customer to arrive: P(X < 1) = 1 - e^{-0.2 × 1} ≈ 0.1813.
• Probability that it takes more than five minutes for the next customer to arrive: P(X > 5) = e^{-1} ≈ 0.3679.