Collina’s Italian Café in Houston, Texas, advertises that carryout orders take about 25 minutes (Collina’s website, February 27, 2008). Assume that the time required for a carryout order to be ready for customer pickup has an exponential distribution with a mean of 25 minutes.
What is the probability than a carryout order will be ready within 20 minutes (to 4 decimals)?
If a customer arrives 30 minutes after placing an order, what is the probability that the order will not be ready (to 4 decimals)?
A particular customer lives 15 minutes from Collina’s Italian Café. If the customer places a telephone order at 5:20 P.M., what is the probability that the customer can drive to the café, pick up the order, and return home by 6:00 P.M. (to 4 decimals)?
a.) .5507
b.).3012
c.) .6321
To find the probability that a carryout order will be ready within 20 minutes, we need to calculate the cumulative distribution function (CDF) of the exponential distribution. The formula for the CDF of an exponential distribution is:
CDF(x) = 1 - e^(-λx)
Where λ is the rate parameter, which is the reciprocal of the mean (in this case, λ = 1/25).
So, to calculate the probability that a carryout order will be ready within 20 minutes (P(X ≤ 20)), we substitute x = 20 into the CDF formula:
CDF(20) = 1 - e^(-λ * 20)
= 1 - e^(-20/25)
≈ 0.4465
Therefore, the probability that a carryout order will be ready within 20 minutes is approximately 0.4465.
Now, let's calculate the probability that a carryout order will not be ready after 30 minutes, given that the mean is 25 minutes. This can be calculated using the complement of the CDF:
P(X > 30) = 1 - CDF(30)
= 1 - (1 - e^(-λ * 30))
= e^(-λ * 30)
= e^(-30/25)
≈ 0.2231
Therefore, the probability that a carryout order will not be ready after 30 minutes is approximately 0.2231.
To calculate the probability that a customer can drive to the café, pick up the order, and return home within 40 minutes, we can use the same formula for the CDF. However, we need to adjust the value of x to account for the total time available (40 minutes) minus the time it takes for the customer to drive to the café and back home (15 minutes each way). So, x = 40 - 2(15) = 10 minutes.
P(X ≤ 10) = 1 - e^(-λ * 10)
= 1 - e^(-10/25)
≈ 0.3297
Therefore, the probability that the customer can drive to the café, pick up the order, and return home within 40 minutes is approximately 0.3297.