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.
a. What is the probability than a carryout order will be ready within 20 minutes (to 4 decimals)?
b. If a customer arrives 30 minutes after placing an order, what is the probability that the order will not be ready (to 4 decimals)?
c. 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. Well, if the mean time for a carryout order to be ready is 25 minutes, we can assume that the parameter λ for the exponential distribution is 1/25 (since the mean is equal to 1/λ). By plugging in the values into the exponential distribution formula, we can find the probability that a carryout order will be ready within 20 minutes:
P(X < 20) = 1 - e^(-λx) = 1 - e^(-1/25 * 20) ≈ 0.2642
So, the probability that a carryout order will be ready within 20 minutes is approximately 0.2642.
b. If a customer arrives 30 minutes after placing an order, we can find the probability that the order will not be ready using the complementary event. The probability that it will not be ready is equal to 1 minus the probability that it will be ready within 30 minutes:
P(X > 30) = 1 - P(X < 30) = 1 - (1 - e^(-1/25 * 30)) ≈ 0.1922
So, the probability that the order will not be ready when the customer arrives is approximately 0.1922.
c. To calculate the probability that the customer can pick up the order and return home within 40 minutes (from 5:20 P.M. to 6:00 P.M.), we need to find:
P(X < 40)
Using the exponential distribution formula, this becomes:
P(X < 40) = 1 - e^(-1/25 * 40) ≈ 0.3297
So, the probability that the customer can drive to the café, pick up the order, and return home by 6:00 P.M. is approximately 0.3297.
To solve these questions, we can use the properties of the exponential distribution. The exponential distribution is characterized by a parameter called the rate parameter (λ), which is equal to the reciprocal of the mean (1/mean).
In this case, the mean is given as 25 minutes. Therefore, the rate parameter (λ) is 1/25.
a. 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 at 20 minutes. The CDF of the exponential distribution is given by the equation:
CDF(x) = 1 - exp(-λx)
Substituting the values, we have:
CDF(20) = 1 - exp(-(1/25) * 20)
Calculating this expression, we find:
CDF(20) ≈ 0.4866
So, the probability that a carryout order will be ready within 20 minutes is approximately 0.4866.
b. To find the probability that the order will not be ready after 30 minutes, we need to calculate the survival function (1 - CDF) at 30 minutes. The survival function is given by:
Survival(x) = 1 - CDF(x)
Substituting the values, we have:
Survival(30) = 1 - CDF(30)
= 1 - (1 - exp(-(1/25) * 30))
Calculating this expression, we find:
Survival(30) ≈ 0.2231
So, the probability that the order will not be ready after 30 minutes is approximately 0.2231.
c. To find the probability that the customer can drive to the café, pick up the order, and return home within 40 minutes, we need to calculate the CDF at 40 minutes. The CDF is given by:
CDF(x) = 1 - exp(-λx)
Substituting the values, we have:
CDF(40) = 1 - exp(-(1/25) * 40)
Calculating this expression, we find:
CDF(40) ≈ 0.6321
So, the probability that the customer can complete the entire process within 40 minutes is approximately 0.6321.
To calculate the probabilities in this scenario, we'll use the exponential distribution formula:
f(x) = λ * e^(-λx)
where:
- f(x) is the probability density function for the exponential distribution
- λ (lambda) is the rate parameter, which is the reciprocal of the mean (λ = 1/mean)
- e is the base of the natural logarithm (approximately 2.71828)
- x is the time variable
For this problem, we're given that the mean of the exponential distribution is 25 minutes, so we can calculate λ as 1/25.
a. To find the probability that a carryout order will be ready within 20 minutes, we need to calculate the cumulative distribution function (CDF) at x = 20. The CDF represents the probability that a random variable is less than or equal to a certain value.
P(X ≤ 20) = 1 - e^(-λx)
Plugging in the values:
P(X ≤ 20) = 1 - e^(-(1/25) * 20)
P(X ≤ 20) ≈ 0.4866
So, the probability that a carryout order will be ready within 20 minutes is approximately 0.4866.
b. To find the probability that the order will not be ready after 30 minutes, we need to calculate the complementary cumulative distribution function (CCDF) at x = 30. The CCDF represents the probability that a random variable is greater than a certain value.
P(X > 30) = e^(-λx)
Plugging in the values:
P(X > 30) = e^(-(1/25) * 30)
P(X > 30) ≈ 0.2231
However, we're looking for the probability that the order will not be ready, which is the complement of the probability that it will be ready.
P(Order not ready after 30 minutes) ≈ 1 - P(X ≤ 30) = 1 - (1 - e^(-(1/25) * 30)) ≈ 0.2231
So, the probability that the order will not be ready after 30 minutes is approximately 0.2231.
c. To find the probability that the customer can drive to the café, pick up the order, and return home within 40 minutes (since the customer lives 15 minutes away), we need to calculate the CDF at x = 40.
P(X ≤ 40) = 1 - e^(-(1/25) * 40)
Plugging in the values:
P(X ≤ 40) = 1 - e^(-(1/25) * 40)
P(X ≤ 40) ≈ 0.6321
So, the probability that the customer can complete the entire process within 40 minutes is approximately 0.6321.
A.) .55067
B.) .30119
C.).3297