Service time, T, at an auto center has a Weibull distribution with parameters θ=3.7 and β=1.35. What is the probability that at least 8 of the next 10 customers will have to wait more than 4.69 hours? What is the expected number of customers that will have to wait more than 4.69 hours?
To solve this problem, we need to use the cumulative distribution function (CDF) of the Weibull distribution.
1. Probability that at least 8 of the next 10 customers will have to wait more than 4.69 hours:
Step 1: Calculate the CDF of the Weibull distribution at 4.69 hours:
CDF(4.69) = 1 - exp(-(4.69/θ)^β)
Step 2: Calculate the probability that a customer will have to wait more than 4.69 hours:
P(wait time > 4.69) = 1 - CDF(4.69)
Step 3: Calculate the probability that at least 8 out of 10 customers will have to wait more than 4.69 hours using the binomial distribution:
P(at least 8 out of 10 customers wait > 4.69) = sum(k=8 to 10) [C(10, k) * P(wait time > 4.69)^k * (1 - P(wait time > 4.69))^(10-k)]
Calculating P(wait time > 4.69) using the given parameters:
P(wait time > 4.69) = 1 - exp(-(4.69/3.7)^1.35)
P(wait time > 4.69) = 1 - exp(-1.372)
Now we can substitute this value in the formula:
P(at least 8 out of 10 customers wait > 4.69) = sum(k=8 to 10) [C(10, k) * (1 - exp(-1.372))^k * (1 - (1 - exp(-1.372)))^(10-k)]
2. Expected number of customers that will have to wait more than 4.69 hours:
Step 1: Calculate the expected value of the Weibull distribution, which is given by θ * Γ(1 + 1/β), where Γ is the gamma function.
Expected value = θ * Γ(1 + 1/β)
Using the given parameters:
Expected value = 3.7 * Γ(1 + 1/1.35)
Now you can substitute this value in the formula to calculate the expected number of customers:
Expected number of customers = 10 * P(wait time > 4.69) = 10 * (1 - exp(-1.372))
To find the probability that at least 8 of the next 10 customers will have to wait more than 4.69 hours, we can use the cumulative distribution function (CDF) of the Weibull distribution.
The CDF of the Weibull distribution is given by:
CDF(T) = 1 - exp(-((T/θ)^β))
In this case, θ = 3.7 and β = 1.35. We want to find the probability that the service time T is greater than 4.69 hours, so we need to calculate CDF(4.69).
CDF(4.69) = 1 - exp(-((4.69/3.7)^1.35))
Next, we need to calculate the probability that at least 8 of the next 10 customers will have to wait more than 4.69 hours. This can be done using the binomial distribution.
The probability of having exactly k successes in n trials is given by:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
In this case, n = 10 (the number of customers), p = CDF(4.69) (the probability that a customer waits more than 4.69 hours), and we want to find the probability of having at least 8 successes, so we need to calculate P(X >= 8).
P(X >= 8) = P(X = 8) + P(X = 9) + P(X = 10)
To find the expected number of customers that will have to wait more than 4.69 hours, we can use the expected value (mean) of the binomial distribution. The expected value of a binomial distribution is given by:
E(X) = n * p
In this case, n = 10 (the number of customers) and p = CDF(4.69) (the probability that a customer waits more than 4.69 hours), so we need to calculate E(X).
E(X) = 10 * p
Using the formulas above, you can calculate the probability that at least 8 of the next 10 customers will have to wait more than 4.69 hours and the expected number of customers that will have to wait more than 4.69 hours.