Question 1169639: A construction project is at present 30 days away from the scheduled completion date. Depending on the weather condition next month ,the time required for the remaining construction are as follows :
If weather is good, the time required distribution would be: normal, mean=25 ,Standard deviation = 4
If weather is Bad, the time required distribution would be: Log-Normal mean=30,Standard deviation =6
The weather next month would be equally likely to be either good or bad. What is the probability that there will be a delay in the completion of the project
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**Understanding the Problem**
We need to calculate the probability that the construction project will be delayed beyond the scheduled completion date, given two possible weather scenarios and their associated time requirements.
**Given Information**
* Current time until scheduled completion: 30 days
* Weather scenarios:
* Good weather: Normal distribution, mean (μ) = 25 days, standard deviation (σ) = 4 days
* Bad weather: Log-normal distribution, mean (μ) = 30 days, standard deviation (σ) = 6 days
* Weather likelihood: Equally likely (50% good, 50% bad)
**Solution**
1. **Good Weather Scenario:**
* We want to find the probability that the time required is greater than 30 days (delay).
* Since the distribution is normal, we need to calculate the z-score:
* z = (x - μ) / σ = (30 - 25) / 4 = 5 / 4 = 1.25
* We need to find P(Z > 1.25) using a standard normal distribution table or calculator.
* P(Z > 1.25) = 1 - P(Z ≤ 1.25) ≈ 1 - 0.8944 = 0.1056
2. **Bad Weather Scenario:**
* We want to find the probability that the time required is greater than 30 days.
* Since the distribution is log-normal, we need to convert the mean and standard deviation to the parameters of the underlying normal distribution.
* Let Y = ln(X), where X follows a log-normal distribution.
* E(X) = 30, SD(X) = 6.
* Var(X) = 36
* We use the following formulas to find the mean (μy) and standard deviation (σy) of Y:
* σy^2 = ln(1 + (Var(X) / E(X)^2)) = ln(1 + 36 / 900) = ln(1 + 0.04) = ln(1.04) ≈ 0.03922
* σy = √0.03922 ≈ 0.198
* μy = ln(E(X)) - 0.5 * σy^2 = ln(30) - 0.5 * 0.03922 ≈ 3.4012 - 0.01961 ≈ 3.3816
* Now, we need to find P(X > 30), which is equivalent to P(ln(X) > ln(30)) = P(Y > ln(30)).
* ln(30) ≈ 3.4012
* We calculate the z-score:
* z = (ln(30) - μy) / σy = (3.4012 - 3.3816) / 0.198 ≈ 0.0196 / 0.198 ≈ 0.09899
* We need to find P(Z > 0.099) using a standard normal distribution table or calculator.
* P(Z > 0.099) = 1 - P(Z ≤ 0.099) ≈ 1 - 0.5393 = 0.4607
3. **Combined Probability:**
* Since the weather is equally likely, we average the probabilities of delay for each scenario:
* P(delay) = (0.5 * 0.1056) + (0.5 * 0.4607) = 0.0528 + 0.23035 = 0.28315
**Final Answer**
The probability that there will be a delay in the completion of the project is approximately 0.2832.
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