A particular project network has two paths through it: Path A and Path B. Path A has an expected completion time of 15 weeks and a variance in completion time of 8 weeks, while Path B has an expected completion time of 16 weeks and a variance in completion time of 4 weeks. What is the probability that this project is going to take more than 18 weeks?

Probability that this project is going to take more than 18 weeks = 0.99991

Step-by-step explanation:

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

For this distribution, the total time the project will take = A + B

A ~ (15, 8)

B ~ (16, 4)

Combined mean = μ₁ + μ₂ = 15 + 16 = 31

Combined variance = 1²σ₁² + 1²σ₂² = 8 + 4 = 12

Combined Standard Deviation = √(12) = 3.464 weeks

So, with the right assumption that this combined distribution is a normal distribution

Probability that this project is going to take more than 18 weeks

P(x > 18)

We first normalize/standardize 18

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (18 - 31)/3.464 = - 3.75

The required probability

P(x > 18) = P(z > -3.75)

We'll use data from the normal probability table for these probabilities

P(x > 18) = P(z > -3.75) = 1 - P(x ≤ -3.75)

= 1 - 0.00009 = 0.99991

Hope this Helps!!!

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values

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