173k views
2 votes
You have just used the network planning model and found the critical path length is 30 days and the variance of the critical path is 25 days. The probability that the project will be completed in 33 days or less is equal to ______. (2 decimal accuracy)

User Fred
by
7.9k points

2 Answers

6 votes

Answer:

0.73

Explanation:

A Z-score is a statistical measurement that defines the relationship between a value and the mean of a group, given the standard deviations from the mean. If a Z score is 0, it implies that that the value being measured is identical to the mean score. A Z-score of 2.0 indicate that the value being measured is two standard deviation away from the group mean.

Standard deviation = √variance = √25

=5 days

Z value that corresponding to the possibility of completing the project in 33 days or less is

Z=(33-30)/5 = 3/5 = 0.6

probability that the project will be completed in 33 days or less = P(Z ≤ 0.6)

from Z table

P(Z ≤ 0.6) = 0.72575 ≈ 0.73 (2 decimal accuracy)

You have just used the network planning model and found the critical path length is-example-1
User Fault
by
7.6k points
3 votes

Answer: The probability that the project will be completed in 33 days or less is equal to 0.73 .

Explanation:

We assume that this is a normally distribution.

Given : The critical path length is 30 days .

i.e. Population mean =
\mu=30

Variance :
\sigma^2=25

Then, standard deviation :
\sigma=√(25)=5

Z-score :
z=(x-\mu)/(\sigma)

The value of z corresponding to 33 =
z=(33-30)/(5)=0.6

Now, the p-value =
P(x\leq33)=P(z\leq0.6)=0.7257469\approx0.73

Hence, the probability that the project will be completed in 33 days or less is equal to 0.73.

User Gouda Elalfy
by
6.8k points