Question

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)

Answer 1

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

Step-by-step 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 : $\sigma=\sqrt{25}=5$

Z-score :  $z=\dfrac{x-\mu}{\sigma}$

The value of z corresponding to 33 = $z=\dfrac{33-30}{5}=0.6$

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.

Answer 2

Answer:

0.73

Step-by-step 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)