A project is found to have expected time T = 35.33 days and variance V = 3.22.
a. What value of z is needed to find the probability that the project will take at least 40 days? (round final answer to 2 decimals as needed)
b. What value of z is needed to find the probability that the project will take at most 40 days? (round final answer to 2 decimals as needed)
c. What value of z is needed to find the probability that the project will take at most 30 days? (round final answer to 2 decimals as needed)

For this case we know that the mean for the random variable of interest is [tex]\mu = 35.33[/tex] and the variance [tex]\sigma^2 = 3.22[/tex] so then the deviation would be [tex]\sigma = \sqrt{3.22}= 1.794[/tex]

The z score is given by thsi formula:

[tex] z = \frac{X -\mu}{\sigma}[/tex]

Part a

We want this probability:

[tex] P(X>40)[/tex]

And if we find the z score we got:

[tex] z = \frac{40-35.33}{1.794}= 2.60[/tex]

And we can find this probability: [tex] P(Z>2.60)[/tex]

Part b

We want this probability:

[tex] P(X<40)[/tex]

And if we find the z score we got:

[tex] z = \frac{40-35.33}{1.794}= 2.60[/tex]

And we can find this probability: [tex] P(Z<2.60)[/tex]

Part c

We want this probability:

[tex] P(X<30)[/tex]

And if we find the z score we got:

[tex] z = \frac{30-35.33}{1.794}= -2.97[/tex]

And we can find this probability: [tex] P(Z<-2.97)[/tex]