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The activity completion time of each of the 5 critical activities of a project follows normal distribution. If the mean activity completion time of each activity is 2 weeks, and the standard deviation of the activity completion time of each of the critical activity is 10, then what is the probability that the project would take between 9 and 13 week?

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Answer:

P(x = 9) and P(x = 13) is:

0.6551965

Step-by-step explanation:

By Z score,

[tex]Z = \frac{x - 2}{\frac{\sigma}{\sqrt{n}}}[/tex],

P(x = 9 ) = [tex]P(Z = \frac{9 -2 }{\frac{10}{\sqrt{5}}} )[/tex]

P(x =13) = [tex]P(Z = \frac{13 -2 }{\frac{10}{\sqrt{5}}} )[/tex]

And these give:

P(x = 9) ==>  P(Z = 0.7) = 0.7580363

P(x = 13) ==>  P(Z = 1.1) = 0.8643339

Therefore, the probability that the project would take between 9 and 13 week = P(x = 9) * P(x =13) =  0.6551965.

For replication, see the R codes below:

Z1 = (9 - 2)/(10/sqrt(length(5)))

a = pnorm(Z1)

Z2 = (13 - 2)/(10/sqrt(length(5)))

b = pnorm(Z2)

a*b

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