Given a simple Project like the one pictured above. Optimistic / Most Likely / P

Question

Given a simple Project like the one pictured above. Optimistic / Most Likely / Pessimist Times were assessed and Path Means and Path Standard Deviations were calculated. For Path 1-2-5 we have a Path Mean = 33 and a Path Standard Deviation = 4. For Path 1-3-4-5 we have a Path Mean = 34 and a Path Standard Deviation = 3. Determine both the Probability that Path 1-2-5 will finish in 34 Days or Less and the Probability that the Project will finish in 34 Days or Less.

Question 5 options:

0.84134474 : 0.707860972?

0.77337272 : 0.578101934?

0.598706274 : 0.299353137?

0.987775567 : 0.983991966?

0.84134474 : 0.707860972?

0.77337272 : 0.578101934?

0.598706274 : 0.299353137?

0.987775567 : 0.983991966?

Explanation / Answer

In above case diagram is not visible .

I am just predicting from the given information.

For

1 - 2 - 5

Mean = 33

SD = 4

Probability that Path 1-2-5 will finish in 34 Days or Less

Z= T- Te / Sigma

= 34 - 33/4

=0.25

The value of 0.25 from Normal table is 0.098

Z = 0.5 + 0.098

=0.5987

-------------------------------------------------------------------------------------------------------------------------------------------------------------------

For Path 1-3-4-5

1-3-4-5

Mean = 34

SD = 3

Z= T- Te / Sigma

= 33 - 34/3

= - 0.33 Means 0.33

The value of Z for 0.33 is 0.20

Z = 0.5 - 0.200

=0.299353

So answer

0.598706274 : 0.299353137

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