Recently, you were assigned to manage a project for your company. You have const
ID: 458341 • Letter: R
Question
Recently, you were assigned to manage a project for your company. You have constructed a network diagram depicting the various activities in the project (shown below to the right). In addition you have asked your team to estimate the amount of time that they would expect each of the activities to take. Their responses are shown in the following table: What is the expected completion time of the project? The expected completion time is 20 days. (Enter your response as an integer.) What is the probability of completing the project in 22 days? Refer to the standard normal table. The probability of completing the project in 22 days is_8735. (Enter your response rounded to four decimal places.) What is the probability of completing the project in 18 days? The probability of completing the project in 18 days is. (Enter your response rounded to four decimal places.)Explanation / Answer
Time estimated in days
Activity
Optimistic
Most Likely
Pessimistic
Expected time
Variance
A
6
7
12
8.33
1.00
1.00
B
4
8
12
8.00
1.33
1.78
C
4
6
8
6.00
0.67
0.44
D
1
5
6
4.00
0.83
0.69
E
4
8
11
7.67
1.17
1.36
Expected completion time
20.00
For Path A-D-E
Standard Deviation for the path =SQRT (1+.69+1.36)/6 =
1.75
Days
Values of Z
Probabilities
22
Z
1.14
0.87286
18
Z
-1.14
0.12714
For 22 days Z = ( 22- 20)/1.75 = 1.14
For 18 day Z = (18 – 20)/1.75 = -1.14
By referring standard normal table for the above Z values we can get probabilities mentioned above.
Therefore probability of completing the project in 18 days is 0.12714
Time estimated in days
Activity
Optimistic
Most Likely
Pessimistic
Expected time
Variance
A
6
7
12
8.33
1.00
1.00
B
4
8
12
8.00
1.33
1.78
C
4
6
8
6.00
0.67
0.44
D
1
5
6
4.00
0.83
0.69
E
4
8
11
7.67
1.17
1.36
Expected completion time
20.00
For Path A-D-E
Standard Deviation for the path =SQRT (1+.69+1.36)/6 =
1.75
Days
Values of Z
Probabilities
22
Z
1.14
0.87286
18
Z
-1.14
0.12714
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