Chapter 14 Project Planning and Acquishtion The expected length of the critical path T. for the entire project is obtained simply by adding fhe 302 expected times t, for (only) those activities lying on the critical path. Standard deviations cannot be adlel the total project duration therefore becomes the root mean square of the standard deviations of activities νΣσ) in the same way-only variances (the squares of standard deviations) can. The standard deviationC lying along the critical path: (14-2 OT average on According to the central limit theorem of statistics, the probability distribution ec though te sum of a set of variables tends toward (approaches) the symmetrical normal distribution, even t original variables fit other distributions. Knowing the expected time (mean) and standard deviation for te critical path permits us to draw a normal distribution fitting those two criteria. If the mean duration of the critical path Te were calculated as 58.0 weeks and its standard deviet as 3.0 weeks, the critical path length would have the probability distribution shown in Figure 144 Then, if you had a contract to complete the project in 61.0 weeks (which is (61.0 – 58.0)/3.0. or 1. standard deviation longer than the mean of 58.0 weeks), you would estimate an 84 percent probabilin (50 percent + 34 percent) of completing the project within that time. The probability of complein within a 52-week year, 2.0 standard deviations (6.0 weeks) less than the mean Te, on the other hand would only be (50.0 – 34.1 – 13.6) = 2.3 percent. Example -99%- -95%- 68%* | 2.3% 2.3% | 13.6% 34.1% 34.1% 13.6% Weeks 70 46 Standard deviation 49 52 55 58 61 64 67 -30 -20 -lo +20 +30 Cumulative +lo 2.3% 15.9% 50.0% 84.1% 977% 99.9% percent Figure 14-8 Normal probability distrihur Probability

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If a project has an expected time of completion Te of 45 weeks witha standard deviation of 7 weeks, what is the probability of completing it (a) within one year (52 weeks)? (b) within 38 weeks.

Chapter 14 Project Planning and Acquishtion
The expected length of the critical path T. for the entire project is obtained simply by adding fhe
302
expected times t, for (only) those activities lying on the critical path. Standard deviations cannot be adlel
the total project duration therefore becomes the root mean square of the standard deviations of activities
νΣσ)
in the same way-only variances (the squares of standard deviations) can. The standard deviationC
lying along the critical path:
(14-2
OT
average on
According to the central limit theorem of statistics, the probability distribution ec
though te
sum of a set of variables tends toward (approaches) the symmetrical normal distribution, even t
original variables fit other distributions. Knowing the expected time (mean) and standard deviation for te
critical path permits us to draw a normal distribution fitting those two criteria.
If the mean duration of the critical path Te were calculated as 58.0 weeks and its standard deviet
as 3.0 weeks, the critical path length would have the probability distribution shown in Figure 144
Then, if you had a contract to complete the project in 61.0 weeks (which is (61.0 – 58.0)/3.0. or 1.
standard deviation longer than the mean of 58.0 weeks), you would estimate an 84 percent probabilin
(50 percent + 34 percent) of completing the project within that time. The probability of complein
within a 52-week year, 2.0 standard deviations (6.0 weeks) less than the mean Te, on the other hand
would only be (50.0 – 34.1 – 13.6) = 2.3 percent.
Example
-99%-
-95%-
68%*
| 2.3%
2.3% |
13.6% 34.1% 34.1% 13.6%
Weeks
70
46
Standard deviation
49
52
55
58
61
64
67
-30
-20
-lo
+20
+30
Cumulative
+lo
2.3% 15.9% 50.0% 84.1% 977% 99.9%
percent
Figure 14-8 Normal probability distrihur
Probability

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