Activity A Activity B Optimistic time ($t_o$) = 4 days Optimistic time ($t_o$) = 4 days Most likely time ($t_l$) = 7 days Most likely time ($t_l$) = 6 days Pessimistic time ($t_p$) = 16 days Pessimistic time ($t_p$) = 22 days
PERT Standard Deviation Calculation
The standard deviation for project activities is calculated using the PERT (Program Evaluation and Review Technique) formula, which estimates the dispersion of activity durations.
The formula for standard deviation ($\sigma$) is:
$ \sigma = \frac{t_p - t_o}{6} $
Where:
- $t_p$ = Pessimistic time estimate
- $t_o$ = Optimistic time estimate
Activity A Standard Deviation
Given values for Activity A:
- Optimistic time ($t_o$) = 4 days
- Pessimistic time ($t_p$) = 16 days
Calculation:
$ \sigma_A = \frac{16 \text{ days} - 4 \text{ days}}{6} = \frac{12 \text{ days}}{6} = 2 \text{ days} $
Activity B Standard Deviation
Given values for Activity B:
- Optimistic time ($t_o$) = 4 days
- Pessimistic time ($t_p$) = 22 days
Calculation:
$ \sigma_B = \frac{22 \text{ days} - 4 \text{ days}}{6} = \frac{18 \text{ days}}{6} = 3 \text{ days} $
Conclusion
The standard deviation for Activity A is 2 days, and the standard deviation for Activity B is 3 days.
Important Questions from Basics of Probability
If the data are skewed, which option of central tendency measure is the most unreliable indicator?
Events A, Band C are mutually exclusive events such that \(P(A) = \dfrac{3x + 1}{3}, P(B) = \dfrac{1-x}{4}\)and \(P(C) = \dfrac{1-2x}{4}\)The set of possible values of x are in the interval
Let A, B be two events in a discrete probability space with ℙ(A) > 0 and ℙ(B) > 0. Which of the following are necessarily true?
The Excess Kurtosis of the Geometric distribution with parameter p is:
For a distribution, the mean is 10, variance is 16, γ 1is + 1 and β 2is 4. The distribution is:
