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Question

Considering A and B as two activities of a project, what are the standard deviation of Activity A and B ?
Activity AActivity B
Optimistic time ($t_o$) = 4 daysOptimistic time ($t_o$) = 4 days
Most likely time ($t_l$) = 7 daysMost likely time ($t_l$) = 6 days
Pessimistic time ($t_p$) = 16 daysPessimistic time ($t_p$) = 22 days

The correct answer is
Standard deviation of Activity A = 2 and B = 3

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.

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Important Questions from Basics of Probability

  1. If the data are skewed, which option of central tendency measure is the most unreliable indicator?

  2. In a negatively skewed distribution

  3. If the distribution is negatively skewed, then the:

  4. The first four moments about the mean of distribution are 0, μ 2, 0.7 and 18.75. If the distribution is mesokurtic, the value of μ 2, is

  5. If Mean > Median > Mode, the distribution is:

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