Question

The pessimistic time and optimistic time of completion of an activity are given as 10 days and 4 days respectively. Then the variance of the activity is

• A. 14
• B. 6
• C. 2
• D. 1

Hint:

Be careful not to confuse standard deviation with variance! - $\text{Standard Deviation} = \frac{t_p - t_o}{6}$ - $\text{Variance} = \left(\frac{t_p - t_o}{6}\right)^2$ Always make sure to square your result when a question asks for the variance.

Answer 1

The Correct Option is D

Step 1: Recall the PERT spread rule.
PERT treats the six unit spread between the pessimistic time \( t_p \) and the optimistic time \( t_o \) as covering the practical range of an activity's duration, so the standard deviation is \[ \sigma = \frac{t_p - t_o}{6} \]

Step 2: Plug in the given times and square the result.
Here \( t_p = 10 \) days and \( t_o = 4 \) days, so \[ \sigma = \frac{10 - 4}{6} = 1 \text{ day} \] and the variance is simply this value squared, \[ \boxed{\sigma^2 = 1^2 = 1} \] which matches option 4.