Question

In a PERT network, the time estimate of an activity is as per the following: Optimistic time is \(4\) days, most likely time is \(6\) days and pessimistic time is \(14\) days. The expected time and standard deviation of the activity (in days) are respectively

• A. \(6.0\) & \(2.76\)
• B. \(6.0\) & \(3.78\)
• C. \(7.0\) & \(1.67\)
• D. \(7.0\) & \(2.76\)

Hint:

PERT formulas: \[ \boxed{ t_e=\frac{a+4m+b}{6} } \] \[ \boxed{ \sigma=\frac{b-a}{6} } \] \[ \boxed{ \text{Variance}=\sigma^2=\left(\frac{b-a}{6}\right)^2 } \]

Answer 1

The Correct Option is C