Consider the following work breakdown structure: $$\begin{array}{r}
\text { Duration Estimates (Days) }\quad\quad\quad\
\begin{array} { l c c c c }
\text { Activity } & \text { Precedes } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \
\hline \text { Start } & \text { U, V } & - & - & - \
\text { U } & \text { W } & 35 & 50 & 65 \
\text { V } & \text { W, X } & 28 & 40 & 52 \
\text { W } & \text { Z } & 26 & 35 & 44 \
\text { X } & \text { Y } & 28 & 40 & 52 \
\text { Y } & \text { Z } & 26 & 29 & 38 \
\text { Z } & \text { End } & 36 & 60 & 84
\end{array}
\end{array}$$ What is the estimated standard deviation in the time for activity Z?
A) 3 days
B) 2 days
C) 4 days
D) 8 days
E) 5 days

Question

Consider the following work breakdown structure: $$\begin{array}{r}
\text { Duration Estimates (Days) }\quad\quad\quad\
\begin{array} { l c c c c } 
\text { Activity } & \text { Precedes } & \text { Optimistic } & \text { Most Likely } & \text { Pessimistic } \
\hline \text { Start } & \text { U, V } & - & - & - \
\text { U } & \text { W } & 35 & 50 & 65 \
\text { V } & \text { W, X } & 28 & 40 & 52 \
\text { W } & \text { Z } & 26 & 35 & 44 \
\text { X } & \text { Y } & 28 & 40 & 52 \
\text { Y } & \text { Z } & 26 & 29 & 38 \
\text { Z } & \text { End } & 36 & 60 & 84
\end{array}
\end{array}$$ What is the estimated standard deviation in the time for activity Z?
A) 3 days
B) 2 days
C) 4 days
D) 8 days
E) 5 days

Quizplus · Accepted Answer

For activity Z, using the beta distribution formula, we have:

$E(Z) = \frac{26 + 4(29) + 38}{6} = \frac{176}{6} \approx 29.3 $

$SD(Z) = \frac{P - O}{6} = \frac{84-26}{6} = \frac{58}{6} \approx 9.7$

Therefore, the estimated standard deviation in the time for activity Z is about 9.7 days, which rounds to 10 days. However, since we are given answer choices in days, we choose the closest answer, which is 8 days (choice D).

Consider the Following Work Breakdown Structure What Is the Estimated Standard Deviation in the Time for (Days)