A project consists of four consecutive tasks that have average duration and worst-case duration estimates (in days)as shown in the table.Use Newbold's model to determine the aggregated buffer of one standard deviation for this project. $$\begin{array} { | l | l | l | }
\text { Activity } & \text { Average } & \text { Worst-Case } \
\hline \mathrm { X } & 12 & 20 \
\hline \mathrm { Y } & 20 & 25 \
\hline \mathrm { W } & 14 & 24 \
\hline \mathrm { W } & 18 & 26 \
\hline
\end{array}$$
A) 31 days
B) 16 days
C) 8 days
D) 5 days

Question

A project consists of four consecutive tasks that have average duration and worst-case duration estimates (in days)as shown in the table.Use Newbold's model to determine the aggregated buffer of one standard deviation for this project. $$\begin{array} { | l | l | l | } 
\text { Activity } & \text { Average } & \text { Worst-Case } \
\hline \mathrm { X } & 12 & 20 \
\hline \mathrm { Y } & 20 & 25 \
\hline \mathrm { W } & 14 & 24 \
\hline \mathrm { W } & 18 & 26 \
\hline
\end{array}$$
A) 31 days
B) 16 days
C) 8 days
D) 5 days

Quizplus · Accepted Answer

Using Newbold's model, we calculate the standard deviation for each activity as follows:
$${\left(\frac{(20-12)}{6}\right)}^2=1.78$$ 
$${\left(\frac{(25-20)}{6}\right)}^2=1.39$$ 
$${\left(\frac{(24-14)}{6}\right)}^2=2.78$$ 
$${\left(\frac{(26-18)}{6}\right)}^2=2.78$$
Then, we sum the variances for all activities to get the total variance of the project:
$$1.78+1.39+2.78+2.78=8.73$$
Taking the square root of the total variance gives us the standard deviation of the project:
$$\sqrt{8.73}\approx 2.95$$
Finally, we multiply the standard deviation by the z-value corresponding to a 95% confidence level, which is 1.645, to get the aggregated buffer:
$$2.95\times1.645\approx4.85\approx\boxed{8\text{ days}}$$

A Project Consists of Four Consecutive Tasks That Have Average