Question

The optimistic time $\left(t_o\right)$, most likely time $\left(t_m\right)$ and pessimistic time $\left(t_p\right)$ (in days) of the activities in the critical path of a network are given in the format $t_o-t_m-t_p$ in the figure below.


What is the probability (P%) of completing the project in 38 days if the non-exceedance probability factor (z) is tabulated below

Z	1.0	1.04	1.1
(P%)	84.13	85.25	86.43

• A. 85.25
• B. 14.75
• C. 90.32
• D. 9.68

Answer 1

The correct option is B 14.75

Total standard deviation, $\left({\sigma }_t\right)=\sqrt{{\sigma }_{p\to Q}^2+{\sigma }_{Q\to R}^2+{\sigma }_{R\to S}^2+{\sigma }_{S\to T}^2}$

$=\sqrt{{\left(\frac{15-9}{6}\right)}^2+{\left(\frac{11-7}{6}\right)}^2+{\left(\frac{10-6}{6}\right)}^2+{\left(\frac{15-7}{6}\right)}^2}$

$=\sqrt{1+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2}=1.915$

$\therefore z=\frac{T_s-T_E}{\sigma }=\frac{38-40}{1.915}=-1.04$

When z= 1.04, P = 85.25%

Therefore, for $z=-1.04;P\%=1-0.8525=14.75\%$