The optimistic, most likely and pessimistic time of an activity are 6, 9 and 12 days respectively. The standard deviation and variance of the activity are ________ respectively.

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Question

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JKSSB JE CE 2021 Official Paper Shift 3 (Held on 27 Oct 2021)

Answer 1

Concept:

Optimistic time (to), Pessimistic time (tp), Most likely time (tm):

Expected time: \({t_E} = \frac{{{t_o} + 4{t_m} + {t_p}}}{6}\)

Standard deviation: \(\sigma = \frac{{{t_p} - {t_o}}}{6}\)

For a PERT network,\(\;{\sigma ^2} = {\left( {\frac{{{t_p} - {t_o}}}{6}} \right)^2}\)

The variance is given by:

\(V = {\sigma ^2} = {\left( {\frac{{{t_p} - {t_o}}}{6}} \right)^2}\)

Calculation:

Given:

t_o = 6, tm = 9, tp = 12

Standard deviation: \(\sigma = \frac{{{t_p} - {t_o}}}{6}\)

\(\sigma = \frac{{{12} - {6}}}{6}=1\)

\(V = {\left( {\frac{{{t_p} - {t_o}}}{6}} \right)^2} = {\left( {\frac{{12 - 6}}{6}} \right)^2} = 1\)