The correct answer is $\frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$.
The expected time of an activity is the weighted average of the
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The correct answer is $\frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$.
The expected time of an activity is the weighted average of the
213.5V175.2l142.7 81.2-142.7 81.2z"/> Subscribe on YouTubeThe formula for the expected time of an activity is:
$E = \frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6}$
where:
For example, if the optimistic time for an activity is 1 day, the most likely time is 3 days, and the pessimistic time is 5 days, then the expected time of the activity is:
$E = \frac{{{\text{a}} + 4{\text{m}} + {\text{b}}}}{6} = \frac{{1 + 4 \times 3 + 5}}{6} = 3$ days