
How to calculate approximate probability
Submitted by dearhari on Wed, 12/19/2007 - 02:28.
Activity 1 has an optimistic eestimate of 7 days, a most likely estimate of 16 days and a pessimistic estimate of 25 days. What is the approximate calculated probability that the activity will be finished in 10 to 22 days?
A. 95.4%
B. 50.0%
C. 99.7%
D. 68.3%
How to arrive at the correct answer? Appreciate if any of you can help me on this.


95.47%
Standard deviation determination
Std Deviation
As explained in the previous comment, P-O/6=3, so every multiple of 3 is a one std. deviation. So given the range 10-22, and if you draw the distribution curve, it would be like 10(2 sigma),13 (1 sigma),16(mean), 19(1 sigma), 22(2 sigma) i.e., 2 Standard Deviation.
Hope this explains.
Std. dev - more complete explanation
More specifically:
sd = (P-O)/6 = (25-7)/6 = 18/6 = 3.
mean is (7+16+25)/3 = 16.
So,
+ or - 1 sd around mean is (16-(1*3)) and (16+(1*3)) or 13 to 19 days
+ or - 2 sd around mean is (16-(2*3)) and (16+(2*3)) or 10 to 22 days
+ or - 3 sd around mean is (16-(3*3)) and (16+(3*3)) or 7 to 25 days
and so on.
What is the probability that the activity will complete in these ranges? (Another way to say this is, "how confident are we in these estimated ranges?") That's where "area under curve" comes in from assuming standard normal distribution:
+ or - 1 sd or 13 to 19 days: we are 68.26% confident
+ or - 2 sd or 10 to 22 days: we are 95.44% confident
+ or - 3 sd or 7 to 25 days: we are 99.73% confident