How to calculate approximate probability
Submitted by Anonymous (not verified) 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.
Forums:
tonsi (not verified)
Tue, 12/25/2007 - 16:04
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95.47%
This is 2 standard deviation. (p-o)/6 = 3. So 2 stad deviation would be 6 days which is 10-22 days (mostlikely is 16 days)
thoekwater (not verified)
Sat, 03/22/2008 - 03:05
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Standard deviation determination
How do you interpret from the question what the standard deviation is? You indicate it is 2 standard deviations but it is not clear how you figured that out. Please help.
vrnekkanti (not verified)
Thu, 02/14/2008 - 04:16
Permalink
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.
jdixon63 (not verified)
Sun, 03/23/2008 - 21:39
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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