Mechanical Engineering - Industrial Engineering and Production Management - Discussion
Discussion Forum : Industrial Engineering and Production Management - Section 2 (Q.No. 6)
6.
A PERT network has three activities on critical path with mean time 3, 8 and 6 and standard deviations 1, 2 and 2 respectively. The probability that the project will be completed in 20 days is
Discussion:
27 comments Page 1 of 3.
UCHIHA said:
7 years ago
Explanation for z=1.
Remember normal curve is symmetrical hence for z=0 the probability is 50%
now if you see chart for 6σ, it is 99.97%, for 4σ it is 95.95% and for 2σ it is 68%.
For 6σ it implies 3σ on both sides of z=0 line on a normal curve and that implies z=3.
Hence for z=1, it implies 1σ and for 1σ it is 68/2=34%,
Hence we go here for z=1 as 50%+34%=84%,
Henceforth remember it!
Remember normal curve is symmetrical hence for z=0 the probability is 50%
now if you see chart for 6σ, it is 99.97%, for 4σ it is 95.95% and for 2σ it is 68%.
For 6σ it implies 3σ on both sides of z=0 line on a normal curve and that implies z=3.
Hence for z=1, it implies 1σ and for 1σ it is 68/2=34%,
Hence we go here for z=1 as 50%+34%=84%,
Henceforth remember it!
(3)
Yawer abbas said:
1 decade ago
how we calculate project completion time probability through mean time of critical activities and standard deviation eaxmple:
A PERT network has three activities on critical path with mean time 3, 8 and 6 and standard deviations 1, 2 and 2 respectively. The probability that the project will be completed in 20 days is.
A PERT network has three activities on critical path with mean time 3, 8 and 6 and standard deviations 1, 2 and 2 respectively. The probability that the project will be completed in 20 days is.
Debaduyti Nath said:
5 years ago
Agree @Chigumbura and @Uchiha.
In PERT analysis, the time estimates of activities and probability of their occurrence follow Bet Distribution Curve.
But Expected project duration generally follows the Normal Distribution Curve.
The answer is 0.84.
In PERT analysis, the time estimates of activities and probability of their occurrence follow Bet Distribution Curve.
But Expected project duration generally follows the Normal Distribution Curve.
The answer is 0.84.
Chigumbura said:
9 years ago
Standard deviation = square root of (variance) = square root of (1 * 1 + 2 * 2 + 2 * 2) = square root of (9) = 3.
D = 20.
S = 3 + 8 + 6 = 17.
Probability = (D - S)/standard deviation = 1.
Now, from Z table probability = 0.84.
D = 20.
S = 3 + 8 + 6 = 17.
Probability = (D - S)/standard deviation = 1.
Now, from Z table probability = 0.84.
Midhunlal said:
6 years ago
Standard deviation equal to square root of variance. Sahir standard deviation is given directly so no need to take root.
Amrit kshetri said:
6 years ago
But PERT network follows beta probability distribution. So can concept of normal distribution curve be applied here?
Ashok kumar said:
1 decade ago
Sigma is standard deviation = Under root of 1^2+2^2+2^2 means it will 3.
Z = (t-te)/under root of variance.
Z = (t-te)/under root of variance.
Kapilkotadiya said:
6 years ago
It is working on the function of Z which is denoted by Z(1)=0.84, therefore the answer is 0.84.
Thanks.
Thanks.
Amal said:
10 years ago
Z = (ts-te)/sigma.
Where sigma = deviation, sqrt(variance).
Ts = 20 days.
Te = mean time = 3+8+6 = 17.
Where sigma = deviation, sqrt(variance).
Ts = 20 days.
Te = mean time = 3+8+6 = 17.
Nir said:
7 years ago
3+8+6=17.
Standard deviation 1+2+2=5,
20-17/5=0.6.
So, I think the answer is 0.66.
Standard deviation 1+2+2=5,
20-17/5=0.6.
So, I think the answer is 0.66.
(1)
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