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
0.50
0.66
0.84
0.95
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
27 comments Page 2 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!
(3)

Harsh said:   8 years ago
I am not getting it, please explain in detail.

Dishant said:   8 years ago
Why 1*1?

Nerellakarthik said:   8 years ago
What is Z table?

Jayesh Chavda said:   9 years ago
Which Z table? @Chigumbura.

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.

Sushant Bhalla said:   9 years ago
Check the table @Jerrin.

Jerrin said:   9 years ago
Then how it become 84%?

Sachin said:   10 years ago
(20-3-8-6)/(1^2+2^2+2^2)^1/2 = 1.

Amal said:   10 years ago
Z = (ts-te)/sigma.
Where sigma = deviation, sqrt(variance).

Ts = 20 days.
Te = mean time = 3+8+6 = 17.


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