## PMP question

- anuragca24
**Posts:**3**Joined:**Tue Aug 25, 2015 5:26 pm

### PMP question

37. A project manager has been asked by the client to meet the promise date of the project. The project manager analyzes the schedule before promising a date to the customer. The project manager uses the program evaluation and review technique to evaluate the project schedule. She decides that based on the PERT calculations she can promise a delivery date of June 30. The expected value of the project completion date is May 30. If the project manager is willing to accept a 5% probability that the project will be delivered later than June 30, what is the standard deviation of the duration of the activities on the critical path? Assume a five-day workweek.

a. Ten days

b. Fifteen days

c. One-half month

d. One month

what should be the answer and why? thanks in advance!

a. Ten days

b. Fifteen days

c. One-half month

d. One month

what should be the answer and why? thanks in advance!

### Re: PMP question

Expected date = May 30

probability of completing on or before June 30 = 95% confidence level

probability of completing after June 30 = 5% confidence level

To get 95% probability of a date less than June 30, you will need to add 50% of the area on the left side of the bell (normal) curve and 45% of the area on the right side of the bell curve.

PERT = [O+4M+P] / 6

O = 30 days * 95% = 28.5 days

M = 30 day * 50% = 15 days

P = 30 days * 5% = 1.5 day

So PERT = [ 28.5 + (4*15) +1.5 ] / 6 = 90 / 6 = 15 days

So correct answer is C - 1/2 month

probability of completing on or before June 30 = 95% confidence level

probability of completing after June 30 = 5% confidence level

To get 95% probability of a date less than June 30, you will need to add 50% of the area on the left side of the bell (normal) curve and 45% of the area on the right side of the bell curve.

PERT = [O+4M+P] / 6

O = 30 days * 95% = 28.5 days

M = 30 day * 50% = 15 days

P = 30 days * 5% = 1.5 day

So PERT = [ 28.5 + (4*15) +1.5 ] / 6 = 90 / 6 = 15 days

So correct answer is C - 1/2 month

- anuragca24
**Posts:**3**Joined:**Tue Aug 25, 2015 5:26 pm

### Re: PMP question

Thanks!

However, I miss the part of distribution on the bell curve and how did you identify the % to apply. Can you refer the source or thread where I can get the understanding of this. thanks!

However, I miss the part of distribution on the bell curve and how did you identify the % to apply. Can you refer the source or thread where I can get the understanding of this. thanks!

### Re: PMP question

Yes. I didn't understand it too. please suggest for reasons of distribution.

Regards

Regards

### Re: PMP question

I learnt bell curve in my stats class so this was helpful. you can refer to this link to learn more about bell curve

http://www.dummies.com/how-to/content/b ... certi.html

BTW what's the right answer?

http://www.dummies.com/how-to/content/b ... certi.html

BTW what's the right answer?

### Re: PMP question

Is it 15 days or 1/2 month? Both the options are given there.

### Re: PMP question

I would go with 1/2 month instead of 15 days as the question mentions about month names (30 May - 30 June) rather than days (30 days).

### Re: PMP question

Is not an ambiguous question then? First of all it talks about dates. Then it talks about 5% probability of delay in delivering late of 30 June, which itself is 1 month delay from what customer expects.

Again it does not say how much delay from 30th June. 5% is probability and not % delay. So may be we cannot say that since there is 5% probability of delay, there is 95% probability of early delivery. We don't know how much 'early' or 'late'. So we do not have optimistic and pessimistic estimate.

I did not get the crux of the question itself. May be I have to revisit time estimation T&T.

Can you please explain it in more detail?

Again it does not say how much delay from 30th June. 5% is probability and not % delay. So may be we cannot say that since there is 5% probability of delay, there is 95% probability of early delivery. We don't know how much 'early' or 'late'. So we do not have optimistic and pessimistic estimate.

I did not get the crux of the question itself. May be I have to revisit time estimation T&T.

Can you please explain it in more detail?

### Re: PMP question

Agree...we don't' come across such questions in exam...not sure the source of this question. Some online pmp practice questions are of not good quality and make candidates dejected.

After solving even I was debating between picking 15 days vs. 1/2 months but ended up with latter one.. I would not worry about such questions but would concentrate on pmp exam preparation and be thorough with everything as much as possible..

After solving even I was debating between picking 15 days vs. 1/2 months but ended up with latter one.. I would not worry about such questions but would concentrate on pmp exam preparation and be thorough with everything as much as possible..

### Re: PMP question

I got some more clarification on this. It requires good amount of statistics knowledge.

In order for the project to have a 5% probability of being late, there is a

I found this statistical formula -

In terms of the PERT calculation this means that

So one standard deviation = (30-1)/6 = 4.83 i.e., 5 days

Based on above table, n = 2 should be taken.

So standard deviation is 2* 5 = 10 days (second sigma), which is one-half month.

This is a better answer than fifteen days because on the basis of a five-day work week this is close to three weeks.

In order for the project to have a 5% probability of being late, there is a

**95% probability that the project will be delivered on time or earlier.**I found this statistical formula -

Code: Select all

`We can determine the Probability associated with a Range. This Probability is also called the Confidence Level. The Statistical Distribution Theory provides Probability for different values of ‘n’.`

If n=1 then Probability is 68.27%

If n=2 then Probability is 95.45%

If n=3 then Probability is 99.73%

If n=6 then Probability is 99.999999%

In terms of the PERT calculation this means that

**2 standard deviations**should be added to the expected value date of May 30. Since there is a one month difference between promise date of June 30 and the expected value of May 30, we can take optimistic day = 0 (first day - 30 May), pessimistic day = 30 (end day - 30 June).So one standard deviation = (30-1)/6 = 4.83 i.e., 5 days

Based on above table, n = 2 should be taken.

So standard deviation is 2* 5 = 10 days (second sigma), which is one-half month.

This is a better answer than fifteen days because on the basis of a five-day work week this is close to three weeks.

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