Required: Answers with explanation

 Can someone provide the answers with explanations  to the following questions in the Quality Management

You are setting up a factory making nails and the quality plan specifies the pro 
cess will adhere to 3 sigma standards. While inspecting the first lot of 500,000 
, there were 1000 defective pieces. Are you meeting the quality goals and how ma 
ny defective pieces you can have for a lot of 500,000 if you are at 6 sigma? 

You are with in quality goals now and you can have 50 defective pieces when yo 
u are 6 sigma 
You are with in quality goals now and at six sigma you can have only 300 defec 
t pieces 
You are not with in quality goals now and at 6 sigma you can have 1350 defecti 
ve pieces 
You are not with in quality goals and you can have 300 defective pieces at 6 s 
igma 

 

For an activity, the optimistic, most likely and pessimistic estimates are 12, 1 
6 and 18 respectively. Which of the following statement is correct?
 

There is 68.26% chance that the activity will be completed in 15 to 17 days
There is 95.44% chance that the activity will be completed in 15 to 17 days
There is 68.26% chance that the activity will be completed in 14.67 to 16.67 d 
ays
 
There is 93.73% chance that the activity will be completed in 15 to 17 days


 

If the optimistic estimate for an activity is 18 and standard deviation is 3. If 
the Most likely estimate is 30, the chance the activity will be finished betwee 
n 23 and 35 days is 

95.44
68.26
99.73
99.999

My answers below


1) A


2) C


3) C (I think we are missing pessimistic days estimation in this question but if not then I would guess B as  3 std deviation can give you 12 day range)

 Thanks... can you please provide an explataion for the selections?

Q1: ANS (A)

1000/500,000 = 0.002 or 0.2% defects per unit, it seems like within a quality goal beacuse 3 sigma is 99.73%

Six sigma is 99.99% it should be close to 50 i believe. Therefore answer is A. (Not sure the answer and calculation though).

Q2: ANS (C)

By utilizing PERT and standard deviaition formulas P+4*MP+O/6 and P-O/6 , we can calculate the following

PERT Calc = 15.667 or 15.67 and SD = 1 or +/- 1, therefore option c should be correct.

 

Q3: ANS (A)

Range: 23days  - 35 days = 12 days (mean value = 29days)

Pessimistic value is missing here but based on given st, dev. SD=3 at 68.26% means +3 to -3 days 

therefore range = 6 or we can say 26 - 32 days with one standard dev,

hence for two stand. deviation sd = 3x 2 = 6 or +/- 6 range 

so, there is a 95.44% chance that activity will be finished between the 23days - 35days, the answer should be "A" not "C" - calculated based on the value 3 for SD.

Dear Mr. Karthik

This question is manipulated,

  1. In first question, questioner is asking 2 questions-  1st is for 3sigma and then for 6sigma.

           As per Normal distribution and 6 sigma theory - this question can be answered like this-

          3sigma - accuracy required - 99.73%  and for 500000 samples defective pieces should not be more     than     0.27% =1350.

Hence if 1000 defective pieces are there, then process is meeting to the 3 sigma accuracy.

6sigma - acuracy required - 99.9996% and for 1million not more than 3.4defects are be there. (0.00034%) hence for 500,000 defects should not be more than  = 1.7~ 2. Means not more than 2 pieces should be defective to meet 6sigma accuracy.

when we go for options - all are vague.

 

2.   Question 2, is ok answer is C,  the mean value will (12+4x16+18)/6  = 15.7, and sd  =(18-12)/6 = 1

hence for 1sigma  = 68.26 % chance are there for 14.7 to 16.7 days is correct.

Here little confusion may arise with = most likely or realstic chance of duration is 16 days may be considered as mean(myu). But mean value  after calculation by 3 point formula is correct. Most likely value is raw data.

 

3.

let see:

OP = 18

SD = 3

Most likely is 30

what is chance of 23 to 35 days

now SD =( ps-op)/6 = (ps -18)/6 = 3

hence ps = 36days

Now mean (expected duration) value is  = (18+4x30+36)/6 = 29days

from this value - 23 days  have -6 days variance,

while 35 days have  +6days variance

It ranges between  - 6/3 = - 2sigma to +6/3 = 2sigma

as per this information answer would be  = 95.44%- Option A

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Friends,

I like this forum, but  I  could come very rare here, because of my busy schedule.

Regards

 

 

 

 

 

 Thank you very much for the explamnation. ITs clear now