How Probability of Acceptance of a Part is Calculated

Rameshwar25

Quite Involved in Discussions
#1
Can Someone explain me the methodology adopted in determining the probability of accepting a part with given ref value.
This has been explained on Page 177, 178 of MSA manual 4th Ed.
My confusion is how Ø(9.0)-Ø(1.0)=1.0-0.84=0.16 -Pg 178, Line-12
and how Ø(5.0)-Ø(-3.0)=0.999 -Pg 178, Line-20

regards
rameshwar
 
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B

Barbara B

#2
The greek symbol Phi (uppercase) indicates the standard normal distribution, that is the normal distribution with mean 0 and standard deviation 1.

To calculate the values in the MSA manual in Excel you could use the function NORMDIST
Phi(9.0): =NORMDIST(9.0, 0, 1, 1) value: 1.000000000
Phi(1.0): =NORMDIST(1.0, 0, 1, 1) value: 0.841344746
Phi(5.0): =NORMDIST(5.0, 0, 1, 1) value: 0.999999713
Phi(-3.0): =NORMDIST(-3.0, 0, 1, 1) value: 0.001349898

The difference between Phi(9.0) and Phi(1.0) is 0.16=16%. This equals the probability of acceptance and therefore the probability of rejection is 100%-prob_acceptance = 100%-16%=84%. (The same goes for the third and fourth value.)

Best Regards,

Barbara
 

Rameshwar25

Quite Involved in Discussions
#3
Thank you Barbara
Further, i would ask that the syantax for normdist is: NORMDIST(x, mu, sigma, cumulative). It means that the value is dependent on Mean and Std Dev also.

But in your answer (Phi(9.0): =NORMDIST(9.0, 0, 1, 1) value: 1.000000000), you have assumed mean=0 and std dev=1.

Kindly refer Example given on Pg 178 (sample pages attached) and show the calculation with real std dev given in example.

Thanks and regards
Rameshwar
 

Attachments

B

Barbara B

#4
The values in the MSA manual are based on the standard normal distribution (mean=0, stddev=1), because they are standardized by the formula itself.

On page 177 the distribution of the actual value is given as normal distribution with mean (X_T+bias) and standard deviation sigma^2, so the probability for accepting a part P_alpha is the integral form LSL (lower specification limit, LL in MSA manual formula) to USL (upper specification limit, UL in MSA manual formula). This could be directly calculated with

P_alpha = NORMDIST(LL, X_T+bias, sigma, 1) - NORMDIST(UL, X_T+bias, sigma, 1)

For the example given in the MSA manual (see p.178):
LL=LSL=0.6, UL=USL=1.0, b=0.05, sigma=0.05, X_T=0.5 (reference torque value)
NORMDIST(UL, X_T+bias, sigma, 1) = NORMDIST(0.6, 0.5+0.05, 0.05, 1) = 0.841344746
NORMDIST(LL, X_T+bias, sigma, 1) = NORMDIST(1.0, 0.5+0.05, 0.05, 1) = 1.000000000
P_alpha = 1.0 - 0.841344746 = 0.16 = 16%

But instead of using the values directly in the MSA manual, the values are first standardized:
subtract mean (X_t+b), divide result by standard deviation (sigma)

yielding 9.0 and 1.0 in the first example, but with equal result as in the direct calculation of the acceptance probability:
Phi( ( UL - (X_t+b) )/sigma ) - Phi( LL - (X_t+b) )/sigma )
= Phi( ( 1.0 - (0.5+0.05) )/0.05 ) - Phi( 0.6 - (0.5+0.05) )/0.05 )
= Phi(9.0) - Phi(1.0)
= NORMDIST(9.0, 0, 1, 1) - NORMDIST(1.0, 0, 1, 1)
= 1.0 - 0.841344746 = 0.16 = 16%

The benefit of first standardizing the real values and than using the standard normal distribution Phi (mean 0 and stddev 1) like in the MSA manual proposed is, that you can look up the values in textbooks without access to Excel or a statistics software package. The result remains the same (and it is IMHO questionable if this approach is an advantage compared to the direct calculation not shown in the MSA manual as today nearly everyone has access to at least Excel).

I hope this clarifies your question further.

Best regards,

Barbara
 

Rameshwar25

Quite Involved in Discussions
#5
Hello Barbara
Thank you very much for such a detailed answer. I am very much clear now on the subject. This has helped me in making a GPC (gauge performance Curve) for the example given in MSA manual.

The GPC has been given in my another thread posted today just to share with other readers.

Thank you once again.:thanx:

Best Regards
Rameshwar
 
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