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