Assume that the dimensions X and Y are normally distributed with means ?x and ?y and standard deviations sigma x and sigma y, respectively. The parts are produced on different machines and are assembled at random. Based on actual measurements ?x = 0.40 and ?y = 0.17. If it is desired that the probability of a smaller clearance (i.e., X - Y) than 0.09 should be 0.006, what distance between the mean or nominal dimensions (i.e., mue x - mue y) should be specified? Solution lets consider z=X-Y since X and Y are normally distributed then X-Y is also normally distributed so (X-Y-(mue x-mue y))/sqrt(square(sigma(x))+square(sigma(y))) is a standard normal distribution lets denote it by Z=X-Y-(mue x-mue y)/.4346 P(X-Y<.09)=.06 so Z value corresponding to .06 is .26 so z should be less then .26 .09-(mue x-muey)/.4346=.26 mue x -mue y=mod(.09-.113)=.023 for Z table use this http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm.

1. Assume that the dimensions X and Y are normally distributed with means ?x and ?y and standard
deviations sigma x and sigma y, respectively. The parts are produced on different machines and
are assembled at random. Based on actual measurements ?x = 0.40 and ?y = 0.17. If it is desired
that the probability of a smaller clearance (i.e., X - Y) than 0.09 should be 0.006, what distance
between the mean or nominal dimensions (i.e., mue x - mue y) should be specified?
Solution
lets consider z=X-Y
since X and Y are normally distributed then X-Y is also normally distributed
so (X-Y-(mue x-mue y))/sqrt(square(sigma(x))+square(sigma(y))) is a standard normal
distribution
lets denote it by Z=X-Y-(mue x-mue y)/.4346
P(X-Y<.09)=.06
so Z value corresponding to .06 is .26 so z should be less then .26
.09-(mue x-muey)/.4346=.26
mue x -mue y=mod(.09-.113)=.023
for Z table use this http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm