Question 384378
Assume that the diameters are normally distributed.  The z-score formula is {{{z = (X - mu)/sigma}}}, or vice-versa, {{{X = mu + z*sigma}}}. Within 1 sd, he can accept only  diameter values within the interval [12.0 - 0.012, 12.0 + 0.012] = [11.988, 12.012].  There is 100% - 68.26%  = 31.74% chance that a CD disc diameter is outside this interval.  (The z-score tells us how many sd's X is away from {{{mu}}}.)