50/1000 = 0.05

So the subscription process fails to meet the three day requirement in 5%

of cases. Now we do an inverse table look up on the standard normal table to see what z score cuts of 5% of the distribution on the top end, that is we

find z0 such that:

p(z>z0)=0.05.

doing this we find that z0=1.645 which is the sigma level that the process

is operating at.

RonL