Please read the following question:

A lightbulb has a normally distributed light ouput with a mean

5000 end foot-candles and standard deviation of 50 end foot candles.

Find a lower specification limit such that only 0.5% of the bulbs

will not exceed this limit.

In order to solve this question I used Z = Mean - LSL / Sigma

Z = .05 or .69146

M = 5000

Sigma = 50

.69146 = 5000 - LSL / 50

LSL = 5000 - ( 50 x .69146 )

LSL = 4965.43

The next question I believe uses Z = USL - Mean / Sigma

If x is normally distributed with mean u and standard deviation four,

and given the probability that x is less, than 32 is 0.0228, find the

value of u

Your comments please,

Thanks