# Z - Values

• October 3rd 2005, 06:21 PM
Captain Soames
Z - Values
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

Thanks
• October 6th 2005, 04:29 PM
hpe
[QUOTE=Captain SoamesIn order to solve this question I used Z = Mean - LSL / Sigma
[/quote]
So far it's OK.
Quote:

Z = .05 or .69146
That doesn't seem right. You want z s.th. Prob(Z > z) = 0.5% = 0.005. That leads to z = 2.57. Use this value of z in your calculation
Quote:

M = 5000
Sigma = 50

.69146 = 5000 - LSL / 50
LSL = 5000 - ( 50 x .69146 )
LSL = 4965.43
... and you'll get the correct LSL.
Quote:

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
.0228 is the probability of being more than 2 standard deviations below the mean. So the mean minus 2 std dev equals 32. From that you can find the mean $\mu$.