Question 1099083: A process window is set so that 99.74% of product passes the process spec.
In order to pass, the product must be close to average.
How many standard deviations away does 99.74% represent
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if the distribution is normal, than 99.74% of the area under the normal distribution curve needs to be within specs.
99.74% is equal to .9974
that means that 1 - .9974 = .0026 is outside of the limits.
since you need to be in between the limits, than half the area outside the limits to get .0013 is outside of the limits on either end of the normal distribution curve.
to find the z-score that represents this on either end of the normal distribution curve, then take 1 - .0013 to get .9987 of the normal distribution curve will be to the left of the high z-score.
that gets you a z-score of 3.0115 on the high end.
since the normal distribution curve is symmetrical, that also gets you a z-score of -3.0115 on the low end.
therefore, the area under the normal distribution curve needs to be within the z-scores of -3.115 and 3.115.
that means that the extreme scores can't be more than 3.115 standard deviations from the mean.
whether that's a stringent or lenient requirement depends heavily on what the standard deviation is.
the following pictures show you what i mean.
in these pictures,the required z-scores are shown as -3.012 and 3.012.
there's a slight discrepancy which may have to do with rounding or may have to do with the formula used to calculate the actual z-score.
the discrepancy, however, is not important here, so we'll go with the values shown as -3.012 and 3.012.
the first example uses the mean of 0 and the standard deviation of 1, which is what we started with.
the second example shows the raw scores needed to meet those requirements if the mean is 100 and the standard deviation is 20.
the third example shows the raw scores needed to meet those requirements if the mean is 100 and the standard deviation is 5.
the fourth example shows the raw scores needed to meet those requirements is the mean is 100 and the standard deviation is 1.
the bottom line is that all these displays require no more than 3.012 standard deviations above or below the mean.
it's the size of the standard deviation that determines whether these requirements are stringent or not.
on the loose side, a standard deviation of 20 on a mean of 100 gives you lots of slack.
a raw score of between roughly 40 and 160 will pass the test.
on the tight side, standard deviation of 1 on a mean of 100 gives you much less slack.
a raw score of between roughly 97 and 103 will pass the test.
the difference is what the size of the standard deviation is.
the z-score for any of the examples will be the same, and any of the examples will be plus or minus 3.12 standard deviations above or below the mean.
if this is not what you are asking, then send me an email explaining exactly what it is that you are asking and i'll look at it again.
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