The width of a slot of a duralumin forging is (in inches)...

The width of a slot of a duralumin forging is (in inches) normally distributed with $\mu=.9000$ and $\sigma=.0030 .$ The specification limits were given as $.9000 \pm .0050$ (a) What percentage of forgings will be defective? (b) What is the maximum allowable value of $\sigma$ that will permit no more than 1 in 100 defectives when the widths are normally distributed with $\mu=.9000$ and $\sigma ?$

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00:01 So i believe this is your question.

00:03 We have the mean of this distribution is 0 .9 with lots of zeros.

00:08 And the standard deviation is 0 .003.

00:12 And these are in inches.

00:13 And i believe that interpreting what you wrote, that it's supposed to be 0 .9 plus or minus.

00:20 I think that's what your p and m stands for.

00:23 0 .0050.

00:26 That that is supposed to be what's acceptable.

00:29 If it's in within that range.

00:31 And so if we have values that are down here or down up here, these values would be where in part a, we would want to find the probability that we would be having defects or we would get rid of them.

00:48 So we need to find what that z value is.

00:52 And so this number here, that cutoff point, is that a z value of 0 .0 .0.

00:59 We would take the 9 plus that and subtract way 9.

01:03 So that difference for the z value would be this divided by the standard deviation.

01:08 Again, this would be the x minus the mean with this value.

01:11 And it's also negative that value, and then divided by that standard deviation.

01:17 And when we do that calculation, our z value will come out to be 1 .67...

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