A machine has been programmed to bore holes with a mean diameter = 0.15 mm and with a standard deviation = 0.005 mm. Parts will be rejected if the hole diameters are >= 0.14 mm or <= 0.16 mm. What fraction of the total parts would you expect to be rejected if the data is normally distributed?

My thought is that the mean is .15 and 1 standard deviation going to the right is .155 and 2 standard deviations is .16. Since 68% of the data lies within two standard deviations, I think the answer is 32% would be rejected, but that doesn't account for the data to the "left" of the mean.

Input? Advice? Comments?

Thanks!

lmw,

Good job. You're on the right track, and we like to see students try out problems before posting.

Draw a picture of the normal curve, centered at 0.15mm, with tick marks going out for each std dev +1, +2, +3 and -1, -2, -3. You should see that the rejection region for the parts is <= 0.14 and >=0.16, which correspond to z scores of -2.00 and +2.00.

Now, just find the proportion of the area under the normal curve outside of -2 and +2. Hint - it's pretty small.

JohnM

Thanks John!
The p(z) is .97725, meaning that the rejection rate is .0228 or 2.28% of the holes will be <=14 or >=16??

thanks again

0.97725 is the proportion between negative infinity and z=2, so the proportion beyond 2.0 = 1-0.97725 = 0.02275

Due to symmetry, the proportion to the left of -2 is 1-0.97725 = 0.02275.

You need to add these up to get the total proportion outside of +2 and -2.