# probability--standard deviation question??

#### lmw

##### New Member
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.

Thanks!

#### JohnM

##### TS Contributor
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

#### lmw

##### New Member
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

#### JohnM

##### TS Contributor
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.