1. ## Normal Distribution

A cylinder is to be machined to a diameter of 8cm. The Upper specifications limit is 8.2cm and the lower specifications limit is 7.9cm. A certain machine produces cylinders with a mean of 8.1cm with a standard deviation of 0.1cm.

What is the probability of being outside the tolerances for the cylinder?

I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
I get stuck from there.

Any help would be greatly appreciated

2. Originally Posted by mortalcyrax

I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
I get stuck from there.
This sounds good, you have to use the 68%-95%-99.7% rules here or look up a table. This online calculator will help.

Z table - Normal Distribution

3. Originally Posted by mortalcyrax
A cylinder is to be machined to a diameter of 8cm. The Upper specifications limit is 8.2cm and the lower specifications limit is 7.9cm. A certain machine produces cylinders with a mean of 8.1cm with a standard deviation of 0.1cm.

What is the probability of being outside the tolerances for the cylinder?

I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
I get stuck from there.

Any help would be greatly appreciated
Maybe your z-tables only deal with positive values of z.

The probability of being outside the tolerances is P(z<-2)+P(z>1), since $\displaystyle \frac{7.9-8.1}{0.1}=-2$

The values given by the z-tables are P(z<value) or $\displaystyle P(z\ \le\ value)$
which are basically the same as the curve is continuous.

However P(z<-2)=P(z>2) due to the symmetry of the bell-shaped curve.

and P(z>2)=1-P(z<2)

Also P(z>1)=1-P(z<1)

Hence the probability of being machined outside of specifications is

$\displaystyle 1-P(z<2)+1-P(z<1)=2-P(z<2)-P(z<1)$

4. So $\displaystyle P(X>8.2)= P(Z>1)$

$\displaystyle = 0.5 - P(Z>1)$

$\displaystyle = 0.5 - 0.3413$

$\displaystyle = 0.1587$ <--- probability of being machined above the specifications

$\displaystyle P(X<7.9)=P(Z< -2)$

$\displaystyle = 0.5 - 0.4772$

$\displaystyle = 0.0228$ <--- probability of being machined lower than the specifications

Then add the two probabilities to get the probability of being machined outside of the specifications.

$\displaystyle 0.1587 + 0.0228 = 0.1815$
Is this right???

5. Yes,
that's what I get also.

You're calculating the probability of z being between 0 and 1
then subtracting from 0.5 to get the probability of being above 1....
and z being between 0 and -2
and subtracting from 0.5 to find the probability of z being below -2.

Excellent.