# Normal distribution random variable engine blocks

The diameters of cylinders drilled into an engine block vary slightly, being normally distributed with a mean of 12.500 cm and a standard deviation of 0.002 cm. If the diameter of a given cylinder is within 0.003 cm of the target value of 12.500 cm, it is acceptable.
a) What is the probability that a randomly selected cylinder will be acceptable?

My work:

$P(12.497 < x < 12.503) = [ \frac{ 12.497 - 12.5} {0.11} < z < \frac{12.503 - 12.5} {0.11} ]$

which then equals $(-0.0273 < z < 0.0273) = (0.4880 < z < 0.5120)$

I then did $0.5120-0.4880 = 0.0240$

The final answer is 0.8664, but I'm not sure how to arrive at it. All help is appreciated!

There shouldn't be a $0.11$ in that denom. In fact it should be your given SD because you are selecting only one cylinder. This gives a z-score of $-1.5$ for the left boundary. Looking in the cumulative z-table I read $0.0668$ Due to symmetry the region that falls outside the required measurement is then $2*0.0668$ and when you subtract that from $1$ (total probability) you arrive at \$0.8664