Finding the percentage of bolts likely to be defective

Question

2000 bolts have a mean width of 10mm and a standard deviation of 0.2mm, a bolt is defective if it is less than 9.5mm. Find the percentage of bolts that are likely to be defective.

So far I have;

z = (9.5mm - 10mm) / 0.2mm z = -2.5mm

Comparing this to a normal distribution table I get 0.0062

Adding 0.5 from the other half of the distribution i get 0.5062 or 50.62% of bolts being above the 9.5mm

meaning 1 - 0.5062 = 0.4938 or 49.38% of bolts being defective.

This seems extremely high so I'm not sure if or where I have gone wrong, any help is appreciated.

Answer 1

Note that you never use (and never need to use) the fact that you have $2000$ bolts because you're asked about the percentage of bolts that are defective.

No need to add the "other half" of bolts. (Why do you think that you should?)

$$\int\limits_{-\infty}^{9.5} \frac{1}{\sqrt{2 \pi\ 0.2}} e^{-\frac{(x - 10)^2}{{2\ (0.2)^2}}}\ dx = 0.00620967.$$

or $z = \frac{9.5-10}{.2} = -2.5$ and a table gives $p = 0.0062.$

Answer 2

$z$ does not have units. You have divided mm by mm, so there are none. It is the number of standard deviations you are away from the mean. You are then misinterpreting the table. The $0.0062$ is the fraction of the distribution outside $2.5 \sigma$, which is the percent defective. You should not add the other $50\%$ because they are above the mean and therefore above $9.5$ mm