Question

The lengths of 3-inch nails manufactured on a machine are normally distributed with a mean of 3.0 inches and a standard deviation of .009 inch. The nails that are either shorter than 2.98 inches or longer are than 3.02 inches are unusable. What percentage of all the nails produced by this machine are unusable?

Answer 1

Using the normal distribution, it is found that 2.64% of all the nails produced by this machine are unusable.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of 3 inches, hence $\mu = 3$.

• The standard deviation is of 0.009 inches, hence $\sigma = 0.009$.

Nails that are more than 0.02 inches from the mean are unusable, hence:

$Z = \frac{0.02}{\sigma}$

$Z = \frac{0.02}{0.009}$

$Z = 2.22$

The proportion is P(|Z| > 2.22), which is 2 multiplied by the p-value of Z = -2.22.

Z = -2.22 has a p-value of 0.0132.

2 x 0.0132 = 0.0264

0.0264 x 100% = 2.64%

2.64% of all the nails produced by this machine are unusable.

You can learn more about the normal distribution at brainly.com/question/24663213