The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 175 and a variance of 4. The material is considered defective if the breaking strength is less than 169 pounds. What is the probability that a single, randomly selected piece of material will be defective? (Give the answer to two decimal places.)

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

To find the probability that a single, randomly selected piece of material will be defective, we need to calculate the area under the normal distribution curve to the left of the breaking strength value of 169 pounds.

First, we need to standardize the breaking strength value using the formula:
\(Z = \frac{{X - \mu}}{{\sigma}}\)

Where:
- \(Z\) is the standard score or z-score,
- \(X\) is the breaking strength value (169 pounds),
- \(\mu\) is the mean (175 pounds),
- \(\sigma\) is the standard deviation (square root of the variance, which is 2 pounds in this case).

Substituting the values, we get:
\(Z = \frac{{169 - 175}}{{2}} = -3\)

Next, we need to find the area under the standard normal distribution curve to the left of \(Z = -3\), which represents the probability of the breaking strength being less than 169 pounds. We can use a standard normal distribution table or a statistical calculator to find this value.

Using a standard normal distribution table, we find that the area to the left of \(Z = -3\) is approximately 0.0013.

Therefore, the probability that a single, randomly selected piece of material will be defective (i.e., breaking strength less than 169 pounds) is approximately 0.0013, or 0.13% (rounded to two decimal places).