The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 165 and a variance of 9. The material is considered defective if the breaking strength is less than 156 pounds. What is the probability that a single, randomly selected piece of material will be defective? (Give the answer to two decimal places.)

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 156 pounds.

First, let's calculate the standard deviation (σ), which is the square root of the variance:

σ = √(variance) = √(9) = 3

Next, we will calculate the z-score, which measures how many standard deviations an observation is from the mean. The z-score formula is:

z = (x - μ) / σ

where x is the breaking strength value (156 pounds), μ is the mean (165 pounds), and σ is the standard deviation (3 pounds).

z = (156 - 165) / 3
= -9 / 3
= -3

Now, we need to find the probability (p) corresponding to the z-score of -3. We can use a standard normal distribution table or a calculator to find this probability.

Using a standard normal distribution table, the probability corresponding to a z-score of -3 is approximately 0.0013.

Therefore, the probability that a single, randomly selected piece of material will be defective is 0.0013 (or 0.13%).