The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 175 and a variance of 9. The material is considered defective if the breaking strength is less than 166 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 z-score and then use the standard normal distribution.
The z-score can be calculated using the formula: z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, x = 166, μ = 175, and σ = square root of the variance = √9 = 3.
Now we can find the z-score: z = (166 - 175) / 3 = -3.
To find the probability corresponding to this z-score, we need to look up the z-score in the standard normal distribution table or use a calculator.
Using a standard normal distribution table, we can find that the area to the left of a z-score of -3 is approximately 0.0013.
However, we want to find the probability that the breaking strength is less than 166 pounds, so we need to find the area to the left of the z-score.
Since the normal distribution is symmetric, we can subtract this probability from 0.5 to get the probability that the breaking strength is less than 166 pounds.
Probability = 0.5 - 0.0013 = 0.4987.
Therefore, the probability that a single randomly selected piece of material will be defective is approximately 0.50 (or 50% if rounded to two decimal places).