The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 167 and a variance of 9. The material is considered defective if the breaking strength is less than 158 pounds. What is the probability that a single, randomly selected piece of material will be defective? (Give the answer to two decimal places.)
1 %
Z = (score-mean)/SD
Variance = SD^2
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability 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 probability that the breaking strength is less than 158 pounds.
First, let's standardize the value of 158 pounds using the Z-score formula. The Z-score is calculated as:
Z = (X - μ) / σ
Where X is the value we want to standardize (158 pounds), μ is the mean (167 pounds), and σ is the standard deviation (square root of the variance, which is 3 in this case).
Plugging in the values, we get:
Z = (158 - 167) / 3
Z = -3
Next, we need to find the probability corresponding to this Z-score using a standard normal distribution table or a calculator.
The probability of a Z-score of -3 or lower can be looked up from the standard normal distribution table, or calculated as follows:
P(Z < -3) ≈ 0.0013
This means that the probability that a single, randomly selected piece of material will be defective (breaking strength less than 158 pounds) is approximately 0.0013, or 0.13% when rounded to two decimal places.