The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 135 and a variance of 16. The material is considered defective if the breaking strength is less than 126.2 pounds. What is the probability that a single, randomly selected piece of material will be defective? (You may need to use the standard normal distribution table. Round your answer to four decimal places.)
Z = (score -mean) /sd
z = (126.2-135)/4
z = -9.8/4 = -2.45
z-table
To find the probability that a single, randomly selected piece of material will be defective, we need to use the standard normal distribution table.
First, we need to calculate the z-score for the breaking strength of 126.2 pounds. The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the observation, μ is the mean, and σ is the standard deviation.
In this case, x = 126.2 pounds, μ = 135 pounds, and σ = √16 = 4.
z = (126.2 - 135) / 4
z = -8.8 / 4
z = -2.2
Once we have the z-score, we can use the standard normal distribution table to find the corresponding probability. The z-score of -2.2 corresponds to a probability of 0.0139.
Therefore, the probability that a single, randomly selected piece of material will be defective is approximately 0.0139.