The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 115 and a variance of 4. The material is considered defective if the breaking strength is less than 110.6 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.)
To find the probability that a randomly selected piece of material will be defective, we need to calculate the area under the normal distribution curve to the left of the lower threshold.
First, we need to standardize the threshold value 110.6 using the z-score formula:
z = (x - μ) / σ
where x is the threshold value, μ is the mean, and σ is the standard deviation.
In this case, the mean (μ) is 115 and the standard deviation (σ) is the square root of the variance, which is 2.
z = (110.6 - 115) / 2
z = -2.2 / 2
z = -1.1
Next, using the standard normal distribution table or a calculator that provides standard normal probabilities, we can find the probability corresponding to a z-score of -1.1.
Looking up the z-score (-1.1) in the standard normal distribution table, we find that the probability is approximately 0.1357.
Therefore, the probability that a single randomly selected piece of material will be defective is 0.1357 (rounded to four decimal places).