The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 172 and a variance of 4. The material is considered defective if the breaking strength is less than 168 pounds. What is the probability that a single, randomly selected piece of material will be defective?
To find the probability that a randomly selected piece of material will be defective, we need to find the area under the normal distribution curve to the left of the value 168.
Given that the breaking strength of the synthetic material is normally distributed with a mean of 172 and a variance of 4, we first need to standardize the value 168 using the z-score formula:
z = (x - μ) / σ
Where:
- z is the standard score (z-score)
- x is the value we want to standardize
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
Plugging in the values:
z = (168 - 172) / √4
z = -4 / 2
z = -2
The resulting z-score is -2. Now, we need to find the probability associated with this z-score. We can do this by looking up the z-score in a standard normal distribution table or by using a statistical calculator.
Using a standard normal distribution table, we find that the probability associated with a z-score of -2 is approximately 0.0228.
Therefore, the probability that a single, randomly selected piece of material will be defective is approximately 0.0228 or 2.28%.