Question

A machining operation produces bearings with diameters that are normally distributed with mean 3.0002 inches and standard deviation 0.0013 inch. Specifications require the bearing diameters to lie in the interval 3.000±0.0020 inches. Those outside the interval are considered scrap and must be remachined. Suppose that five bearings are randomly drawn from production. What is the probability that at least one is defective? (Round your answer to four decimal places.) You may need to use the appropriate appendix table or technology to answer this question.

Answer 1

To find the mean diameter and standard deviation for the sample, use the properties of the normal distribution. The mean diameter for the sample is 3.0002 inches and the standard deviation is 0.000579 inches. The probability that at least one bearing is defective is approximately 0.9861.

Step-by-step explanation:

To find the mean diameter and standard deviation for the sample, we can use the properties of the normal distribution. Since the sample is randomly drawn, we can assume that the sample mean will be the same as the population mean, which is 3.0002 inches. The standard deviation for the sample can be calculated using the formula:

standard deviation for the sample = population standard deviation / square root of sample size

Substituting the given values, we get:

standard deviation for the sample = 0.0013 / sqrt(5) ≈ 0.000579

Therefore, the mean diameter for the sample is 3.0002 inches and the standard deviation is 0.000579 inches.

The probability that at least one of the five bearings is defective can be found using the complement rule. The probability of no defective bearings can be calculated by finding the probability that each bearing is within the specified interval and multiplying them together:

Probability of no defective bearings = (probability of bearing 1 within interval) × (probability of bearing 2 within interval) × (probability of bearing 3 within interval) × (probability of bearing 4 within interval) × (probability of bearing 5 within interval)

Since the bearings are normally distributed, we can use the z-score to find the probability of each bearing being within the interval. The formula for the z-score is:

z-score = (x - μ) / σ

Where x is the diameter of the bearing, μ is the population mean, and σ is the population standard deviation.

Substituting the given values, we find the z-scores for the lower and upper limits of the interval:

z-score for lower limit = (3 - 3.0002) / 0.000579 ≈ -0.345

z-score for upper limit = (3 + 3.002) / 0.000579 ≈ 0.345

Using a z-table or technology, we can find the probability of each bearing being within the interval:

Probability of each bearing being within interval ≈ 0.365 for both lower and upper limits

Therefore, the probability of no defective bearings = 0.365^5 ≈ 0.0139

Finally, the probability of at least one defective bearing can be found by subtracting the probability of no defective bearings from 1:

Probability of at least one defective bearing ≈ 1 - 0.0139 = 0.9861