The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?
A).4602
B).8413
C).1587
D).5397

Question

Quizplus · Accepted Answer

We need to use the central limit theorem since we are dealing with sample means. The formula to find z-score is:
z = (x̄ - μ) / (σ / sqrt(n))
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Substituting the given values:
z = (29.95 - 30.05) / (0.3 / sqrt(9))
z = -1.0
Using a standard normal table, we find that the probability of getting a z-score of -1.0 or less is 0.1587. Since we are interested in the probability of getting a sample mean more than 29.95 inches, this is the same as the probability of getting a z-score of 1.0 or more. Using symmetry, we can subtract 0.1587 from 1 to get the final answer:
P(x̄ > 29.95) = 1 - 0.1587 = 0.8413, which is option B.

The Population of Lengths of Aluminum-Coated Steel Sheets Is Normally