Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 12.6 inches, and standard deviation of 1.1 inches.

If 47 items are chosen at random, what is the probability that their mean length is greater than 12.9 inches?

Answer 1

To solve this problem, we can use the Central Limit Theorem, which says that the sample mean approaches a normal distribution as the sample size increases.

First, we find the standard error of the mean (SEM) by dividing the population standard deviation by the square root of the sample size.

Given:
- Population standard deviation (σ) = 1.1 inches
- Sample size (n) = 47

SEM = 1.1 / √47 ≈ 0.159 inches

Next, we standardize the value 12.9 inches using the SEM to find the z-score.

z = (12.9 - 12.6) / 0.159 ≈ 1.887

Then, we find the probability corresponding to this z-score in the standard normal distribution table, which is approximately 0.9699.

So, the probability that the mean length of 47 items chosen at random is greater than 12.9 inches is approximately 0.9699, or 96.99%.