explain please.

Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 9.5 inches, and standard deviation of 1.8 inches.

If 48 items are chosen at random, what is the probability that their mean length is greater than 10.3 inches?

Answer 1

The standard deviation of a sample mean is equal to Ó/sqrt(N), where Ó is equal to the standard deviation of the population and N is the size of the sample. We begin by asserting that the sampling distribution of the sample mean is approximately normal by the Central Limit Theorem, since N is 48 (greater than or equal to 30). Since Ó is given to be 1.8 and N is given to be 48, we calculate the standard deviation of this sample mean to be 1.8/sqrt(48), or .260. We then calculate the Z-score of 10.3. Z-scores are (Observed Value - Mean)/(Standard Deviation), which in this case is (10.3-9.5)/(.260), or 3.077. Converting the Z-score to a percentile, we find that the percentile for a mean length of 10.3 inches is 99.896, so the probability of a mean length of 10.3 inches or greater is 1-.999=.001.