Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 17.3 inches, and standard deviation of 4.3 inches.

If 50 items are chosen at random, what is the probability that their mean length is greater than 18.3 inches?

Answer 1

The probability that the mean length of 50 items is greater than 18.3 inches is approximately 0.24%.

Step-by-step explanation:

To find the probability that the mean length of 50 items is greater than 18.3 inches, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sample mean of a large sample size will be approximately normally distributed, regardless of the shape of the population distribution.

In this case, the sample mean length will be normally distributed with a mean of 17.3 inches (same as the population mean) and a standard deviation of 4.3/sqrt(50) inches (which is the population standard deviation divided by the square root of the sample size).

We can then use the z-score formula to calculate the area under the normal curve to the right of 18.3 inches:

z = (x - mean) / standard deviation = (18.3 - 17.3) / (4.3 / sqrt(50))

Using a standard normal distribution table or a calculator, we can find that the z-score is approximately 2.767. The probability of the mean length being greater than 18.3 inches is the area under the normal curve to the right of 2.767, which is approximately 0.0024 (or 0.24%).