QUESTION 8
Prob 1 118:
Use your TI83 (or Excel):
A Test has scores that are normally distributed with a mean of 76 and a standard deviation of 16 . Determine the probability that a random sample of 38 test scores has an average greater than 75 .

Round to four decimal places.

Step 1 ：The problem is asking for the probability that the average of a random sample of 38 test scores is greater than 75. The scores are normally distributed with a mean of 76 and a standard deviation of 16.

Step 2 ：We can use the Central Limit Theorem to solve this problem. The Central Limit Theorem states that the distribution of sample means approximates a normal distribution as the sample size gets larger, regardless of the shape of the population distribution.

Step 3 ：We calculate the z-score for the sample mean of 75. The z-score is the number of standard deviations the sample mean is from the population mean. The formula for the z-score is: \(z = \frac{X - \mu}{\sigma / \sqrt{n}}\), where \(X\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size.

Step 4 ：Substituting the given values into the formula, we get: \(z = \frac{75 - 76}{16 / \sqrt{38}} = -0.385\)

Step 5 ：We then use the standard normal distribution (z-distribution) to find the probability that a z-score is greater than the calculated value. The probability is approximately 0.65.

Step 6 ：Thus, the probability that a random sample of 38 test scores has an average greater than 75 is approximately 0.65.

Step 7 ：Final Answer: \(\boxed{0.65}\)