Questions: The average grade in a statistics course has been 74, with a standard deviation of 9. If a random sample of 69 is selected from this population, what is the probability that the average grade is more than 77? Use Appendix B. 1 for the z values. (Round your z value to 2 decimal places and the final answer to 4 decimal places.) Probability

Solution Steps

To find the probability that the average grade is more than 77, we first need to calculate the Z-score for the sample mean. The Z-score is given by the formula:

\[ Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \]

Where:

• \(\bar{X} = 77\) (the sample mean we are testing)
• \(\mu = 74\) (the population mean)
• \(\sigma = 9\) (the population standard deviation)
• \(n = 69\) (the sample size)

Substituting the values, we calculate:

\[ Z = \frac{77 - 74}{9 / \sqrt{69}} \approx 2.7689 \]

Next, we need to find the probability that the sample mean is greater than 77, which can be expressed as:

\[ P(\bar{X} > 77) = 1 - P(\bar{X} \leq 77) = 1 - \Phi(Z) \]

Using the Z-score we calculated:

\[ P(\bar{X} > 77) = 1 - \Phi(2.7689) \]

From the standard normal distribution table, we find:

\[ \Phi(2.7689) \approx 0.9972 \]

Thus, the probability becomes:

\[ P(\bar{X} > 77) = 1 - 0.9972 = 0.0028 \]

Final Answer