There are 300 students enrolled in Business Statics . Historically,exam scores are normally distributedwith a standard deviation of 30.9. Your instructor randomly selected a sample of 30 examinations and finds a mean of 74.2. Determine a 90% confidence interval for the mean score for all students taking the course.

To determine the confidence interval for the mean score of all students taking the course, we can use the formula:

Confidence Interval = Mean ± (Z * (Standard Deviation / √(Sample Size)))

In this case, we have a sample size of 30, a mean score of 74.2, and a standard deviation of 30.9. We want to calculate a 90% confidence interval, which means that we need to find the corresponding value of Z.

The Z-value for a 90% confidence interval can be found using a standard normal distribution table or a statistical software. For a 90% confidence level, the Z-value is approximately 1.645.

Now, let's plug the values into the formula:

Confidence Interval = 74.2 ± (1.645 * (30.9 / √30))

Calculating the expression within the parentheses:

Confidence Interval = 74.2 ± (1.645 * (30.9 / 5.48))

Simplifying further:

Confidence Interval = 74.2 ± (1.645 * 5.64)

Confidence Interval = 74.2 ± 9.28

Therefore, the 90% confidence interval for the mean score of all students taking the course is approximately 64.92 to 83.48.

See process in your later post.