In one statistics course, students reported studying an average of 9.92 hours a week, with a standard deviation of 4.54.Mark studies 16 hours per week.What percent of students are at least as extreme as Mark, in both directions? (Round z score to two decimal places.)
A) 9.01
B) 10.89
C) 18.02
D) 22.64

Question

Quizplus · Accepted Answer

To find the percentage of students who study at least as extreme as Mark, we first calculate the z-score for Mark's study hours. The z-score formula is $z = \frac{(X - \mu)}{\sigma}$, where $X$ is Mark's study hours, $\mu$ is the mean study hours, and $\sigma$ is the standard deviation. Plugging in the values, we get $z = \frac{(16 - 9.92)}{4.54} = \frac{6.08}{4.54} \approx 1.34$. Looking up 1.34 in a z-table or using a calculator for the normal distribution, we find that the area to the right of $z = 1.34$ is approximately 0.0901 (or 9.01%). Since the question asks for students at least as extreme in both directions, we double this percentage (to account for both tails of the distribution), resulting in approximately 18.02%.

In One Statistics Course, Students Reported Studying an Average of 9.92