A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11. Students who made 59.99 or lower on the exam failed the course. What percent of students failed the course?
A) 11.85%
B) 88.15%
C) 1.80%
D) 98.20%

Question

Quizplus · Accepted Answer

To find the percentage of students who failed the course, we calculate the z-score for a grade of 59.99 using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the grade, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the values, we get $z = \frac{(59.99 - 73)}{11} = -1.1827$. Looking up this z-score in a standard normal distribution table or using a calculator, we find that approximately 11.85% of the area under the curve falls to the left of this z-score, indicating that 11.85% of students failed the course.

A Professor at a Local University Noted That the Exam