Question

Exam grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of 11. Use the normal distribution to answer the following questions.
(a) What percent of students scored above a 91?
(b) What percent of students scored below a 63?
(c) If the lowest 4% of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?
(d) If the highest 10% of students will be given a grade of A, what is the cutoff to get an A?

Answer 1

Answer:

a) 4.18% of students scored above a 91

b) 20.61% of students scored below a 63

c) A grade of 52.75 is the cutoff for being required to attend these sessions

d) The cutoff to get an A is a grade of 86.08.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 72, \sigma = 11$

(a) What percent of students scored above a 91?

This is 1 subtracted by the pvalue of Z when X = 91. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{91 - 72}{11}$

$Z = 1.73$

$Z = 1.73$ has a pvalue of 0.95682

1 - 0.9582 = 0.0482

4.18% of students scored above a 91

(b) What percent of students scored below a 63?

This is the pvalue of Z when X = 63. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{63 - 72}{11}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061.

20.61% of students scored below a 63

(c) If the lowest 4% of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?

This is the value of X when Z has a pvalue of 0.04. So it is X when Z = -1.75.

$Z = \frac{X - \mu}{\sigma}$

$-1.75 = \frac{X - 72}{11}$

$X - 72 = -1.75*11$

$X = 52.75$

A grade of 52.75 is the cutoff for being required to attend these sessions

(d) If the highest 10% of students will be given a grade of A, what is the cutoff to get an A?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 72}{11}$

$X - 72 = 1.28*11$

$X = 86.08$

The cutoff to get an A is a grade of 86.08.