Question

Suppose the grades of a certain exam are normally distributed with a mean of 72 and a standard deviation of 8. The instructor is going to award bonus points to all students whose grade is in the top 5%. What is the minimum grade a student needs to have to qualify for the bonus points

Answer 1

Answer:

The minimum grade a student needs to have to qualify for the bonus points is of 85.16.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 72 and a standard deviation of 8.

This means that $\mu = 72, \sigma = 8$

What is the minimum grade a student needs to have to qualify for the bonus points?

The 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

$1.645 = \frac{X - 72}{8}$

$X - 72 = 1.645*8$

$X = 85.16$

The minimum grade a student needs to have to qualify for the bonus points is of 85.16.