Question

The mean grade in this class last semester was 78.3, and the variance was 81. The distribution of grades was unimodal and symmetrical. Using this information determine the probability that “Joe”, a random student you know nothing about, other than the fact that they are taking this class next semester, what is the probability that this student will receive a grade that is above 90

Answer 1

Using the normal distribution, it is found that there is a 0.0968 = 9.68% probability that this student will receive a grade that is above 90.

-------------------

Normal Probability Distribution  

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

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

• It measures how many standard deviations the measure is from the mean.
• Each z-score has an associated p-value, which is it's percentile.
• The probability of finding a value greater than X is 1 subtracted by the p-value of Z.

In this problem:

• Mean of 78.3 means that $\mu = 78.3$
• Standard deviation of 81 means that $\sigma = \sqrt{81} = 9$
• Probability above 90 is 1 subtracted by the p-value of Z when X = 90, thus:

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

$Z = \frac{90 - 78.3}{9}$

$Z = 1.3$

$Z = 1.3$ has a p-value of 0.9032.

1 - 0.9032 = 0.0968.

0.0968 = 9.68% probability that this student will receive a grade that is above 90.

A similar problem is given at https://brainly.com/question/24342706