Question

A professor gives her 100 students an exam; scores are normally distributed. The section has an average exam score of 80 with a standard deviation of 6.5. What percentage of the class has an exam score of A- or higher (defined as at least 90)? Type your calculations along with your answer for full credit; round your final percentage to two decimal places.

Answer 1

Answer:

6.18% of the class has an exam score of A- or higher.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 80, \sigma = 6.5$

What percentage of the class has an exam score of A- or higher (defined as at least 90)?

This is 1 subtracted by the pvalue of Z when X = 90. So

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

$Z = \frac{90 - 80}{6.5}$

$Z = 1.54$

$Z = 1.54$ has a pvalue of 0.9382

1 - 0.9382 = 0.0618

6.18% of the class has an exam score of A- or higher.