Question

Professor Halen teaches a College Mathematics class. The scores on the midterm exam are normally distributed with a mean of 72.3 and a standard deviation of 8.9. What is the probability that a student scores between 82 and 90

Answer 1

Answer:

14.63% probability that a student scores between 82 and 90

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:

$\mu = 72.3, \sigma = 8.9$

What is the probability that a student scores between 82 and 90?

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So

X = 90

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

$Z = \frac{90 - 73.9}{8.9}$

$Z = 1.81$

$Z = 1.81$ has a pvalue of 0.9649

X = 82

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

$Z = \frac{82 - 73.9}{8.9}$

$Z = 0.91$

$Z = 0.91$ has a pvalue of 0.8186

0.9649 - 0.8186 = 0.1463

14.63% probability that a student scores between 82 and 90