This problem also requires you to use the z score
like you did in Problem 16. This problem also comes from Section. You will need to use a table of z scores (supplied on the Final Exam) to do part c.
Problem 17 Suppose the mean score on a test is 76 with a standard deviation of 8.
a. What is the z-score for a student who scored a 90 on the test?
Solution In part a, you were given a data value and asked to find the z score. This problem does the opposite. Put the values into the z score formula:
c. If the test scores are normally distributed, then what percent of the students would score between 80 and 89 on the test?
Solution Graphically, we are trying to determine the portion of area under a bell curve from 80 to 89.
To get areas, we need to calculate the z values and read the area from the z table formula.
For a score of x = 80,
And for x = 89,
A z score of 0.5 corresponds to an area of 0.192 from the mean of 76 to 80. A z value of 1.63 corresponds to an area of .449 from the mean of 76 to 89. The area between 80 and 89 is the difference between these areas or .449 – .192 or .257.