Exercises on Measures of Central Tendency

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Given each of the questions below, select the most appropriate response. Label your answers A, B, C or D.

Please enter your name

1. Given the data set {0, 3, 3, 4, 11, 0, 9, 0, 5}, identify the mode.

The mode is 0.
The mode is 3.
The mode does not exist.


2. The data set {6, 2, 8, 2, 5, 2, 2, 3, 7, 5, 9, 5, 8, 7, 7, 2, 2, 1}, represents a list of baseball scores for games played on Monday, September 1, 2000. Identify the (arithmetic) mean number of runs scored that day.
The mean is 83.
The mean is 4.7.
The mean is 4.6.
The mean does not exist.


3. Given the data set of baseball scores above, identify the median.

The median is 3.
The median is 4.
The median is 5.
The median does not exist.


4. Given the data set {8, 12, 0, 37, 12}, identify the median.

The median is 0.
The median is 12.
The median is 37.


5. If a data set has outliers, which measure of central tendency is most appropriate.

The mean.
The median.
The mode.
It doesn't matter one way or the other.


6. Which is the only measure of central tendency that can be used with nominal data types.

The mean.
The median.
The mode.
Measures of central tendency do not apply to qualitative data types.


7. In grading an elementary Statistics course, an instructor gives a weight of 20% for each exam, 10% for computer projects and 10% for class participation . If a student received grades of 75, 88, 82 and 94 on each of four exams, a grade of 100 for computer projects and a class participation grade of 87, which of the following letter grades will the student receive for the course?

For a grading scheme, consider 90+ to be an A, 80 - 89 a B, 70 - 79 a C, 60 - 69 a D and below 60 an F.

A letter grade of A.
A letter grade of B.
A letter grade of C.
A letter grade of D.
A letter grade of F.


8. One of the problems with the mean as a measure of central tendency was the fact that it can easily be swayed by outliers. One way around this problem is by using a trimmed mean where we take off a certain percentage of the scores from both the high and low end of the data set, thereby eliminating the outliers, and calculate the mean of what is left. Taking a 10% trimmed mean (that is take 10% of the scores off the high end and 10% of the scores off the low end, determine the mean of the following data set {5, 11, 15, 27, 30, 38, 41, 42, 57, 63}.

The mean is 32.6.
The mean is 26.1.
The mean is 29.
None of the above.


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Copyright © 2004 Donna Tupper.