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6. Descriptive Statistics
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Integrated Math 3

6.02 Measures of center

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Concept summary

A measure of center, sometimes called a measure of central tendency, describes the center, or typical, value in a data set. Measures of center include the mean, median, and mode.

Mean

The average of the numbers in a data set.

To calculate the mean, add up all the values in the data set and divide by the number of terms.

For the data set \{2, 3, 5, 5\} the mean is: \dfrac{2+3+5+5}{4}=\dfrac{15}{4} = 3.75

The weighted average can be found when not all numbers in the data set have equal frequency, or weight. To calculate the weighted average, add up each value multipled by its weight and then divide by the sum of the weights.

For a data set with unequal frequencies:

ValueFrequency
21
31
52

The weighted average is: \dfrac{2+3+5(2)}{1+1+2}=\dfrac{15}{4}=3.75

For a data set with weights:

EventValueWeight
\text{Style}8260\%
\text{Difficulty}6630\%
\text{Accuracy}9510\%

The weighted average is: \dfrac{82(0.6)+66(0.3)+95(0.1)}{0.6+0.3+0.1}=\dfrac{78.5}{1}=78.5

Median

The middle value of an ordered data set. For a set with an even number of data points, the median is the average of the two middle values.

For the data set \{2, 3, 5, 5\} the median is: 4
For the data set \{2, 3, 5, 5, 7\} the median is: 5
Mode

The most frequently occuring data value. A data set could have more than one mode or no mode.

For the data set \{2, 3, 5, 5\} the mode is 5. However, for the data set \{2, 3, 5, 6\} there is no mode.

Extreme data point

A value that differs greatly from the rest of the data set. Often called an "outlier."

The mean is largely effected by the addition or removal of an extreme data point in a data set, while the median and mode will have little to no change.

Worked examples

Example 1

On the first three tests of the semester Kobe scored 77, 72, and 83 out of 100 points.

a

Determine the score out of 100 that Kobe needs on the next test to have an average of 80 over the four tests.

Approach

We can let x represent Kobe's score on the next test and find the average as the sum of all four scores divided by 4.

Solution

\displaystyle \dfrac{77+72+83+x}{4}\displaystyle =\displaystyle 80The average of Kobe's four test scores
\displaystyle \dfrac{232+x}{4}\displaystyle =\displaystyle 80Simplify the addition
\displaystyle {232+x}\displaystyle =\displaystyle 320Multiplication property of equality
\displaystyle {x}\displaystyle =\displaystyle 88Subtraction property of equality

Kobe needs to score an 88 on his next test to have a test average of 80 over the four tests.

b

Compare what would happen to Kobe's mean and median test score if he scored a 50 out of 100 points on his fourth test.

Approach

We need to find the current mean and median of Kobe's test scores, then we can calculate the new mean and median with the score of 50 from his fourth test included to make a comparison.

Solution

Before the fourth test:

The mean is \dfrac{77+72+83}{3}=\dfrac{232}{3}\approx 77.3.

To find the median, we first put the scores in order: \{72,77,83\}. Then, we can identify the middle data point.

The median is 77.

After the fourth test:

The mean is \dfrac{77+72+83+50}{4}=\dfrac{282}{4}\approx 70.5.

The scores in order are now: \{50, 72,77,83\} and the median is in between 72 and 77.

The median is \dfrac{72+77}{2}=74.5.

The mean test score drops from 77.3 to 70.5. The median score drops from 77 to 72. Both the mean and median are lowered by the low score of 50, but the mean drops by 6.8 points and the median only changes by 2.5 points.

Reflection

The score of 50 would be an extreme data point since 50 is very different from the value of Kobe's first three scores.

Example 2

The dot plot shows the air quality index (AQI) rating for 20 of the world's most polluted countries measured in micrograms per cubic meter (µ\text{g}/\text{m}^3).

a

Identify the mean, median, and mode.

Approach

The mean is the average of all 20 data points. The median is the middle data point, or average of the two middle data points, and the mode is the most common value.

Solution

The sum of all 20 data points is 832 so the mean is \dfrac{832}{20}=41.6.

The middle two data points are 40 and 41 so the median is \dfrac{40+41}{2}=40.5.

The mode is 44 since it's the only value with three data points.

Reflection

If the data set has an even number of data points then the median will be the average of the two middle values. If the data set is odd, then the median is the middle value.

b

Interpret the meaning of the mean in the context of the data set.

Approach

The measures of center describe a "typical" value.

Solution

The typical air quality index of a polluted country is 40.6 \,µ\text{g}/\text{m}^3

Reflection

All statements about the mean and median should be sentences phrased in terms of the variable of interest, and include the correct units of measurement.

c

Explain and interpret the meaning of the median in the context of the data set.

Approach

Remember, about half the data points will be more than the median and about half the data points will be less.

Solution

Half of the 20 most polluted countries have an air quality rating less than 40.5\, µ\text{g}/\text{m}^3 and half of them have an air quality rating better than 40.5\, µ\text{g}/\text{m}^3. The typical air quality rating is 40.5\, µ\text{g}/\text{m}^3 for the 20 most polluted countries.

Example 3

In Lu's History class, the grading system is as follows:

CategoryWeight
\text{Homework}20\%
\text{Projects}30\%
\text{Tests}40\%
\text{Quizzes}10\%

Find Lu's grade average if they have a 95\% on homework, an 80\% on projects, a 75\% on tests and a 92\% on quizzes.

Approach

The weighted average is the \dfrac{\text{sum of the weighted values}}{\text{sum of the weights}}. To find the sum of weighted terms, multiply each value by its weight and add them together.

Solution

The weighted average is \dfrac{0.20(0.95)+0.30(0.80)+0.40(0.75)+0.10(0.92)}{0.20+0.30+0.40+0.10}=\dfrac{0.822}{1}=0.822=82.2\%

Lu's current grade average is 82.2\%

Outcomes

M3.N.Q.A.1

Use units as a way to understand real-world problems.*

M3.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.

M3.N.Q.A.1.D

Choose an appropriate level of accuracy when reporting quantities.

M3.S.ID.A.1

Use measures of center to solve real-world and mathematical problems.*

M3.S.ID.A.3

Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points.*

M3.MP2

Reason abstractly and quantitatively.

M3.MP3

Construct viable arguments and critique the reasoning of others.

M3.MP4

Model with mathematics.

M3.MP5

Use appropriate tools strategically.

M3.MP6

Attend to precision.

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