When we are given a big set of data, we would like to summarize it. The most common way to summarize a set of data is using a measure of center or a measure of central tendency.
There are three main measures of central tendency:
We will look at the mean for now.
The mean can be thought of as the balance point for a data set. Below are two applets which will allow you to explore what this means.
This applet has three data points that you can slide around. Consider the questions below.
Credit: Steve Phelps |
We should see that the mean is the balancing point, so the sum of the distances from the mean are 0.
For example, if we have the data set $4$4, $6$6 and $11$11, the mean is $7$7.
The applet below will allow us to explore how one value which is far away might change the mean. There is now only one point to slide, but you can get a new data set by clicking the button. Consider the questions below.
Credit: Steve Phelps |
The mean is the most appropriate for sets of data where there are no values which are far away from the rest of the data set. Later we will call these far away values outliers.
The mean is the balance point of a data set and is best when there are not any values which are far away from the rest.
To calculate the mean we need to add up all of the values and divide by the number of values.
For example, the mean of $2,3,5,8$2,3,5,8 is $\frac{2+3+5+8}{4}=\frac{18}{4}$2+3+5+84=184 which is $4.5$4.5 .
Calculate the mean temperature for the following temperature in degrees Fahrenheit $3,-5,6,8,-2$3,−5,6,8,−2.
Think: We need to add up all of the temperatures and divide by the number of values, $5$5.
Do:
$\frac{3+\left(-5\right)+6+8+\left(-2\right)}{5}$3+(−5)+6+8+(−2)5 | $=$= | $\frac{10}{5}$105 |
Add up all of the values |
$=$= | $2$2 |
Divide by $5$5 |
The average, mean, temperature is $5$5$^\circ$°F.
Reflect: Does an average of $2$2 make sense based on the data set?
Calculate the mean of the data set $1,1,1,1,2,2,2,3,3,3,3,3,3,9$1,1,1,1,2,2,2,3,3,3,3,3,3,9. Round to two decimal places.
Think: We need to add up all of the values and divide by the number of values, $14$14. Since we are adding the same numbers over and over, we can use multiplication to help us.
Do:
$\frac{1+1+1+1+2+2+2+3+3+3+3+3+3+9}{14}$1+1+1+1+2+2+2+3+3+3+3+3+3+914 |
The long way to calculate the mean |
|
$=$= | $\frac{4\times1+3\times2+6\times3+9}{14}$4×1+3×2+6×3+914 |
Use multiplication to be more efficient |
$=$= | $\frac{4+6+18+9}{14}$4+6+18+914 |
Multiply |
$=$= | $\frac{37}{14}$3714 |
Add |
$=$= | $2.64$2.64 |
Dividing and rounding |
Reflect: How would the mean change if the $9$9 was removed from the data set?
Find the mean of the following scores:
$-14$−14, $0$0, $-2$−2, $-18$−18, $-8$−8, $0$0, $-15$−15, $-1$−1.
In each game of the season, a basketball team recorded the number of 'three-point shots' they scored. The results for the season are represented in the given line plot.
What was the total number of points scored from three-point shots during the season?
Considering the total number of points, what was the mean number of points scored from three-point shots each game? Round to 2 decimal places if necessary.
What was the mean number of three point shots per game this season? Leave your answer to two decimal places if necessary.
The 5 numbers $11,13,9,13,9$11,13,9,13,9 have a mean of $11$11. If a new number is added that is smaller than $9$9, will the mean be higher or lower?
Higher
Lower