The mean is often referred to as the average. To calculate the mean, add all the scores in a data set, then divide this by number of scores.
To find the mean from a graphical representation, we can use a frequency table to list out the values of on the graph. Consider the histogram below:
We can construct a frequency table like the one below:
Score ($x$x) | Frequency ($f$f) | $xf$xf |
---|---|---|
$1$1 | $3$3 | $3$3 |
$2$2 | $8$8 | $16$16 |
$3$3 | $5$5 | $15$15 |
$4$4 | $3$3 | $12$12 |
$5$5 | $1$1 | $5$5 |
The mean will be calculated by dividing the sum of the last column by the sum of the second column, $\frac{51}{20}=2.55$5120=2.55.
The median is one way of describing the middle or the centre of a data set using a single value. The median is the middle score in a data set.
Suppose we have five numbers in our data set: $4$4, $11$11, $15$15, $20$20 and $24$24.
The median would be $15$15 because it is the value right in the middle. There are two numbers on either side of it.
$4,11,\editable{15},20,24$4,11,15,20,24
If we have an even number of terms, we will need to find the average of the middle two terms. Suppose we wanted to find the median of the set $2,3,6,9$2,3,6,9, we want the value halfway between $3$3 and $6$6. The average of $3$3 and $6$6 is $\frac{3+6}{2}=\frac{9}{2}$3+62=92, or $4.5$4.5, so the median is $4.5$4.5.
$2,3,\editable{4.5},6,9$2,3,4.5,6,9
If we have a larger data set, however, we may not be able to see right away which term is in the middle. We can use the "cross out" method.
Once a data set is ordered, we can cross out numbers in pairs (one high number and one low number) until there is only one number left. Let's check out this process using an example. Here is a data set with nine numbers:
Note that this process will only leave one term if there are an odd number of terms to start with. If there are an even number of terms, this process will leave two terms instead, if you cross them all out, you've gone too far! To find the median of a set with an even number of terms, we can then take the mean of these two remaining middle terms.
The idea behind the cross out method can be used in graphical representations by cross off data points from each side.
The mode describes the most frequently occurring score.
Suppose that $10$10 people were asked how many pets they had. $2$2 people said they didn't own any pets, $6$6 people had one pet and $2$2 people said they had two pets.
In this data set, the most common number of pets that people have is one pet, and so the mode of this data set is $1$1.
A data set can have more than one mode, if two or more scores are equally tied as the most frequently occurring.
The range of a numerical data set is the difference between the smallest and largest scores in the set. The range is one type of measure of spread.
For example, at one school the ages of students in Year $7$7 vary between $11$11 and $14$14. So the range for this set is $14-11=3$14−11=3.
As a different example, if we looked at the ages of people waiting at a bus stop, the youngest person might be a $7$7 year old and the oldest person might be a $90$90 year old. The range of this set of data is $90-7=83$90−7=83, which is a much larger range of ages.
The range of a numerical data set is given by:
Range$=$=maximum score$-$−minimum score
Mean
Median
Mode
Find the range of the following set of scores:
$20,19,3,19,18,3,16,3$20,19,3,19,18,3,16,3
Answer the following given this set of scores:
$9,4,14,19,20,15,12$9,4,14,19,20,15,12
Sort the scores in ascending order.
Find the total number of scores.
Find the median.
Find the range.
Assess how various changes to data sets alter their characteristics.
Consider the set of data:
$1,2,2,4,4,5,6,6,8,8,8,9,9$1,2,2,4,4,5,6,6,8,8,8,9,9
If one score of $8$8 is changed to a $9$9, which two of the following would be altered?
Median
Mean
Range
Mode
Consider this set of data that represents the number of apps on six people’s phones.
$11,12,15,17,19,19$11,12,15,17,19,19
If each person downloads another $7$7 apps, which one of the following would not change?
Mode
Mean
Range
Median