IQR and MAD
Interquartile Range
The interquartile range (IQR) is the difference between the upper quartile and the lower quartile.
In other words, $IQR=\text{upper quartile}-\text{lower quartile}$IQR=upper quartile−lower quartile.
The IQR is best seen in box plots, like the one below.
Calculating the IQR of a data set
Calculate the IQR of $-6,6,2,-1,10,-8,-2,6,9$−6,6,2,−1,10,−8,−2,6,9.
1. Sort the scores in ascending order.
$-8,-6,-2,-1,2,6,6,9,10$−8,−6,−2,−1,2,6,6,9,10
2. Find the number of scores.
There are $9$9 scores in this set.
3. Find the median.
The median is the $\frac{n+1}{2}$n+12th score, where $n$n is the number of scores.
$\frac{9+1}{2}=5$9+12=5
The fifth score in our list is $2$2, so the median is $2$2.
4. Find the first quartile of the set of scores.
The first (lower) quartile is the $\frac{n+1}{4}$n+14th score, where $n$n is the number of scores.
$\frac{9+1}{4}=2.5$9+14=2.5
The $2.5$2.5th score is the average of the second and third scores.
$\frac{-6+\left(-2\right)}{2}=-4$−6+(−2)2=−4
The first quartile is $-4$−4.
5. Find the third quartile of the set of scores.
The third (upper) quartile is the $\frac{3\left(n+1\right)}{4}$3(n+1)4th score, where $n$n is the number of scores.
$\frac{3\left(9+1\right)}{4}=7.5$3(9+1)4=7.5
The $7.5$7.5th score is the average of the seventh and eighth scores.
$\frac{6+9}{2}=7.5$6+92=7.5
So the third quartile is $7.5$7.5.
6. Find the interquartile range.
$7.5-\left(-4\right)=11.5$7.5−(−4)=11.5
Mean Absolute Deviation (MAD)
The Mean Absolute Deviation (MAD) of a set of data is the average distance between the scores and the mean.
Calculating the MAD of a data set
Let's use an example to help explain this.
Find the mean absolute deviation of $23,18,31,28,20$23,18,31,28,20.
1. Find the mean.
$Mean=\frac{23+18+31+28+20}{5}$Mean=23+18+31+28+205$=$=$24$24
2. Find the difference between each individual score and the mean.
$23-24=-1$23−24=−1
$18-24=-6$18−24=−6
$31-24=7$31−24=7
$28-24=4$28−24=4
$20-24=-4$20−24=−4
3. Take the absolute value of each difference.
$\left|-1\right|=1$|−1|=1
$\left|-6\right|=6$|−6|=6
$\left|7\right|=7$|7|=7
$\left|4\right|=4$|4|=4
$\left|-4\right|=4$|−4|=4
4. Find the mean of these differences.
$Mean=\frac{1+6+7+4+4}{5}$Mean=1+6+7+4+45$=$=$4.4$4.4
This means that, on average, scores in this data set are $4.4$4.4 units above or below the mean.
Worked Examples
Question 1
Calculate the mean absolute deviation of the following numbers.
$41$41, $27$27, $82$82, $81$81, $54$54, $123$123.
First, state the mean and complete the following table for the data.
Mean $=$= $\editable{}$
Data Absolute difference $41$41 $\editable{}$ $27$27 $\editable{}$ $82$82 $\editable{}$ $81$81 $\editable{}$ $54$54 $\editable{}$ $123$123 $\editable{}$ Calculate the mean absolute deviation for your data to two decimal places.
Question 2
Answer the following, given this set of scores:
$33,38,50,12,33,48,41$33,38,50,12,33,48,41
Sort the scores in ascending order.
Find the number of scores.
Find the median.
Find the first quartile of the set of scores.
Find the third quartile of the set of scores.
Find the interquartile range.