Worked examples
The question
Question: find the outliers in 2, 15, 17, 18, 20, 21, 23 and 45.
2, 15, 17, 18, 20, 21, 23, 45
An outlier is a number far away from the others.
outlier: far from the others
1 Put them in order
First we put them in order, smallest first: 2, 15, 17, 18, 20, 21, 23 and 45.
= 2, 15, 17, 18, 20, 21, 23, 45
2 Find the quartiles
8 numbers: two in the middle, 18 and 20.
The median goes between them.
It cuts the list in two halves, 4 numbers each.
The median is halfway between them: we add them and divide by 2.
18 plus 20 is 38.
Divided by 2 is 19.
median = 18 + 202 = 382 = 19
The lower half has two in the middle: 15 and 17.
Q1 is halfway between them: we add them and divide by 2.
15 plus 17 is 32.
Divided by 2 is 16.
Q1 = 15 + 172 = 322 = 16
The upper half has two in the middle: 21 and 23.
Q3 is halfway between them: we add them and divide by 2.
21 plus 23 is 44.
Divided by 2 is 22.
Q3 = 21 + 232 = 442 = 22
3 Find the IQR
The IQR is Q3 minus Q1: how wide the middle half is.
22 minus 16 is 6.
IQR = Q3 − Q1 = 22 − 16 = 6
4 Find the fences
The rule: an outlier is more than 1.5 IQRs outside the box.
1.5 times 6 is 9.
1.5 × IQR = 1.5 × 6 = 9
Careful!
We step out from the ends of the box, Q1 and Q3.
Not from the median.
Careful: from Q1 and Q3, not from the median
The low fence: Q1 minus 9.
16 minus 9 is 7.
low fence = Q1 − 9 = 16 − 9 = 7
The high fence: Q3 plus 9.
22 plus 9 is 31.
high fence = Q3 + 9 = 22 + 9 = 31
5 Find the outliers
On a number line: the fences in red, and a dot for each number.
2 and 45 are outside the fences.
So the outliers are 2 and 45.
= outliers: 2 and 45
6 Check
Let's check each one against its fence.
2 is less than 7.
45 is more than 31.
So it is right!
2 < 7, 45 > 31
Let's look back.
First: put them in order.
Then: find the quartiles.
Then: find the IQR.
Then: find the fences.
Then: find the outliers.
Last: check.
Answer = outliers: 2 and 45
▶ Watch it solved
The question
Question: find the outliers in 12, 14, 15, 16, 18, 19, 20 and 22.
12, 14, 15, 16, 18, 19, 20, 22
An outlier is a number far away from the others.
outlier: far from the others
1 Put them in order
First we put them in order, smallest first: 12, 14, 15, 16, 18, 19, 20 and 22.
= 12, 14, 15, 16, 18, 19, 20, 22
2 Find the quartiles
8 numbers: two in the middle, 16 and 18.
The median goes between them.
It cuts the list in two halves, 4 numbers each.
The median is halfway between them: we add them and divide by 2.
16 plus 18 is 34.
Divided by 2 is 17.
median = 16 + 182 = 342 = 17
The lower half has two in the middle: 14 and 15.
Q1 is halfway between them: we add them and divide by 2.
14 plus 15 is 29.
Divided by 2 is 14.5.
Q1 = 14 + 152 = 292 = 14.5
The upper half has two in the middle: 19 and 20.
Q3 is halfway between them: we add them and divide by 2.
19 plus 20 is 39.
Divided by 2 is 19.5.
Q3 = 19 + 202 = 392 = 19.5
3 Find the IQR
The IQR is Q3 minus Q1: how wide the middle half is.
19.5 minus 14.5 is 5.
IQR = Q3 − Q1 = 19.5 − 14.5 = 5
4 Find the fences
The rule: an outlier is more than 1.5 IQRs outside the box.
1.5 times 5 is 7.5.
1.5 × IQR = 1.5 × 5 = 7.5
Careful!
We step out from the ends of the box, Q1 and Q3.
Not from the median.
Careful: from Q1 and Q3, not from the median
The low fence: Q1 minus 7.5.
14.5 minus 7.5 is 7.
low fence = Q1 − 7.5 = 14.5 − 7.5 = 7
The high fence: Q3 plus 7.5.
19.5 plus 7.5 is 27.
high fence = Q3 + 7.5 = 19.5 + 7.5 = 27
5 Find the outliers
On a number line: the fences in red, and a dot for each number.
Every dot is between the fences.
No number is outside the fences.
So there are no outliers.
= no outliers
6 Check
Let's check the ends: 12 is above 7, and 22 is below 27.
So it is right!
7 < 12 and 22 < 27
Let's look back.
First: put them in order.
Then: find the quartiles.
Then: find the IQR.
Then: find the fences.
Then: find the outliers.
Last: check.
▶ Watch it solved
The question
Question: find the outliers in 100, 104, 106, 108, 110, 111, 113, 150 and 60.
100, 104, 106, 108, 110, 111, 113, 150, 60
An outlier is a number far away from the others.
outlier: far from the others
1 Put them in order
First we put them in order, smallest first: 60, 100, 104, 106, 108, 110, 111, 113 and 150.
= 60, 100, 104, 106, 108, 110, 111, 113, 150
2 Find the quartiles
9 numbers: 4 on each side of the middle one.
The median is 108.
The median cuts the list in two halves, 4 numbers each.
The median itself is in neither half.
Careful!
The median is not counted in the halves.
Each half has 4 numbers.
Careful: the median is in neither half
The lower half has two in the middle: 100 and 104.
Q1 is halfway between them: we add them and divide by 2.
100 plus 104 is 204.
Divided by 2 is 102.
Q1 = 100 + 1042 = 2042 = 102
The upper half has two in the middle: 111 and 113.
Q3 is halfway between them: we add them and divide by 2.
111 plus 113 is 224.
Divided by 2 is 112.
Q3 = 111 + 1132 = 2242 = 112
3 Find the IQR
The IQR is Q3 minus Q1: how wide the middle half is.
112 minus 102 is 10.
IQR = Q3 − Q1 = 112 − 102 = 10
4 Find the fences
The rule: an outlier is more than 1.5 IQRs outside the box.
1.5 times 10 is 15.
1.5 × IQR = 1.5 × 10 = 15
The low fence: Q1 minus 15.
102 minus 15 is 87.
low fence = Q1 − 15 = 102 − 15 = 87
The high fence: Q3 plus 15.
112 plus 15 is 127.
high fence = Q3 + 15 = 112 + 15 = 127
5 Find the outliers
On a number line: the fences in red, and a dot for each number.
60 and 150 are outside the fences.
So the outliers are 60 and 150.
= outliers: 60 and 150
6 Check
Let's check each one against its fence.
60 is less than 87.
150 is more than 127.
So it is right!
60 < 87, 150 > 127
Let's look back.
First: put them in order.
Then: find the quartiles.
Then: find the IQR.
Then: find the fences.
Then: find the outliers.
Last: check.
Answer = outliers: 60 and 150
▶ Watch it solved
