Outliers in 30, 31, 33, 34, 35, 37, 60 Step by Step

Step by step

The question

Question: find the outliers in 30, 31, 33, 34, 35, 37 and 60.

30, 31, 33, 34, 35, 37, 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: 30, 31, 33, 34, 35, 37 and 60.

= 30, 31, 33, 34, 35, 37, 60

2 Find the quartiles

7 numbers: 3 on each side of the middle one.

The median is 34.

The median cuts the list in two halves, 3 numbers each.

The median itself is in neither half.

Careful!

The median is not counted in the halves.

Each half has 3 numbers.

Careful: the median is in neither half

Q1 is the middle of the lower half: 31.

Q3 is the middle of the upper half: 37.

3 Find the IQR

The IQR is Q3 minus Q1: how wide the middle half is.

37 minus 31 is 6.

IQR = Q3 − Q1 = 37 − 31 = 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

The low fence: Q1 minus 9.

31 minus 9 is 22.

low fence = Q1 − 9 = 31 − 9 = 22

The high fence: Q3 plus 9.

37 plus 9 is 46.

high fence = Q3 + 9 = 37 + 9 = 46

5 Find the outliers

On a number line: the fences in red, and a dot for each number.

60 is outside the fences.

So the outlier is 60.

= outlier: 60

6 Check

Let's check each one against its fence.

60 is more than 46.

So it is right!

60 > 46

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