Box and Whiskerswith Outliers

Outlier…… An extremely high or an extremely low

value in the data set when compared with the rest of the values.

The IQR is used to identify outliers.

There can be NO outliers, one outlier, or more than one outlier.

13 QQIQR

Steps to Finding the Outliers…… 1. Find Q1 and Q3.

2. Find the IQR: IQR = Q3 – Q1

3. Find Q1 – 1.5(IQR) – Low Boundary #

4. Find Q3 + 1.5(IQR) – High Boundary #

5. Check for numbers outside of this range of numbers.

5. Check the data set for any value which is smaller than Q1-1.5IQR or larger than Q3 + 1.5IQR.

Example 1 Check for outliers.

2, 7, 8, 8, 9, 10, 12, 14

The 5-number summary……

Check for a low outlier…… Q1 – 1.5(IQR) = 7.5 – 5.25 = 2.25 This is the absolute lowest value that

I can accept in my set. Anything below 2.25 would be an outlier.

2, 7, 8, 8, 9, 10, 12, 14

Therefore, 2 is an outlier.

Check for a high outlier…… Q3 + 1.5IQR = 11 + 1.5(3.5) = 16.25 This is the absolute highest value

that I can accept in my set.

2, 7, 8, 8, 9, 10, 12, 14

There is no outlier on the upper end.

Note…… Any number that lies outside the

interval between 2.25 and 16.25 is an outlier.

Therefore, 2 is an outlier.

Example 2…… Check the following set for outliers.

5, 6, 12, 13, 15, 18, 22, 50

Q1 = 9 and Q3 = 20……

Check for Low…… Q1 – 1.5IQR = 9 – 1.5(11) = -7.5

5, 6, 12, 13, 15, 18, 22, 50

Our lowest value was 5, therefore, there is no outlier on the bottom.

Check for High…… Q3 + 1.5IQR = 20 + 1.5(11) = 36.5

5, 6, 12, 13, 15, 18, 22, 50

There is 1 value that is bigger than 36.5 …….

There is one outlier: 50

How can we let the calculator draw the box plot for us?3, 8, 15, 20, 22, 23, 23, 24, 28, 29, 29, 30, 35, 38, 46

Press 2nd y= Hit enter to go to plot one and make sure it is on. Highlight the 4th graph. Set x-list for L1. Set frequency to 1. Press zoom 9. To read numbers press trace and use the cursor

keys.

Now check for outliers using the calculators….. The outliers will be shown as separate boxes.

Example 3…… Draw the box plot (with outliers) and name

the outliers.

9, 12, 15, 27, 33, 45, 63, 72

Answer…… There are not separate boxes showing.

Therefore, there are NO outliers.

Example 4…… Draw the box plot (with outliers) and name

the outliers.

400, 506, 511, 514, 517, 521

Answer…… There is a separate box showing on the left side.

Therefore, there is an outlier at 400.

Empirical Rule

Normal Distribution models give us an idea of how extreme a value is by telling us how likely it is to find one that far from the mean

We need one simple rule…..The Empirical Rule or the 68-95-99.7 Rule.

It turns out that………..

Empirical Rule…… You can only use when the

variable is normally distributed.

Most values are within 3 st. deviations of the mean.

Memorize

St. Dev %

 

1 68%

2 95%

3 99.7%

Example 1…… In a normal

distribution, 95% of the data will fall between what 2 values if

518 sandx

St. Dev %

 

1 68%

2 95%

3 99.70%

Using the empirical rule, 95% is within 2 st. dev.

St. Dev %

 

1 68%

2 95%

3 99.70%

Range of Values…… Mean – 2 St. Dev. 18 – 2(5) = 8

Mean + 2 St. Dev. 18 + 2(5) = 28

Range = 8 - 28

Example 2…… In a normal

distribution, 99.7% of the data will fall between what 2 values if

5.118 sandx

St. Dev %

 

1 68%

2 95%

3 99.70%

Answer……. Mean – 3 St. Dev.

18 – 3(1.5) = 18 – 4.5 = 13.5

Mean + 3 St. Dev. 18 + 3(1.5) = 22.5

Range: 13.5 – 22.5

Now You Try….. In a normal

distribution, 68% of the data will fall between what 2 values if

220 sandx

St. Dev %

 

1 68%

2 95%

3 99.70%

Answer……. Mean – 1 St. Dev.

20 – 1(2) = 20 – 2 = 18

Mean + 1 St. Dev. 20 + 1(2) = 22

Range: 18 - 22

Finding the Percentage….. The mean value of a distribution is 70 and

the st. dev. is 5. What % of the values falls between

a. 60 and 80 b. 65 and 75

a. z = (80-70)/5 =

z = 10/5 = 2

2 st. deviations = 95%

b. Z = (75-70)/5 z = 5/5 z = 1

1 st. deviation =68%

Now You Try. Find the Percentage. The mean value of a distribution is 70 and

the st. dev. is 5. What % of the values falls between 55 and 85?

Answer z = (85-70)/5 =

z = 15/5 = 3

3 st. deviations = 99.7%