Unlocking the Secrets: Spread of Data and Outliers

Name	Height	Height in inches
Jahmani Hot Shot Swanson	$4^{'} 5^{''}$	$12 \cdot 4 + 5 = 53$
Jonte Too Tall Hall	$5^{'} 2^{''}$	$12 \cdot 5 + 2 = 62$
Donald Ducky Moore	$6^{'} 0^{''}$	$12 \cdot 6 + 0 = 72$
Solomon Bam Bam Bamiro	$6^{'} 5^{''}$	$12 \cdot 6 + 5 = 77$
Sean Elevator Williams	$6^{'} 1 0^{''}$	$12 \cdot 6 + 10 = 82$
Paul Tiny Sturgess	$7^{'} 8^{''}$	$12 \cdot 7 + 8 = 92$

Name

Height

Height in inches

Jahmani Hot Shot Swanson

4^{'} 5^{''}

12 \cdot 4 + 5 = 53

Jonte Too Tall Hall

5^{'} 2^{''}

12 \cdot 5 + 2 = 62

Donald Ducky Moore

6^{'} 0^{''}

12 \cdot 6 + 0 = 72

Solomon Bam Bam Bamiro

6^{'} 5^{''}

12 \cdot 6 + 5 = 77

Sean Elevator Williams

6^{'} 1 0^{''}

12 \cdot 6 + 10 = 82

Paul Tiny Sturgess

7^{'} 8^{''}

12 \cdot 7 + 8 = 92

	Mean	Median	Standard Deviation	Interquartile Range
Data A	$20.15$	$20$	$5.13$	$7$
Data B	$25.36$	$28$	$8.54$	$10$
Data C	$20.58$	$21$	$9.92$	$19$
Data D	$7.26$	$6$	$7.52$	$3$
Data E	$21.05$	$22$	$11.57$	$20$

Mean

Median

Standard Deviation

Interquartile Range

Data A

20.15

20

5.13

7

Data B

25.36

28

8.54

10

Data C

20.58

21

9.92

19

Data D

7.26

6

7.52

3

Data E

21.05

22

11.57

20

Data's Distribution	Some Observations	Preferred Statistics
Symmetric Distribution	Both the mean and the median are close to the center.	If the histogram has one peak, then both mean and median describe a typical data value. In this case, the preferred statistic is the mean, since it also considers the actual data values, not just the order.
Skewed Distribution or With Outliers	The extreme values can distort the mean and increase the standard deviation.	In this case, the preferred statistic to describe a typical data value is the median, since it only considers the order of the data. Furthermore, it is less sensitive to extreme values.

Data's Distribution

Some Observations

Preferred Statistics

Symmetric Distribution

Both the mean and the median are close to the center.

If the histogram has one peak, then both mean and median describe a typical data value. In this case, the preferred statistic is the mean, since it also considers the actual data values, not just the order.

Skewed Distribution or With Outliers

The extreme values can distort the mean and increase the standard deviation.

In this case, the preferred statistic to describe a typical data value is the median, since it only considers the order of the data. Furthermore, it is less sensitive to extreme values.

Possible Reason	Example
It can be a result of a data recording error. If this is obvious, then this data can be removed or modified.	Suppose a scientist records the length of some leaves from a tree in centimeters, measured to the nearest millimeter. In that case, a typical data entry has one decimal place. Yet, if the value $104$ appears in the data, then it is likely that it was mistakenly recorded and meant to be recorded as $10.4 .$
The nature of the data is such that unusual entries can occur. In this case, this entry should also be considered. Still, usually, the median better describes a typical data value than the mean.	When looking at a data set of the heights of Star Wars characters, one should expect to see a few extremely low and high values.

Possible Reason

Example

It can be a result of a data recording error. If this is obvious, then this data can be removed or modified.

Suppose a scientist records the length of some leaves from a tree in centimeters, measured to the nearest millimeter. In that case, a typical data entry has one decimal place. Yet, if the value

104

appears in the data, then it is likely that it was mistakenly recorded and meant to be recorded as

10.4 .

The nature of the data is such that unusual entries can occur. In this case, this entry should also be considered. Still, usually, the median better describes a typical data value than the mean.

When looking at a data set of the heights of Star Wars characters, one should expect to see a few extremely low and high values.

The mean age of five employees at a hardware store is

24

years old. A

48

year old woman is hired. By how many years did the mean age increase?

The mean of the employees' ages is calculated by dividing the sum of their ages by the number of employees. Mean=Sum of values/Number of values We know that the mean age is 24 for the 5 employees. With this information, we can determine the sum of their ages.

When the new woman is hired, the sum of the employees' ages increases by 48 and the number of people increases by 1. Let's calculate the new sum and number of values. New sum of values:& 120+48=168 New number of values:& 5+1=6 Now, we have enough information to determine the new mean of the the ages.

The new mean age is 28 years, which means that the mean increased by 28-24=4 years.

When the number

7

is added to a data set, the mean does not change. What was the mean before the observation was added?

Let's assume that the sum of the values before we add 7 is S. If the number of values in the data set is n, we can write an expression for the mean of the values. Mean_(Old) = S/n Next, if we add 7 to the values, we get a new sum of S+7 and the new number of values is n+1. With this information, we can write a second expression describing the mean of the new data set. Mean_(New) = S+7/n+1 We know that the mean before and after did not change. Therefore, we can set these expressions equal to each other.

Consequently, the mean before adding the new observation was 7.

Jordan has recorded the temperature outside her house over the course of a week. Unfortunately, two of the temperatures are covered by a coffee stain. Tadeo looks at the paper and says that they have enough information to figure out the missing values.

What was the temperature on Tuesday?

What was the temperature on Monday?

Examining the note, we see a box plot at the bottom. Since Jordan measured the temperatures over a week, she wrote 7 observations — an odd number. This means the lower quartile, median, and upper quartile must all be observations from the data set.

If we look at the note, we see that a few values from the box plot are also written in the list — the minimum value, the lower quartile, and the upper quartile.

We know that the median is represented by a value in the data set, but we cannot find it among the listed observations. Therefore, it must be one of the observation covered by the coffee stain. Also, because Jordan noted the observation on Monday as an outlier and 33^(∘)F is not an outlier, the temperature on Tuesday must have been 33^(∘) F.

In order to find the value of the outlier, we will use the fact that the mean of the 7 values is 35. If we call the value of the outlier x, we can write the following equation. 35=x+33+34+37+30+28+32/7 Let's solve this equation for x.

The temperature on Monday was 51^(∘) F. We can see that it is indeed an outlier of the data set.

Outliers

Catch-Up and Review

Effects of One Data Value's Change on a Data Set

Outlier

Extra

Identifying Outliers in a Data Set

Hint

Solution

Extra

Analyzing the Effect of Outliers on the Mean

Answer

Hint

Solution

Extra

Statistics of Data with Different Histogram Shapes

Hint

Solution

Extra

Connection Between Different Data Distributions and Best Statistics

Choosing the Best Statistics for Data Sets

Why Do Outliers Appear?

Extra

Outliers

Recommended exercises

	9 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions