Introduction Percentiles

Chapter 3.4Measures of position and outliers

Julian Chan

Department of MathematicsWeber State University

September 11, 2011

Introduction Percentiles

Intro

1 We will talk about how to measure the position of an observation whichdescribes it’s relative position with in the entire data set.

2 The object which allows us to do this is called the Z-score.

3 The Z-score measures how far away your observed data point is from themean.

Introduction Percentiles

Intro

1 We will talk about how to measure the position of an observation whichdescribes it’s relative position with in the entire data set.

2 The object which allows us to do this is called the Z-score.

3 The Z-score measures how far away your observed data point is from themean.

Introduction Percentiles

Intro

1 We will talk about how to measure the position of an observation whichdescribes it’s relative position with in the entire data set.

2 The object which allows us to do this is called the Z-score.

3 The Z-score measures how far away your observed data point is from themean.

Introduction Percentiles

Intro

1 The difference x − µ is a measure of how far away your observed value isfrom the mean.

2 If we divide this by our standard deviation we have expressed how faraway our observation is from the mean in terms of standard deviation.

Introduction Percentiles

Intro

1 The difference x − µ is a measure of how far away your observed value isfrom the mean.

2 If we divide this by our standard deviation we have expressed how faraway our observation is from the mean in terms of standard deviation.

Introduction Percentiles

Z-score

1 This is called the Z-score and is given in two forms

2 Population Z-score:

z =x − µ

σ.

3 Sample Z-score:

z =x − x̄

S.

4 The Z-score is unitless. It has a mean of 0 and standard deviation of 1.

Introduction Percentiles

Z-score

1 This is called the Z-score and is given in two forms

2 Population Z-score:

z =x − µ

σ.

3 Sample Z-score:

z =x − x̄

S.

4 The Z-score is unitless. It has a mean of 0 and standard deviation of 1.

Introduction Percentiles

Z-score

1 This is called the Z-score and is given in two forms

2 Population Z-score:

z =x − µ

σ.

3 Sample Z-score:

z =x − x̄

S.

4 The Z-score is unitless. It has a mean of 0 and standard deviation of 1.

Introduction Percentiles

Z-score

1 This is called the Z-score and is given in two forms

2 Population Z-score:

z =x − µ

σ.

3 Sample Z-score:

z =x − x̄

S.

4 The Z-score is unitless. It has a mean of 0 and standard deviation of 1.

Introduction Percentiles

Examples

1 In 2007 the Yankee’s lead the American league with 968 runs scored,while the Phillies led the national league with 892 runs.

2 A natural question is which team is better at scoring run?

Introduction Percentiles

Examples

1 In 2007 the Yankee’s lead the American league with 968 runs scored,while the Phillies led the national league with 892 runs.

2 A natural question is which team is better at scoring run?

Introduction Percentiles

Examples

1 It is tempting to say that the Yankees are the better team at scoring runs,but the Phillies are in the National league where there is no designatedhitter. Instead the pitcher (usually a weak hitter) has to bat in place of astronger batter.

2 We will determine which team is better at scoring runs by determiningwhich team is better in their respective leagues according to their Z-scoreor in terms of their respive standard deviation. This is equivalent todetermining to proportion of teams in their respective leagues which scoreat least the specified number of runs.

Introduction Percentiles

Examples

1 It is tempting to say that the Yankees are the better team at scoring runs,but the Phillies are in the National league where there is no designatedhitter. Instead the pitcher (usually a weak hitter) has to bat in place of astronger batter.

2 We will determine which team is better at scoring runs by determiningwhich team is better in their respective leagues according to their Z-scoreor in terms of their respive standard deviation. This is equivalent todetermining to proportion of teams in their respective leagues which scoreat least the specified number of runs.

Introduction Percentiles

Examples

1 We can compute the population mean and standard deviation for eachleague.

2 We find that µA = 793.9, µN = 763, and for the standard deviationσA = 73.5, σN = 58.9.

Introduction Percentiles

Examples

1 We can compute the population mean and standard deviation for eachleague.

2 We find that µA = 793.9, µN = 763, and for the standard deviationσA = 73.5, σN = 58.9.

Introduction Percentiles

Examples

1 We can compute the Z-score for the Yankees as:

z =x − µA

σA=

968 − 793.9

73.5= 2.73

2 We can compute the Z-score for the Phillies as:

z =x − µN

σN=

892 − 763

58.9= 2.19

3 The Yankees scored 2.37 standard deviations above the mean of runsscored while the Phillies only scored 2.19 standard deviations above themean number of runs scored. We conclude that the Yankees are better atscoring runs.

Introduction Percentiles

Examples

1 We can compute the Z-score for the Yankees as:

z =x − µA

σA=

968 − 793.9

73.5= 2.73

2 We can compute the Z-score for the Phillies as:

z =x − µN

σN=

892 − 763

58.9= 2.19

3 The Yankees scored 2.37 standard deviations above the mean of runsscored while the Phillies only scored 2.19 standard deviations above themean number of runs scored. We conclude that the Yankees are better atscoring runs.

Introduction Percentiles

Examples

1 We can compute the Z-score for the Yankees as:

z =x − µA

σA=

968 − 793.9

73.5= 2.73

2 We can compute the Z-score for the Phillies as:

z =x − µN

σN=

892 − 763

58.9= 2.19

3 The Yankees scored 2.37 standard deviations above the mean of runsscored while the Phillies only scored 2.19 standard deviations above themean number of runs scored. We conclude that the Yankees are better atscoring runs.

Introduction Percentiles

Examples

1 Suppose the the population mean of cell phone calls is 15 minutes, andthe standard deviation is 5 minutes.

2 You have two friends John and Lisa. John talks to you on the phone for20 minutes, and Lisa for 25.

Introduction Percentiles

Examples

1 Suppose the the population mean of cell phone calls is 15 minutes, andthe standard deviation is 5 minutes.

2 You have two friends John and Lisa. John talks to you on the phone for20 minutes, and Lisa for 25.

Introduction Percentiles

Examples

1 We can compute the Z-score for John’s phone call as:

z =x − µA

σA=

20 − 15

5= 1

2 We can compute the Z-score for Lisa as:

z =x − µN

σN=

25 − 15

5= 2

3 The 68 percent rule says that 68 percent of phone calls will last between10 to 20 minutes.

4 The 95 percent rule says that 95 percent of phone calls will last between5 to 25 minutes.

Introduction Percentiles

Examples

1 We can compute the Z-score for John’s phone call as:

z =x − µA

σA=

20 − 15

5= 1

2 We can compute the Z-score for Lisa as:

z =x − µN

σN=

25 − 15

5= 2

3 The 68 percent rule says that 68 percent of phone calls will last between10 to 20 minutes.

4 The 95 percent rule says that 95 percent of phone calls will last between5 to 25 minutes.

Introduction Percentiles

Examples

1 We can compute the Z-score for John’s phone call as:

z =x − µA

σA=

20 − 15

5= 1

2 We can compute the Z-score for Lisa as:

z =x − µN

σN=

25 − 15

5= 2

3 The 68 percent rule says that 68 percent of phone calls will last between10 to 20 minutes.

4 The 95 percent rule says that 95 percent of phone calls will last between5 to 25 minutes.

Introduction Percentiles

Examples

1 We can compute the Z-score for John’s phone call as:

z =x − µA

σA=

20 − 15

5= 1

2 We can compute the Z-score for Lisa as:

z =x − µN

σN=

25 − 15

5= 2

3 The 68 percent rule says that 68 percent of phone calls will last between10 to 20 minutes.

4 The 95 percent rule says that 95 percent of phone calls will last between5 to 25 minutes.

Introduction Percentiles

Examples

1 The 68 percent rule implies that 16 percent of phone calls will last morethan 20 minutes or more. Hence only 16 percent of the population talksas long as John does or more.

2 The 95 percent rule implies that only 2.5 percent of phone calls will last25 minutes or more. Hence only 2.5 percent of the population will talk aslong as Lisa does or more.

Introduction Percentiles

Examples

1 The 68 percent rule implies that 16 percent of phone calls will last morethan 20 minutes or more. Hence only 16 percent of the population talksas long as John does or more.

2 The 95 percent rule implies that only 2.5 percent of phone calls will last25 minutes or more. Hence only 2.5 percent of the population will talk aslong as Lisa does or more.

Introduction Percentiles

Percentiles

1 The Median divides the lower 50 percent of the data set from the upper50 percent of the data set.

2 The K th percentile of a data set denoted by Pk is a value suck that kpercent of the observations are less than or equal to to the value.

Introduction Percentiles

Percentiles

1 The Median divides the lower 50 percent of the data set from the upper50 percent of the data set.

2 The K th percentile of a data set denoted by Pk is a value suck that kpercent of the observations are less than or equal to to the value.

Introduction Percentiles

Percentiles

1 The 15th percentile of the head circumference of males 3 to 5 months ofage is 41 centimeters.

2 This means that 15 percent of males 3 to 5 months have a headcircumference of less than 41 centimeters, and 85 percent have a headcircumference of larger than 41 centimeters.

Introduction Percentiles

Percentiles

1 The 15th percentile of the head circumference of males 3 to 5 months ofage is 41 centimeters.

2 This means that 15 percent of males 3 to 5 months have a headcircumference of less than 41 centimeters, and 85 percent have a headcircumference of larger than 41 centimeters.

Introduction Percentiles

Quartiles

1 Here is how to find the quartiles of your data set

2 Arrange your data in ascending order.

3 Determine the median.

4 Determine the first and third quartiles Q1 and Q3, by dividing the datasets into two halves. The bottom half will be the observations below themedian and the top the observations above the median. The median ofthe lower half is Q1 and the median of the upper half is Q3.

Introduction Percentiles

Quartiles

1 Here is how to find the quartiles of your data set

2 Arrange your data in ascending order.

3 Determine the median.

4 Determine the first and third quartiles Q1 and Q3, by dividing the datasets into two halves. The bottom half will be the observations below themedian and the top the observations above the median. The median ofthe lower half is Q1 and the median of the upper half is Q3.

Introduction Percentiles

Quartiles

1 Here is how to find the quartiles of your data set

2 Arrange your data in ascending order.

3 Determine the median.

4 Determine the first and third quartiles Q1 and Q3, by dividing the datasets into two halves. The bottom half will be the observations below themedian and the top the observations above the median. The median ofthe lower half is Q1 and the median of the upper half is Q3.

Introduction Percentiles

Quartiles

1 Here is how to find the quartiles of your data set

2 Arrange your data in ascending order.

3 Determine the median.

4 Determine the first and third quartiles Q1 and Q3, by dividing the datasets into two halves. The bottom half will be the observations below themedian and the top the observations above the median. The median ofthe lower half is Q1 and the median of the upper half is Q3.

Introduction Percentiles

Example

1 The data .97, 1.14, 1.85, 2.34, 2.47, 2.78, 3.41 3.48 represent the 8months of rain in Chicago.

2 We already have the data arranged in order so we find the median. Wecompute that the median is 2.34+2.47

2= 2.405

3 We find that the lower data set is .97, 1.14, 1.85, 2.34 which has amedian of Q1 = 1.14+1.85

2= 1.495.

We find that the upper data set is 2.47, 2.78, 3.41 3.48, and the medianof this data set is Q3 = 2.78+3.41

2= 3.095.

Introduction Percentiles

Example

1 The data .97, 1.14, 1.85, 2.34, 2.47, 2.78, 3.41 3.48 represent the 8months of rain in Chicago.

2 We already have the data arranged in order so we find the median. Wecompute that the median is 2.34+2.47

2= 2.405

3 We find that the lower data set is .97, 1.14, 1.85, 2.34 which has amedian of Q1 = 1.14+1.85

2= 1.495.

We find that the upper data set is 2.47, 2.78, 3.41 3.48, and the medianof this data set is Q3 = 2.78+3.41

2= 3.095.

Introduction Percentiles

Example

1 The data .97, 1.14, 1.85, 2.34, 2.47, 2.78, 3.41 3.48 represent the 8months of rain in Chicago.

2 We already have the data arranged in order so we find the median. Wecompute that the median is 2.34+2.47

2= 2.405

3 We find that the lower data set is .97, 1.14, 1.85, 2.34 which has amedian of Q1 = 1.14+1.85

2= 1.495.

We find that the upper data set is 2.47, 2.78, 3.41 3.48, and the medianof this data set is Q3 = 2.78+3.41

2= 3.095.

Introduction Percentiles

IQR

1 The interquartile range denoted by IQR is the range of the middle 50percent of the observations in a data set. That is the IQR is given by theformula

IQR = Q3 − Q1

2 In our last example we can find the IQR = 3.095 − 2.405 = .69

Introduction Percentiles

IQR

1 The interquartile range denoted by IQR is the range of the middle 50percent of the observations in a data set. That is the IQR is given by theformula

IQR = Q3 − Q1

2 In our last example we can find the IQR = 3.095 − 2.405 = .69

Introduction Percentiles

Outliers

1 An outlier is an extreem observation.

2 For example if we take a simple random sample of hourly salsries andfound the data collected was given by 15, 20, 21, 23, 25, 100. Once cansee that 100 is an outlier.

Introduction Percentiles

Outliers

1 An outlier is an extreem observation.

2 For example if we take a simple random sample of hourly salsries andfound the data collected was given by 15, 20, 21, 23, 25, 100. Once cansee that 100 is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.

Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)

Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Outliers

1 We can check for outliers using quartiles.

2 First determine the first and third quartiles of the data set.

3 Compute the IQR

4 Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR)Upper fence = Q3 − 1.5(IQR)

5 If a data value is less than the lower or upper fence it is an outlier.

Introduction Percentiles

Example

1 In our last example we had

Determine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR) = 2.405 − 1.5(.69) = 1.37Upper fence = Q3 − 1.5(IQR) = 3.095 + 1.5(.69) = 4.13

2 We thus have two outliers! They are the observations .97 and 1.14.

Introduction Percentiles

Example

1 In our last example we hadDetermine the lower fence and upper fence.

Lower fence = Q1 − 1.5(IQR) = 2.405 − 1.5(.69) = 1.37Upper fence = Q3 − 1.5(IQR) = 3.095 + 1.5(.69) = 4.13

2 We thus have two outliers! They are the observations .97 and 1.14.

Introduction Percentiles

Example

1 In our last example we hadDetermine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR) = 2.405 − 1.5(.69) = 1.37

Upper fence = Q3 − 1.5(IQR) = 3.095 + 1.5(.69) = 4.13

2 We thus have two outliers! They are the observations .97 and 1.14.

Introduction Percentiles

Example

1 In our last example we hadDetermine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR) = 2.405 − 1.5(.69) = 1.37Upper fence = Q3 − 1.5(IQR) = 3.095 + 1.5(.69) = 4.13

2 We thus have two outliers! They are the observations .97 and 1.14.

Introduction Percentiles

Example

1 In our last example we hadDetermine the lower fence and upper fence.Lower fence = Q1 − 1.5(IQR) = 2.405 − 1.5(.69) = 1.37Upper fence = Q3 − 1.5(IQR) = 3.095 + 1.5(.69) = 4.13

2 We thus have two outliers! They are the observations .97 and 1.14.