DESCRIPTIVE STATISTICS FOR ALGEBRA I

Mean, Standard Deviation, Mean Absolute Deviation and Z-scores

Mean

Mean = the average of the numbers. To find the mean of a set of numbers, you must add up the numbers then divide by the number of numbers. The µ symbol is used to represent mean in a formula.

Example: {18 23 10 39 22 17 16 15}

18+23+10+39+22+17+16+15 = 160 160

= 208

Mean Practice

Find the mean of the following sets of numbers.

{20 30 15 40 20 80 60 50}

{34 20 56 82 78 30}

Mean Practice

20 + 30 + 15 + 40 + 20 + 80 + 60 + 50 = 320

320

34 + 20 + 56 +82 + 78 + 30 = 300300

8 = 40

6 = 50

Standard Deviation

The standard deviation is the positive square root of the variance of the data set. The greater the value of the standard deviation, the more spread out the data are about the mean. The lesser (closer to 0) the value of the standard deviation, the closer the data are clustered about the mean. The symbol is used to represent standard deviation in a formula.

σ

Mean and Standard Deviation

Computation of descriptive statistics using the graphing calculator.

From the menu screen select STAT and press EXE.

{40, 30, 50, 30, 50, 50, 60, 50, 30, 40, 50, 40, 60, 50}

In List 1, enter the data set. Enter the first element and press EXE to move to the next line and continue until all the elements have been entered into List 1.

Press F2 (CALC) Press F1 (1VAR) 9x = Mean of the data = 45 x n-1 = Standard deviation = 10.2

σ

Mean Absolute Deviation

The mean absolute deviation is the arithmetic mean of the absolute values of the deviations of elements from the mean of a data set.

Mean Absolute Deviation Computation Method 1

Continuing from the calculation of standard deviation, press EXIT until the function button menus read

“GRAPH CALC TEST INTR DIST” Press ► to move the cursor to List 2. Press ▲ until “List 2” is highlighted. Press OPTN F4 (NUM) F1 (ABS) Press ( OPTN F1 F1 (to get “List”) 1 - VARS F3 F5 F2 )

EXE (List 2 will automatically fill with data). Continue to press EXIT until the function button menus read

“GRAPH CALC TEST INTR DIST” Press F2 (CALC) F6 (SET) F2 (List 2) Press EXE EXIT Press F2 (CALC) F1 (1 VAR)X = mean absolute deviation

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Mean Absolute Deviation Computation Method 2

Continuing from the calculation of standard deviation, press EXIT until the function button menus read

“GRAPH CALC TEST INTR DIST”.

Press ► to move the cursor to List 2. Enter the absolute value of the difference

between the data value and the mean for all the data values in List 1.

Press F2 (CALC) F6 (SET) F2 (List 2) Press EXE EXIT Press F2 (CALC) F1 (1 VAR)X = mean absolute deviation

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Mean Absolute Deviation

The mean absolute deviation for the data{40, 30, 50, 30, 50, 50, 60, 50, 30, 40, 50, 40,

60, 50}is 8.57142857.

Mean and Standard Deviation

What is the mean score?

What is the standard deviation?

Here are 20 test scores from Mr. Barnes algebra class.

83 82 79 72 84 73 94 77 81 74 68 87

89 80 90 84 83 81 75 78

Mean and Standard Deviation

Here are 20 test scores from Mr. Barnes algebra

class.

What is the mean score? 80.7

What is the standard deviation? 6.5

83 82 79 72 84 73 94 77 81 74 68 87

89 80 90 84 83 81 75 78

Mean Absolute Deviation

What is the mean absolute deviation?

Here are 20 test scores from Mr. Barnes algebra class.

83 82 79 72 84 73 94 77 81 74 68 87

89 80 90 84 83 81 75 78

Mean Absolute Deviation

What is the mean absolute deviation? 5.03

Here are 20 test scores from Mr. Barnes algebra class.

83 82 79 72 84 73 94 77 81 74 68 87

89 80 90 84 83 81 75 78

WHERE DO I STAND?

Z-Scores

Z-score

Position of a data value relative to the mean.

Tells you how many standard deviations above or below the mean a particular data point is.

μ

σ

ix

z-score = describes the location of a data value within a distribution

referred to as a standardized value

Population

= Z-SCORE

Z-score

In order to calculate a z-score you must know:

μ

σ

ix• a data value = X

• the mean = µ

• the standard deviation =

i

σ

Z-scores

The bold score is Michael’s. How did he perform relative to his classmates?

Here are 18 test scores from Mr. Barnes algebra class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

The bold score is Michael’s. How did he perform relative to his classmates?

Michael’s score is “above average”, but how much above average is it?

Here are 18 test scores from Mr. Barnes stat class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

If we convert Michael’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean.

What we need:

• Michele’s score

• mean of test scores

• standard deviation

93

80.2

5.8

= 2.2

Therefore, Michael’s standardized test score is 2.2, over two standard deviations above the class mean.

μ

σ

ix

93 – 80.2 5.8

Z=

Z-score Number Line

80.2

x x σ86.0

2x σ91.8

3x σ97.6

x σ74.4

2x σ68.6

3x σ

62.8

0 1 2 3-1-2-3

Michael’s Score = 93

Michael’s z-score = 2.2

Z-scores

The bold score is Bonnie’s. How did she perform relative to her classmates?

Here are 18 test scores from Mr. Barnes algebra class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

The bold score is Bonnie’s. How did she perform relative to her classmates?

Bonnie’s score is “below average”, but how much above average is it?

Here are 18 test scores from Mr. Barnes stat class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

If we convert Michael’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean.

What we need:

• Michele’s score

• mean of test scores

• standard deviation

67

80.2

5.8

= -2.3

Therefore, Bonnie’s standardized test score is -2.3, over two standard deviations below the class mean.

μ

σ

ix

67 – 80.2 5.8

Z=

Z-score Number Line

80.2

x x σ86.0

2x σ91.8

3x σ97.6

x σ74.4

2x σ68.6

3x σ

62.8

0 1 2 3-1-2-3

Bonnie’s Score = 67

Bonnie’s z-score = -2.3

Z-scores

The bold score is Ray’s. How did he perform relative to his classmates?

Here are 18 test scores from Mr. Barnes algebra class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

The bold score is Ray’s. How did he perform relative to his classmates?

Ray’s score is slightly “below average”, but how much below average is it?

Here are 18 test scores from Mr. Barnes stat class.

83 84 89 83 79 83 77 73 67 75

80 78 93 83 77 80 81 79

Z-scores

If we convert Ray’s score to a standardized value, then we can determine how many standard deviations his score is away from the mean.

What we need:

• Ray’s score

• mean of test scores

• standard deviation

80

80.2

5.8

= -0.03

Therefore, Ray’s standardized test score is -0.03, slightly below the class mean.

μ

σ

ix

80 – 80.2 5.8

Z=

Z-score Number Line

80.2

x x σ86.0

2x σ91.8

3x σ97.6

x σ74.4

2x σ68.6

3x σ

62.8

0 1 2 3-1-2-3

Ray’s Score = 80

Ray’s z-score = -0.03

Practice Problem 1

Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

56 54 52 45 59 49 55

Which students’ heights have a z-score greater than 1?

A) All of themB) Ann, Bill, Caryn, Ken and Gary

C) Ken

D) Don and Fred

Practice Problem 1

Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

56 54 52 45 59 49 55

Which students’ heights have a z-score greater than 1?

A) All of themB) Ann, Bill, Caryn, Ken and Gary

C) Ken

D) Don and Fred

Mean = 53

Standard Deviation = 4.7

X – 53 4.7

X > 57.7

> 1

Practice Problem 2

Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

55 44 50 47 56 49 55

Which students have a z-score less than -1?

A) All of them

B) Bill, Don and Fred

C) Only Bill

D) Ann, Caryn, Ken and Gary

Practice Problem 2

Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

55 44 50 47 56 49 55

Which students have a z-score less than -1?

A) All of them

B) Bill, Don and Fred

C) Only Bill

D) Ann, Caryn, Ken and Gary

Mean = 51

Standard Deviation = 4.6

X – 51 4.6

< -1

X < 46

Practice Problem 3Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

56 57 48 46 42 50 51

Which student’s height has a z-score of zero?

A) Bill

B) Fred

C) Caryn

D) None of them

Practice Problem 3

Student Ann Bill Caryn Don Ken Fred Gary

Height(in inches)

56 57 48 46 42 50 51

Which student’s height has a z-score of zero?

A) Bill

B) Fred

C) Caryn

D) None of them

Mean = 50

Standard Deviation = 5.3

X – 50 5.3

= 0

X = 50

Practice Problem 4

Given a data set with a mean of 135 and a standard deviation of 10, describe the z-score of a data value of 100?

A) Less than -5

B) Between -1 and 0

C) Between -5 and -1

D) Greater than 0

Practice Problem 4

Given a data set with a mean of 135 and a standard deviation of 10, describe the z-score of a data value of 100?

A) Less than -5

B) Between -1 and 0

C) Between -5 and -1

D) Greater than 0

Mean = 135

Standard Deviation = 10

100 – 135 10

Z =

Z = - 3.5

Practice Problem 5

Given a data set with a mean of 35 and a standard deviation of 2.5, find the data value associated with a z-score of 3?

A) 46

B) 45

C) 44.5

D) 42.5

Practice Problem 5

Given a data set with a mean of 35 and a standard deviation of 2.5, find the data value associated with a z-score of 3?

A) 46

B) 45

C) 44.5

D) 42.5

Mean = 35

Standard Deviation = 2.5

X – 35 2.5

= 3

X = 42.5

Practice Problem 6

Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}1) List the z-scores of students that were above the mean.

Practice Problem 6

Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}1) List the z-scores of students that were above the mean. 2.5, 1.75, and 0.4

Practice Problem 7

Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}

2) If the mean of the exam is 85, did any of the students selected have an exam score of 85? Explain.

Practice Problem 7Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}

2) If the mean of the exam is 85, did any of the students selected have an exam score of 85? Explain. One student with a z-score of 0.

Practice Problem 8

Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}

3) If the standard deviation of the exam was 4 and the mean is 85, what was the actual test score for the student having a z-score of 1.75?

Practice Problem 8

Suppose the test scores on the last exam in Algebra I are normally distributed. The z-scores for some of the students in the course were:

{2.5, 0, -3.2, -2.1, 1.75, 0.4}

3) If the standard deviation of the exam was 4 and the mean is 85, what was the actual test score for the student having a z-score of 1.75?

X – 85

4= 1.75

X = 92

Practice Problem 9

A set of mathematics exam scores has a mean of 70 and a standard deviation of 8. A set of English exam scores has a mean of 74 and a standard deviation of 16. For which exam would a score of 78 have a higher standing?

Practice Problem 9

A set of mathematics exam scores has a mean of 70 and a standard deviation of 8. A set of English exam scores has a mean of 74 and a standard deviation of 16. For which exam would a score of 78 have a higher standing?

78 – 70 8

= 1

Math English

78 – 74 16

= 14

Math because it has a higher z-score.

Practice Problem 10

Maya represented the heights of boys in Mrs. Constantine’s and Mr. Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements are above the mean?

Practice Problem 10 Maya represented the heights of boys in Mrs. Constantine’s and Mr.

Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements are above the mean? There are 9 elements above the mean value of 68.4.

Practice Problem 11 Maya represented the heights of boys in Mrs. Constantine’s and Mr.

Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements are below the mean?

Practice Problem 11 Maya represented the heights of boys in Mrs. Constantine’s and Mr.

Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements are below the mean? There are 12 elements below the mean value of 68.4.

Practice Problem 12 Maya represented the heights of boys in Mrs. Constantine’s and Mr.

Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements fall within one standard deviation of the mean?

Practice Problem 12 Maya represented the heights of boys in Mrs. Constantine’s and Mr.

Kluge’s classes on a line plot and calculated the mean and standard deviation.

Heights of Boys in Mrs. Constantine’s and Mr. Kluge’s Classes (in inches)

________________________________________________________________ 64 65 66 67 68 69 70 71 72 73

x x x x x x x x x x x x x x x x x x x x

Mean = 68.4 Standard Deviation = 2.3

1. How many elements fall within one standard deviation of the mean? There are 12 elements that fall within one standard deviation. The values of the mean plus one standard deviation (68.4 + 2.3 = 70.7) and the mean minus one standard deviation (68.4 – 2.3 = 66.1).

Practice Problem 13

In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the superintendent asks him how many students scored less than one standard deviation below the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored less than one standard deviation below the mean on the mathematics portion of the SAT?

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 13

In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the superintendent asks him how many students scored less than one standard deviation below the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored less than one standard deviation below the mean on the mathematics portion of the SAT? Add up all the percentages less that -1 (13.6% + 2.1% + 0.1% = 15.8%). 15.8% x 1653 = 261. Therefore, 261 students less than one deviation below the mean.

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 14

In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the superintendent asks him how many students scored between the mean and one standard deviation below the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored between the mean and one standard deviation below the mean on the mathematics portion of the SAT?

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 14

In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the superintendent asks him how many students scored between the mean and one standard deviation below the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored between the mean and one standard deviation below the mean on the mathematics portion of the SAT? 34% scored between the mean and one standard deviation below the mean. 34% x 1653 = 562. Therefore, 562 students scored between the mean and one standard deviation below the mean.

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 15

In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the superintendent asks him how many students scored between one standard deviation below the mean and one standard deviation above the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored between one standard deviation below the mean and one standard deviation above the mean on the mathematics portion of the SAT?

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 15 In a school district, Mr. Mills is in charge of SAT testing. In a meeting, the

superintendent asks him how many students scored between one standard deviation below the mean and one standard deviation above the mean on the mathematics portion of the SAT in 2009. He looks through his papers and finds that the mean of the scores is 525 and 1653 students took the SAT in 2009. He also found a chart with percentages of z-scores on the SAT in 2009 as follows:

How can Mr. Mills determine the number of students that scored between one standard deviation below the mean and one standard deviation above the mean on the mathematics portion of the SAT? Add up the percentages between one standard deviation below the mean and one standard deviation above the mean (34% + 34% = 68%). 68% x 1653 = 1124. Therefore 1124 students scored between one standard deviation below the mean and one standard deviation above the mean.

Z-score (mathematics) Percent of students

z < -3 0.1

-3 < z < -2 2.1

-2 < z < -1 13.6

-1 < z < 0 34.0

0 < z < 1 34.0

1 < z < 2 13.6

2 < z < 3 2.1

z > 3 0.1

Practice Problem 16

Given the following data, find the mean, the standard deviation and the mean absolute deviation:{10 15 20 25 30 35 40 45 50}

Practice Problem 16

Given the following data, find the mean, the standard deviation and the mean absolute deviation:{10 15 20 25 30 35 40 45 50}

µ = 30 = 13.7

Mean absolute deviation = 11.1

σ

Practice Problem 17

Given the following data, find the mean, the standard deviation and the mean absolute deviation:{20 40 60 80 100 120 140 180 200}

Practice Problem 17

Given the following data, find the mean, the standard deviation and the mean absolute deviation:{20 40 60 80 100 120 140 180 200}µ = 110 = 60.6

Mean absolute deviation = 50

σ