Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 1 of 3

Topic 17

Standard Deviation, Z score, and Normal Distribution

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 2 of 3

Standard Deviation

● Compute the standard deviation of6, 1, 2, 11

● Compute the mean first = (6 + 1 + 2 + 11) / 4 = 5

● Now compute the squared deviations(1–5)2 = 16, (2–5)2 = 9, (6–5)2 = 1, (11–5)2 = 36

● Average the squared deviations(16 + 9 + 1 + 36) / 3 = 20.7

• Taking the square root of 20.7 = 4.55 Standard deviation = 4.55Standard deviation is not a resistant measurement of the spread.

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 3 of 3

Empirical Rule

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 4 of 3

Empirical Rule

● The empirical rule● If the distribution is roughly bell shaped, then

● The empirical rule● If the distribution is roughly bell shaped, then

Approximately 68% of the data will lie within 1 standard deviation of the mean

● The empirical rule● If the distribution is roughly bell shaped, then

Approximately 68% of the data will lie within 1 standard deviation of the mean

Approximately 95% of the data will lie within 2 standard deviations of the mean

● The empirical rule● If the distribution is roughly bell shaped, then

Approximately 68% of the data will lie within 1 standard deviation of the mean

Approximately 95% of the data will lie within 2 standard deviations of the mean

Approximately 99.7% of the data (i.e. almost all) will lie within 3 standard deviations of the mean

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 5 of 3

Empirical Rule

● For a variable with mean 17 and standard deviation 3.4

● For a variable with mean 17 and standard deviation 3.4 Approximately 68% of the values will lie between

(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4

● For a variable with mean 17 and standard deviation 3.4 Approximately 68% of the values will lie between

(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 Approximately 95% of the values will lie between

(17 – 2 3.4) and (17 + 2 3.4), i.e. 10.2 and 23.8

● For a variable with mean 17 and standard deviation 3.4 Approximately 68% of the values will lie between

(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 Approximately 95% of the values will lie between

(17 – 2 3.4) and (17 + 2 3.4), i.e. 10.2 and 23.8 Approximately 99.7% of the values will lie between

(17 – 3 3.4) and (17 + 3 3.4), i.e. 6.8 and 27.2

● For a variable with mean 17 and standard deviation 3.4 Approximately 68% of the values will lie between

(17 – 3.4) and (17 + 3.4), i.e. 13.6 and 20.4 Approximately 95% of the values will lie between

(17 – 2 3.4) and (17 + 2 3.4), i.e. 10.2 and 23.8 Approximately 99.7% of the values will lie between

(17 – 3 3.4) and (17 + 3 3.4), i.e. 6.8 and 27.2

● A value of 2.1 and a value of 33.2 would both be very unusual

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 6 of 3

Z score

● z-scores can be used to compare the relative positions of data values in different samples

● z-scores can be used to compare the relative positions of data values in different samples Pat received a grade of 82 on her statistics exam

where the mean grade was 74 and the standard deviation was 12

● z-scores can be used to compare the relative positions of data values in different samples Pat received a grade of 82 on her statistics exam

where the mean grade was 74 and the standard deviation was 12

Pat received a grade of 72 on her biology exam where the mean grade was 65 and the standard deviation was 10

● z-scores can be used to compare the relative positions of data values in different samples Pat received a grade of 82 on her statistics exam

where the mean grade was 74 and the standard deviation was 12

Pat received a grade of 72 on her biology exam where the mean grade was 65 and the standard deviation was 10

Pat received a grade of 91 on her kayaking exam where the mean grade was 88 and the standard deviation was 6

Sullivan – Statistics: Informed Decisions Using Data – 2nd Edition – Chapter 3 Introduction – Slide 7 of 3

Z score

● Statistics Grade of 82 z-score of (82 – 74) / 12 = .67

● Statistics Grade of 82 z-score of (82 – 74) / 12 = .67

● Biology Grade of 72 z-score of (72 – 65) / 10 = .70

● Statistics Grade of 82 z-score of (82 – 74) / 12 = .67

● Biology Grade of 72 z-score of (72 – 65) / 10 = .70

● Kayaking Grade of 81 z-score of (91 – 88) / 6 = .50

● Statistics Grade of 82 z-score of (82 – 74) / 12 = .67

● Biology Grade of 72 z-score of (72 – 65) / 10 = .70

● Kayaking Grade of 81 z-score of (91 – 88) / 6 = .50

● Biology was the highest relative grade