1. Sullivan – Statistics: Informed Decisions Using Data – 2 nd Edition – Chapter 3 Introduction – Slide 1 of 3 Topic 17 Standard Deviation, Z score, and Normal Distribution

2. Sullivan – Statistics: Informed Decisions Using Data – 2 nd Edition – Chapter 3 Introduction – Slide 2 of 3 Standard Deviation ●Compute the standard deviation of 6, 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 = 4.55 Standard deviation is not a resistant measurement of the spread.

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

4. Sullivan – Statistics: Informed Decisions Using Data – 2 nd 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

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

6. Sullivan – Statistics: Informed Decisions Using Data – 2 nd 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  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

7. Sullivan – Statistics: Informed Decisions Using Data – 2 nd 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