1. Chapter 9 Statistics Section 9.2 Measures of Variation

2. Who Has Better Scores? Adam and Bonnie are comparing their quiz scores in an effort to determine who is the “best”. Adam and Bonnie are comparing their quiz scores in an effort to determine who is the “best”. Help them decide by calculating the mean, median, and mode for each. Help them decide by calculating the mean, median, and mode for each. Adam’s ScoresBonnie’s Scores 8581 6085 10586 10586 8585 7290 10080 10080

3. The Winner? AdamBonnie Mean84.584.5 Median8585 Mode8585 So, who has the better quiz scores?

4. Another Way to Compare Sometimes the measures of central tendency (mean, median, mode) aren’t enough to adequately describe the data. Sometimes the measures of central tendency (mean, median, mode) aren’t enough to adequately describe the data. We also need to take into account the consistency, or spread, of the data. We also need to take into account the consistency, or spread, of the data.

5. Range The range of the data is the difference between the largest and smallest number in a sample. The range of the data is the difference between the largest and smallest number in a sample. Find the range of Adam and Bonnie’s scores. Find the range of Adam and Bonnie’s scores. Adam: 105 – 60 = 45 Adam: 105 – 60 = 45 Bonnie: 90 – 80 = 10 Bonnie: 90 – 80 = 10 Based on the range, Bonnie’s scores are more consisent, and some might argue therefore, better than Adam’s. Based on the range, Bonnie’s scores are more consisent, and some might argue therefore, better than Adam’s.

6. Another Measure of Dispersion The most useful measure of variation (spread) is the standard deviation. The most useful measure of variation (spread) is the standard deviation. First, we will look at the deviations from the mean. First, we will look at the deviations from the mean.

7. Deviations from the Mean The deviation from the mean is the difference between a single data point and the calculated mean of the data. The deviation from the mean is the difference between a single data point and the calculated mean of the data. Data point close to mean: small deviation Data point close to mean: small deviation Data point far from mean: large deviation Data point far from mean: large deviation Sum of deviations from mean is always zero. Sum of deviations from mean is always zero. Mean of the deviations is always zero. Mean of the deviations is always zero.

8. Deviations from Mean for Adam and Bonnie Data Point Deviation from Mean 85 85 - 84.5 = 0.5 60 60 – 84.5 = -24.5 105 105 – 84.5 = 20.5 85 85 – 84.5 = 0.5 72 72 – 84.5 = -12.5 100 100 – 84.5 = 15.5 Sum = 0 Adam Data Point Deviation from Mean 81 81 - 84.5 = -3.5 85 85 – 84.5 = 0.5 86 86 – 84.5 = 1.5 85 85 – 84.5 = 0.5 90 90 – 84.5 = 5.5 80 80 – 84.5 = -4.5 Sum = 0 Bonnie

9. Variance Because the average of the mean deviation is always zero, we must modify our approach using the variance. Because the average of the mean deviation is always zero, we must modify our approach using the variance. The variance is the mean of the squares of the deviation. The variance is the mean of the squares of the deviation.

10. Variance for Adam and Bonnie’s Scores Using the deviations from the mean we have already calculated for Adam and Bonnie, we will find the variance for each. Using the deviations from the mean we have already calculated for Adam and Bonnie, we will find the variance for each. Adam : s² = Adam : s² = Bonnie: s² = Bonnie: s² = s² = 1417.5 5 = 283.5 s² = 65.5 5 = 13.1

11. Standard Deviation To find the variance, we squared the deviations from the mean, so the variance is in squared units. To find the variance, we squared the deviations from the mean, so the variance is in squared units. To return to the same units as the data, we use the square root of the variance, the standard deviation. To return to the same units as the data, we use the square root of the variance, the standard deviation. Standard Deviation Variance

12. Adam and Bonnie’s Standard Deviation Adam: Adam: Bonnie: Bonnie: Based on the standard deviation, Bonnie’s scores are better because there is less dispersion. In other words, she is more consistent than Adam. Based on the standard deviation, Bonnie’s scores are better because there is less dispersion. In other words, she is more consistent than Adam.

13. Formulas for Variance and Standard Deviation

14. Example 1 The number of homicide victims in Vermont from 1992 through 2001 is given in the table at right. (Source: http://170.222.24.9/cjs/crime_01/homicide_01.html) The number of homicide victims in Vermont from 1992 through 2001 is given in the table at right. (Source: http://170.222.24.9/cjs/crime_01/homicide_01.html)http://170.222.24.9/cjs/crime_01/homicide_01.html Find the mean, median, mode, and standard deviation of the data. Find the mean, median, mode, and standard deviation of the data. Year Homicide Victims 199221 199315 19945 199513 199611 19979 199812 199917 200012 200111

15. Sample vs. Population The mean, variance, and standard deviation of a random sample is referred to as the sample mean ( ), sample variance (s²), and sample standard deviation (s). The mean, variance, and standard deviation of a random sample is referred to as the sample mean ( ), sample variance (s²), and sample standard deviation (s). The sample mean, variance, standard deviation, etc. can only give us an approximation to the population mean (µ), the population variance (σ²), and the population standard deviation (σ). The sample mean, variance, standard deviation, etc. can only give us an approximation to the population mean (µ), the population variance (σ²), and the population standard deviation (σ). The main difference lies in the denominator of the formulas for standard deviation. When the value of the sample, n, is large, the sample standard deviation gives a good estimate of the population standard deviation. The main difference lies in the denominator of the formulas for standard deviation. When the value of the sample, n, is large, the sample standard deviation gives a good estimate of the population standard deviation.

16. Grouped Distributions

17. Example 2 Mr. Smith recently gave a math test and organized his scores into the table at right. Mr. Smith recently gave a math test and organized his scores into the table at right. Help Mr. Smith determine the class average, the median score, and standard deviation. ScoreFrequency 40 - 49 2 50 - 59 3 60 - 69 6 70 - 79 12 80 - 89 7 90 - 100 5

18. Chebyshev’s Theorem Chebyshev’s Theorem states that for any set of numbers, the fraction (or probability) that will lie within k standard deviations of the mean (for k > 1) is at least Chebyshev’s Theorem states that for any set of numbers, the fraction (or probability) that will lie within k standard deviations of the mean (for k > 1) is at least 1 - __1__ k ²

19. Example 3 Use Chebyshev’s Theorem to find the fraction of all the numbers of a data set that must lie within 4 standard deviations from the mean. Use Chebyshev’s Theorem to find the fraction of all the numbers of a data set that must lie within 4 standard deviations from the mean.

20. Example 4 In a certain distribution of numbers, the mean is 50 with a standard deviation of 6. Use Chebyshev’s Theorem to tell the probability that a number lies in each interval. In a certain distribution of numbers, the mean is 50 with a standard deviation of 6. Use Chebyshev’s Theorem to tell the probability that a number lies in each interval. a.) between 38 and 62 b.) less than 38 or more than 62