1. Objectives The student will be able to: find the variance of a data set. find the standard deviation of a data set.

2. Interquartile range and outliers The interquartile range is Outliers are numbers far from the others. Technically,outliers are data that fall beyond these:

3. outlier A data point that is distinctly separate from the rest of the data. A piece of data that is far away from the other pieces is an outlier. One definition of outlier is any data point more than 1.5 times interquartile range (IQR) below the first quartile or above the third quartilefirst quartile or above the third quartile Outlier Note: The IQR definition given here is widely used but is not the last word in determining whether a given number is an outlier.

4. Interquartile range (IQR) Example: For the data 2, 5, 6, 9, 12, we have the following summary: first quartile = 3.5 median = 6 median = 6 third quartile = 10.5 IQR = 10.5 – 3.5 = 7, so 1.5(IQR) = 10.5. To determine if there are outliers we must consider the numbers that are 1.5·IQR or 10.5 beyond the quartiles. quartiles. first quartile – 1.5·IQR = 3.5 – 10.5 = –7 third quartile + 1.5·IQR = 10.5 + 10.5 = 21 Since none of the data are outside the interval from –7 to 21, there are no outliers.

5. Variance Variance is the average squared deviation from the mean of a set of data. It is used to find the standard deviation.

6. Variance 1. Find the mean of the data. Hint – mean is the average so add up the values and divide by the number of items. 5. Divide the total by the number of items. 4. Find the sum of the squares. 3. Square each deviation of the mean. 2. Subtract the mean from each value – the result is called the deviation from the mean.

7. Variance Formula The variance formula includes the Sigma Notation, which represents the sum of all the items to the right of Sigma. Mean is represented by and n is the number of items.

8. Standard Deviation Standard Deviation shows the variation in data. If the data is close together, the standard deviation will be small. If the data is spread out, the standard deviation will be large. Standard Deviation is often denoted by the lowercase Greek letter sigma,.

9. Standard Deviation Standard Deviation for a sample of the population is S on calculator. Standard Deviation for an entire population is The mean is denoted by on the calculator.

10. Standard Deviation Find the variance. a) Find the mean of the data. b) Subtract the mean from each value. c) Square each deviation of the mean. d) Find the sum of the squares. e) Divide the total by the number of items. Take the square root of the variance.

11. Standard Deviation Formula The standard deviation formula can be represented using Sigma Notation: Notice the standard deviation formula is the square root of the variance.

12. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 1) Find the mean: (92+88+80+68+52)/5 = 76. 2) Find the deviation from the mean: 92-76=16 88-76=12 80-76=4 68-76= -8 52-76= -24

13. 3) Square the deviation from the mean: Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52.

14. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 4) Find the sum of the squares of the deviation from the mean: 256+144+16+64+576= 1056 5) Divide by the number of data items to find the variance: 1056/5 = 211.2

15. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 6) Find the square root of the variance: Thus the standard deviation of the test scores is 14.53.

16. Calculator Go to “stat” edit and enter the data Go to “stat, calc 1, L 1

17. Standard Deviation A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class.

18. Hint: 1.Find the mean of the data. 2.Subtract the mean from each value – called the deviation from the mean. 3.Square each deviation of the mean. 4.Find the sum of the squares. 5.Divide the total by the number of items – result is the variance. 6.Take the square root of the variance – result is the standard deviation.

19. Solve: Answer Now A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class.

20. The math test scores of five students are: 92,92,92,52 and 52. 1) Find the mean: (92+92+92+52+52)/5 = 76 2) Find the deviation from the mean: 92-76=16 92-76=16 92-76=16 52-76= -24 52-76= -24 4) Find the sum of the squares: 256+256+256+576+576= 1920 3) Square the deviation from the mean:

21. The math test scores of five students are: 92,92,92,52 and 52. 5) Divide the sum of the squares by the number of items : 1920/5 = 384 variance 6) Find the square root of the variance: Thus the standard deviation of the second set of test scores is 19.6.

22. Consider both sets of scores. Both classes have the same mean, 76. However, each class does not have the same scores. Thus we use the standard deviation to show the variation in the scores. With a standard variation of 14.53 for the first class and 19.6 for the second class, what does this tell us? Analyzing the data: Answer Now

23. Analyzing the data: Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 With a standard variation of 14.53 for the first class and 19.6 for the second class, the scores from the second class would be more spread out than the scores in the second class.

24. Analyzing the data: Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 Class C: 77,76,76,76,75 Estimate the standard deviation for Class C. a) Standard deviation will be less than 14.53. b) Standard deviation will be greater than 19.6. c) Standard deviation will be between 14.53 and 19.6. d) Can not make an estimate of the standard deviation. Answer Now

25. Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 Class C: 77,76,76,76,75 Estimate the standard deviation for Class C. a) Standard deviation will be less than 14.53. b) Standard deviation will be greater than 19.6. c) Standard deviation will be between 14.53 and 19.6 d) Can not make an estimate if the standard deviation. Analyzing the data: Answer: A The scores in class C have the same mean of 76 as the other two classes. However, the scores in Class C are all much closer to the mean than the other classes so the standard deviation will be smaller than for the other classes.

26. Summary: As we have seen, standard deviation measures the dispersion of data. The greater the value of the standard deviation, the further the data tend to be dispersed from the mean.

27. The bell curve which represents a normal distribution of data shows what standard deviation represents. One standard deviation away from the mean ( ) in either direction on the horizontal axis accounts for around 68.2 percent of the data. Two standard deviations away from the mean accounts for roughly 95 percent of the data with three standard deviations representing about 99.8 percent of the data.

28. Values go under here for each problem. 68.2%1.7+.5 +.1 = 2.3% Use the curve to fill in values underneath: 2.2

29. -2 1 4 10 16 22 28 34 68.2%1.7+.5 +.1 = 2.3% Ex: Mean = 16 and standard deviation = 6: 2.2

30. Mean = 64, standard deviation is 3. What % is less than 58? 2.2 58 61 64 67 70 Less than 58 is 1.7 +.5 +.1 = 2.3% 2.2

31. Mean = 84, standard deviation is 4. How many of 31 days are above 90? 2.2 76 80 84 88 90 Above 90 is 4.4 +1.7 +.5 +.1 = 6.7% 2.2

32. Mean = 84, standard deviation is 4. How many of 31 days are above 90? 2.2 76 80 84 88 90 Above 90 is 6.7% 31(.067)=2.077, so 2 days 2.2

33. Normal Distribution Reference sheet Re-draw for problems

34. Z-Scores above the mean is positive, below is negative Formula: Given the mean is 60 and standard deviation is 6, what is the z- score of 48?

35. Z-Scores: how many standard deviations from the mean Formula: Given the mean is 60 and standard deviation is 6, what is the z- score of 48?

36. Z-Scores: how many standard deviations from the mean Formula: Given the mean is 160 and 186 has a z-score of 2, find the standard deviation.

37. Z-Scores: how many standard deviations from the mean Formula: Given the mean is 160 and 186 has a z-score of 2, find the standard deviation.

38. Summary: Percents are found on the curve. Some problems use the % to answer for a number Some problems ask what % are certain values

39. -2 1 4 10 16 22 28 34 68.2%1.7+.5 +.1 = 2.3% 50 % are above 16, 50% are below 16. 15.9% are above 22, etc. 2.2

40. precentiles.1% 2.3% 15.9% 50% 84.1% 97.7% 99.9% 2.2

41. Vocabulary Survey-collecting and analyzing data. Reliable data is based on random samples with no bias. Bias-choosing people that may have a specific opinion Un-baised- randomly selected

42. Experiments Observational- simply watching what happens without influencing the outcome. Choose, analyze, measure. Controlled experiment involves changing some aspect of the study