1. Chapter 3: Section 2 Measures of Variance

2. Paint Comparison: How many months will they last??? Brand ABrand B 1035 6045 5030 35 40 2025 Average for Brand A: Average for Brand B: 210/6 = 35 months Find the mean of each Brand.

3. Lets Check the Range: Range for Brand A: Range for Brand B: 60 – 10 = 50 Months 45 – 25 = 20 Months

4. Finding Variance and Standard Deviation of a Population Variance Formula for a Population Standard Deviation Formula for a Population X = individual value μ = population mean N = population size

5. μ= 35 N = 6 Finding Variance and Standard Deviation of a Population for Brand A 1750Total 225-1520 25+540 25-530 2251550 6252560 625-2510 (X – μ)^2X - μX

6. Finding Variance and Standard Deviation of a Population for Brand A Variance is 291.7Standard Deviation is 17.1

7. μ= 35 N = 6 Finding Variance and Standard Deviation of a Population for Brand B 250Total 100-1025 540 0035 25-530 1001045 0035 (X – μ)^2X - μX

8. Finding Variance and Standard Deviation of a Population for Brand B Variance is 41.7Standard Deviation is 6.5

9. Which Brand is right for you? Brand A Variance: 291.7 Standard Deviation: 17.1 Brand B Variance: 41.7 Standard Deviation: 6.5

10. Page 137 # 6

11. Finding Variance and Standard Deviation of a Sample Variance Formula for a Sample Standard Deviation Formula for a Sample X = individual value = sample mean n = sample size

12. Page 129 Example 3-23 11.2, 11.9,12.0, 12.8, 13.4, 14.3

13. Finding Variance and Standard Deviation of a Sample Variance is 1.28 Standard Deviation is 1.13

15. = 12.6 n = 6 Finding Variance and Standard Deviation of a Sample using Chart 6.38Total 2.891.714.3.640.813.4.040.212.8.36-0.612.0.49-0.711.9 1.96-1.411.2 (X – )^2X -X

16. Finding Variance and Standard Deviation of a Sample using Chart Variance is 1.28 Standard Deviation is 1.13

17. Page 138 # 10

18. Variance and Standard Deviation from a Frequency Table 13,310490Total: 20 2,8887638235.5-40.5 3,2679933330.5-35.5 3,13611228425.5-30.5 2,64511523520.5-25.5 9725418315.5-20.5 3382613210.5-15.5 648815.5-10.5 Fr *(MP^2)Freq * MPMidpointFrequencyClass

19. Formula Variance is 68.7 Standard Deviation is 8.3

20. Page 139 # 27

21. 91,3822,733Total: 80 3,5288442240.5-43.5 18,252468391237.5-40.5 25,920720362034.5-37.5 34,8481,056333231.5-34.5 8,10027030928.5-31.5 3,64513527525.5-28.5 Fr *(MP^2)Freq * MPMidpointFrequencyClass Page 139 # 27

22. Formula Variance is 68.7 Standard Deviation is 8.3

23. Standard Deviation 1 Standard Deviation above and below the Mean 2 Standard Deviation above and below the Mean

24. Chebyshev’s Theorem The proportion of values from the a data set that fall within k standard deviations of the mean will be at least 1-1/k², where k is a number greater then 1 (k is not necessarily an integer). Basically, it states that: At least 75% of the data will fall within 2 Standard Deviations from the mean At least 88.89% if the data will fall within 3 Standard Deviations of the mean.

25. Proof of 75% and 88.89%: 1-1/k² can be written as: 1-1/2² = 1-1/4 = ¾ = 75% 1-1/3² = 1-1/9 = 8/9 = 88.89%

26. Example: The mean price of a house in a certain neighborhood is $50,000 and the standard deviation is $10,000. Find the price range for which at least 75% of the houses will sell.

27. Page 135 Example 3-27 Since 75% of the data fall with in two standard deviations above and below the mean: Use the following formulas: Above = mean + 2 standard Deviations times the standard deviation Below = mean - 2 standard Deviations times the standard deviation

28. Page 135 Example 3-27 Therefore: 50,000 + 2(10,000) = $70,000 50,000 – 2(10,000) = $30,000 Hence, at least 75% of the houses will sell between $30,000 and $70,000.

29. Chebyshev and Percents: A survey of local companies found that the mean amount of travel allowance for executives was $0.25 per mile. The standard deviation was $0.02. Using Chebyshev’s Theorem, find the minimum percentage of data values that will fall between $0.20 and $0.30.

30. Using Chebyshev Step 1: Subtract the mean from the larger value: $0.30 - $0.25 = $0.05 Step 2: Divide the difference by the standard deviation to get k. k = 0.05/0.02 = 2.5 Step 3: Use Chebyshev Theorem to find percent: 1-1/k² = 1-1/2.5² = 0.84

31. Conclusion: 84% of the data will fall between $0.20 and $0.30.

32. Class/Homework Page: 140-141 #’s 32-40 and 45

33. Finding Q1, Q2, and Q3 89, 47, 164, 296, 30, 215, 138, 78, 48, 39 Step 1: Arrange Data in ascending order 30, 39, 47, 48, 78, 89, 138, 164, 215, 296 Step 2: Find Q2 a.k.a. the MEDIAN 30, 39, 47, 48, 78, 89, 138, 164, 215, 296 (78+89)/2 = 83.5

34. Continued Step 3: Find Q1 30, 39, 47, 48, 78, Q1 = 47 Step 4: Find Q3 Q3 = 164 89, 138, 164, 215, 296 30, 39, 47, 48, 78,89, 138, 164, 215, 296

35. Making a Box Plot (5 pt summary) Step 1: Find the following Values: Min = Q1 = Med = Q3 = Max = 30 47 83.5 164 296

36. Making a Box Plot (5 pt summary) Step 2: Draw a number line Step 3: Draw 5 pts Step 4: Draw box around Q1 and Q3 Step 5: Draw line on Median Step 6: Connect max and min to box 0 100 200300

37. Practice: Page 175 Problems: 21 and 22 Four Box plots

38. Using the z score A student scored 38 on a Calculus test that had a mean of 40 and a standard deviation of 5; she scored 94 on a history test with a mean of 100 and a standard deviation of 10. Compare her relative position on the two tests.

39. Chapter 3 Section 3: From a to z score The z score (standard score) is obtained by subtracting the mean from the value and dividing by result by the standard deviation. For SampleFor Population Words Formula

40. Using the Formula s= 10Mean = 100X = 94Test B s= 5Mean = 40X = 38Test A Test B

41. Try: Page 153 #’s 9-15

42. Finding Your Percentile A teacher gives a 20-pt test to 10 students. The scores are: 18, 15, 12, 6, 8, 2, 3, 5, 20, 10 Find the percentile rank of a score of 12.

43. Steps for Percentiles Step 1: Arrange the Data in accending order 2, 3, 5, 6, 8, 10, 12, 15, 18, 20 Step 2: Use Percentile Formula Percentile = (Number of values below X) + 0.5 Number of Values below X: Total Number of values *100 134562

44. Percentile Continued….. Therefore a score of 12 was better than 65% of the class.

45. Find the score of a given Percentile Step 1: Arrange Data in increasing order Step 2: Use Formula: Where: n = total number of values p = percentile Step 3A: If c is not a whole #, round up to the next number and count over from smallest #. Step 3B: If c is a whole #, count over from the smallest #. Then add that number to the next number and divide by two

46. Example: What score represents the 60 th percentile? 2, 3, 5, 6, 8, 10, 12, 15, 18, 20 Whole # Since 6 is a whole number, count over 6 terms. Then add that # to the next # and divide by 2. The score of 11 represents the 60 th percentile.

47. Try: Page 154 – 155 # 22-28 Even # 23-29 odd