1. Ebola

2. Maths warm-up (2–3 hours)  Recap Stem and Leaf Diagrams  Recap Histograms  Recap Cumulative Frequency and Box Plots  Recap types of data 2

3. Watch the following link. What does it show? http://www.breathingearth.net/ 3

4. Ebola by numbers 4

5. Video clip – Ebola https://www.youtube.com/watch?v=JNi H18JNmqA 5

6. Facts and figures on Ebola 2014 West Africa outbreak:  Guinea – 2339 cases, 1454 deaths  Liberia – 7765 cases, 3222 deaths  Mali – 8 cases, 6 deaths  Nigeria – 20 cases, 8 deaths  Senegal – 1 case, 0 deaths (infection originated in Guinea)  Sierra Leone – 8014 cases, 1857 deaths  Spain – 1 case, 0 deaths  United States – 4 cases, 1 death (two infections originated in the United States, one in Liberia and one in Guinea) 6

7. What is the death rate? Calculate the percentage death rate for each of these countries. CountryTotal casesTotal deaths Guinea23941518 Liberia77973290 Sierra Leone82732033 United States41 United Kingdom10 Total1846913682 7

8. ‘If you catch Ebola, you’ll almost certainly die’  The most widely cited figure about Ebola is that its death rate is ‘up to 90%’. The history of Ebola, prior to this year, is a series of short-lived and very isolated outbreaks of different strains of the disease, and it is true that one of these outbreaks had a fatality rate of 90%.  Is this is a fair representation? 8

9. ‘If you catch Ebola, you’ll almost certainly die’ Does this support the news headline? What can we do with this data? What information do we need to reach a more accurate conclusion? YearCase fatality 201458% 201251% 201257% 201271% 2011100% 200844% 200725% 200771% 200583% 200441% 200390% 200275% 9

10. ‘Ebola is becoming more deadly’ GuineaLiberia Date Frequency of deaths Date Frequency of deaths 1/3/14 – 31/5/14 155 1/3/14 – 31/5/14 11 1/6/14 – 31/7/14 148 1/6/14 – 31/7/14 54 1/8/14 – 31/10/14 407 1/8/14 – 31/10/14 1933 1/11/14 – 30/11/14 331 1/11/14 – 30/11/14 699 1/12/14 – 31/12/14 387 1/12/14 – 31/12/14 480 10

11. ‘Ebola is becoming more deadly’  Create a histogram to compare the number of deaths in 2014 for both Guinea and Liberia.  Is Ebola becoming more deadly? 11

12. ‘40% of people that catch Ebola are below 18 years of age’ AgeMaleFemaleTotal Newborn and infant101424 1–14 years182240 15–29316091 30–495752109 50 or over232649 Unknown235 Total141177318 12

13. ‘40% of people that catch Ebola are below 18 years of age’  Is this statement true?  Create a histogram for males and females to represent this data and calculate what percentage of cases are aged under 18. 13

14. ‘The Ebola death rate is increasing at a rapid pace monthly’  Using the data in the frequency tables following, create a cumulative frequency graph to justify this statement. (Both countries should be on the same diagram.) 14

15. Death rates in Guinea and Liberia Guinea DateFrequency of deaths 01/03/14 – 31/03/140 01/04/14 – 30/04/1480 01/05/14 – 31/05/1475 01/06/14 – 30/06/1438 01/07/14 – 31/07/14110 01/08/14 – 31/08/1443 01/09/14 – 30/09/14171 01/10/14 – 31/10/14193 01/11/14 – 30/11/14331 01/12/14 – 31/12/14387 Liberia DateFrequency of deaths 01/03/14 – 31/03/140 01/04/14 – 30/04/142 01/05/14 – 31/05/149 01/06/14 – 30/06/1421 01/07/14 – 31/07/1433 01/08/14 – 31/08/14162 01/09/14 – 30/09/14862 01/10/14 – 31/10/14909 01/11/14 – 30/11/14699 01/12/14 – 31/12/14480 15

16. Has Guinea or Liberia been more affected by Ebola?  Using your cumulative frequency graph, create a boxplot to compare the death rate in Guinea to that in Liberia.  Which country has been more affected by the outbreak of Ebola? 16

17. The Ebola exponential 17

18. Exponential growth  What’s the same? What’s different?  What gives these graphs their shape? 18

19. Exponential growth  Exponential growth occurs when the growth rate of the value of a mathematical function is proportion to the function’s current value.  Can you think of examples where exponential growth would happen in real life? 19

20. Exponential growth Video on the ‘burial boys’ http://nyti.ms/1pqVE2Thttp://nyti.ms/1pqVE2T 20

21. Exponential growth How does a rumour spread among a population? On day 1, a single person tells someone else a rumour; suppose that on every subsequent day, each person who knows the rumour tells exactly one other person the rumour. How many days until:  50 people have heard the rumour?  The whole school?  The whole country? What is the percentage transmission rate? If we can half the transmission rate (one person every two days), how much longer will it take to infect the whole country? 21

22. Exponential growth Graph showing the number of confirmed cases and the number of deaths 22

23. Exponential growth Confirmed cases each month in 2014 23

24. ‘Ebola is the most deadly virus of our generation’ 24

25. Visual spread of diseases Http://www.washingtonpost.com/wp-srv/special/health/how-ebola-spreads/ 25

26. Comparing how deadly Ebola is to other diseases 26

27. ‘Ebola is the most deadly virus of our generation’ Is this headline correct? You must use all the data given to write a report about your findings. You must compare data from other diseases and finish with a conclusion (minimum 500 words). 27

28. Sampling 28

29. Sampling – what is it? There are five types of sampling: 1. Random 2. Systematic 3. Stratified 4. Quota 5. Cluster 29

30. Starter Methods of sampling RandomSystematicStratified What is it? Advantages Disadvantages 30

31. Which is most effective?  … if you are asking people what shoe size they are in the class.  … if you are looking at the fatality rate when catching a disease. 31

32. Stratified sampling  When the population is composed of different groups of people (e.g. different genders, different ages, different social classes etc.), we may wish to choose our sample so that it contains the same proportion of each group as the entire population.  For example, if 60% of our population is female, a stratified sample would be 60% female too. 32

33. The number of students in each year group at a school is shown in the table: Suppose that a stratified sample of size 10% (90 students) needs to be chosen. Stratified sampling – Example 1 Year groupNumber of students 7180 8200 9170 10190 11160 Total900 33

34. As our sample size is 10% of the entire population of the school, we would choose 10% of each year group. Our sample would therefore be composed as follows: Having decided on the number of students from each year group, the actual students would then be picked at random from those in the year group (e.g. using a random number generator). Stratified sampling – Example 1 Year groupNumber of students 718 820 917 1019 1116 Total90 34

35. The table shows the number of boys and the number of girls in each year group at Springfield Secondary School. There are 500 boys and 500 girls in the school. Aaron took a stratified sample of 10% of the girls, by year group. Work out the number of Year 8 girls in his sample. Have a go at questions 1, 2, and 3… Year group Number of boys Number of girls 10% of girls 7100 10 815050 9100 10 5015015 11100 10 Total500 50 Stratified sampling – Example 2 35

36. The table shows the number of people in each age group who watched the school sports. Martin did a survey of these people. He used a stratified sample of exactly 50 people according to age group. Work out the number of people from each age group that should have been in his sample of 50. Age group0–1617–2930–4445–5960+ Number of people 177111868221 Stratified sampling – Example 3 36

37. We first find the total number of students who watched the school sports: 177 + 111 + 86 + 82 + 21 = 477 The number of people aged 0–16 in the sample needs to be of size: As the number of people chosen for the sample must be a whole number, we would round this to 19. Stratified sampling – Example 3 37

38. We can repeat this approach for all other age groups: Age group0–1617– 2930–4445–5960+Total Number of people in sample 19 = 11.64… = 8.60... = 9.01… = 2.20… = 12= 2= 9 Notice that our total is incorrect. This is due to rounding. To correct this we must reduce the number of students from one of the age groups by 1. The calculation for the number of students in the 0–16 age group resulted in the answer 18.55… (only just big enough to round up). We could reduce the number of students in this age group to 18. 5151 Have a go at questions 4, 5, and 6… Stratified sampling – Example 3 38

39. CountryTotal casesTotal deaths Guinea23941518 Liberia77973290 Sierra Leone82732033 United States41 United Kingdom10 Total1846913682 Case study The table shows the number of Ebola cases and deaths in the main countries affected. 39

40. Case study  Make another column stating the total deaths as a percentage.  A newspaper article stated: ‘25% of people who are diagnosed with Ebola die from it.’ Is this true? What is this data based on? What would be a more accurate statement? 40

41. Case study  The government would like to look further into the relationship between the number of people diagnosed with Ebola and the death rate. They use two different methods. 41

42. Method 1 The government would like to look further into the death rate of Ebola patients. They cannot look at all 13,000 so they take 300 people from each country. 1. What is the problem with this? 2. What would a solution be? 3. If 300 people are asked from Guinea, what would the death rate for these people be? 42

43. Method 2 The government would like to look further into the death rate of Ebola patients. They cannot look at all 13,000 so they take a stratified sample of 1,500. 1. How many people are asked from each country? 2. From this sample, what was the death rate for each country? 43

44. Case study  Which method is more accurate, and why? 44

45. Standard deviation 45

46. Maths warm-up  Recap calculating the mean of grouped data.  Complete the worksheet on calculating mean from a table. 46

47. Two classes took a recent quiz. There were 10 students in each class, and each class had an average score of 81.5 47

48. Since the averages are the same, can we assume that the students in both classes all did pretty much the same in the exam? 48

49. The answer is… No. The average (mean) does not tell us anything about the distribution or variation in the grades. 49

50. Following are Dot-Plots of the grades in each class… 50

51. Mean 51

52. So, we need to come up with some way of measuring not just the average, but also the spread of the distribution of our data. 52

53. Why not just give an average and the range of data (the highest and lowest values) to describe the distribution of the data? 53

54. Well, for example, let’s say from a set of data, the average is 17.95 and the range is 23. But what if the data looked like this: 54

55. Here is the average And here is the range But really, most of the numbers are in this area, and are not evenly distributed throughout the range. 55

56. The standard deviation is a number that measures how far away each number in a set of data is from their mean. 56

57. If the standard deviation is large, it means the numbers are spread out from their mean. If the standard deviation is small, it means the numbers are close to their mean. small, large, 57

58. Here are the scores in the maths quiz for Team A: 72 76 80 81 83 84 85 89 Average: 81.5 58

59. The standard deviation measures how far away each number in a set of data is from their mean. For example, start with the lowest score, 72. How far away is 72 from the mean of 81.5? 72 – 81.5 = - 9.5 - 9.5 Team A Quiz Grades 59

60. Or, start with the highest score, 89. How far away is 89 from the mean of 81.5? 89 – 81.5 = 7.5 - 9.5 Team A Quiz Grades 7.5 60

61. So, the first step to finding the standard deviation is to find all the distances from the mean. 72 76 80 81 83 84 85 89 - 9.5 7.5 Distance from mean 61

62. So, the first step to finding the standard deviation is to find all the distances from the mean. 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 62

63. Next, you need to square each of the distances to turn them all into positive numbers 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 Distances squared 63

64. Next, you need to square each of the distances to turn them all into positive numbers 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 2.25 0.25 2.25 6.25 12.25 56.25 Distances squared 64

65. Add up all of the distances 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 2.25 0.25 2.25 6.25 12.25 56.25 Distances squared Sum: 214.5 65

66. Divide by n where n represents the amount of numbers you have. 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 2.25 0.25 2.25 6.25 12.25 56.25 Distances squared Sum: 214.5 10 = 21.45 66

67. Finally, take the square root of the average distance 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 2.25 0.25 2.25 6.25 12.25 56.25 Distances squared Sum: 214.5 (10 – 1) = 23.8 = 4.88 67

68. This is the standard deviation 72 76 80 81 83 84 85 89 - 9.5 - 5.5 - 1.5 - 0.5 1.5 2.5 3.5 7.5 Distance from mean 90.25 30.25 2.25 0.25 2.25 6.25 12.25 56.25 Distances squared Sum: 214.5 10 = 21.45 = 4.63 68

69. Now find the standard deviation for the other class grades 57 65 83 94 95 96 98 93 71 63 - 24.5 - 16.5 1.5 12.5 13.5 14.5 16.5 11.5 - 10.5 -18.5 Distance from mean 600.25 272.25 2.25 156.25 182.25 210.25 272.25 132.25 110.25 342.25 Distances squared Sum: 2280.5 (10) = 228.05 = 15.1 69

70. Now, let’s compare the two classes again: Team ATeam B Average in the Quiz Standard deviation 81.5 81.5 4.63 15.1 70

71. 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 from 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. 71

72. Solve:  A maths class took a test with these five test scores: 92, 92, 92, 52, 52.  Find the standard deviation for this class. 72

73. The maths 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 3.Square the deviation from the mean: 4.Find the sum of the squares: 256+256+256+576+576 = 1920 73

74. The maths 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. 74

75. Standard deviation practice Complete questions… 75

76. x01234 Frequency13754 We must now complete the table, and calculate the totals of f, fx and fx 2 : xffxfxfx2fx2 01 13 27 35 44 Note: f x 2 means f × x 2 and not (f × x) 2. 0 3 14 15 16 0 3 28 45 64 2048140 x = f xf x f 48 20 = 2.4 = 1.114 x = = σ Example 2 Find the mean and standard deviation for the following distribution: 76

77. x01234 Frequency26478 We must now complete the table, and calculate the totals of f, fx and fx 2 : xffxfxfx2fx2 02 16 24 37 48 Note: f x 2 means f × x 2 and not (f × x) 2. x = f xf x f x = = σ Example 2 Find the mean and standard deviation for the following distribution: 77

78. Standard deviation practice  Find the standard deviation for the averages from your initial worksheet. 78

79. Compare the mean and standard deviation for males and females AgeMaleFemaleTotal 0–10–1 101424 1 – 15 182240 15 – 30 316091 30 – 50 5752109 50 – 70 232649 Total141177318 79

80. The incubation period for Ebola 80

81. Incubation period for Ebola  The incubation period is the time that elapses between exposure to a pathogenic organism, a chemical or radiation, and when symptoms and signs are first apparent.  Depending on the disease, the person may or may not be contagious during the incubation period.  The incubation period, or the time interval from infection to onset of symptoms, is from 2 to 25 days. The patients become contagious once they begin to show symptoms. They are not contagious during the incubation period. 81

82. Incubation period Distribution of Ebola virus incubation period, by days of incubation 82

83. Days of incubation What does this tell us about the incubation period of Ebola? 1.What is the mean? 2.What is the standard deviation? Extension: How can you represent this graphically? Days of IncubationNumber of cases 10 295 3420 4760 5790 6690 7590 8410 9310 10270 11175 12130 1380 1470 1540 1630 1715 1820 1915 2010 2115 2210 2310 245 255 83

84. Starter Mode of –567, 600, 356, 600, 400, 500 Mean of –150, 160, 290 1st quartile of –2, 5, 9, 11, 14, 23, 30 2nd quartile of –2, 5, 9, 11, 14, 23, 30 3rd quartile of –2, 5, 9, 11, 14, 23, 30 Interquartile range of –2, 5, 9, 11, 14, 34, 36 The range of –12, 36, 37, 41, 46, 47 Where have you seen the above before? 84

85. Starter – Answers Mode of –600 Mean of –200 1st quartile of –5 2nd quartile of –11 3rd quartile of –23 Interquartile range of –34 – 5 = 29 The range of –47 – 12 = 35 Where have you seen the above before? – Averages, Cumulative frequency, Box plot… 85

86. Looking at data Look at the data below. Write a 50–100 word comparison on males to females. You may want to represent the data on a graph and use: Averages, Range and IQR AgeMaleFemaleTotal Newborn and infant101424 1 – 14 years 182240 15 – 29 316091 30 – 49 5752109 50 or over232649 Unknown235 Total141177318 86

87. 1.Quentin’s height is between the median and the third quartile for all heights in his school. His height’s percentile rank could be which of the following? (1) 45th(3) 64th (2) 79th (4) 23rd 2.In a certain test, a score of 90 was the 25th percentile. If 20 students took the test, how many received scores of 90 or below? (3) 64th 5 87

88. 3.In a mathematics test, Sal scored at the 90th percentile. Which one of the following statements is true? (1) Ninety per cent of the students who took the test had the same score as Sal had. (2) Ninety per cent of the students who took the test had a score equal to or less than Sal’s score. (3) Sal scored 90% in this test. (4) Sal answer 90 questions correctly. (2) Ninety per cent of the students who took the test had a score equal to or less than Sal’s score. 88

89. Percentiles Here are some scores from a mock test: 17, 12, 5, 36, 24, 13, 33, 27, 21 Find the 16% percentile  Firstly we need to order them: 5, 12, 13, 17, 21, 24, 27, 33, 36  Then we use the formula i = (P/100) x n  i = (16/100) x 9 so i = 1.44; i = 2 so use the 2nd number 16% percentile from the data = 12 Percentage to find Amount of numbers RULES: Always round up. If it becomes a whole number, add 0.5 89

90. Your turn  Find the… 10th and 34th percentile: 13, 85, 99, 69, 56, 42, 42, 99, 101, 157, 76, 58, 57, 140, 142, 70, 132, 77  Find the interpercentile range from the 10th to the 90th percentiles: 16, 63, 55, 32, 89, 70, 49, 89, 19, 92, 86, 17, 52, 27, 39, 30, 87, 21, 15 90

91. Create a cumulative frequency Which class does this lie in? 91

92. ‘21% of females that catch Ebola are aged 14 and under’ AgeFemaleCumulative frequency Newborn and infant14 1 – 14 years 2236 15 – 29 6096 30 – 49 52148 50 or over26174 Unknown3177 Total177 92

93. ‘45% of males that catch Ebola are aged over 20’ AgeMaleCumulative frequency Newborn and infant10 1 – 14 years 1828 15 – 29 3159 30 – 49 57116 50 or over23139 Unknown2141 Total141 93

94. ‘The IQR of people that catch Ebola is between 18 and 44’ AgeTotalCumulative frequency Newborn and infant24 1 – 14 years 4064 15 – 29 91155 30 – 49 109264 50 or over49313 Unknown5318 Total318 94

95. The interpercentile range from the 20th to the 80th percentile of people that catch Ebola is 38 years. AgeMaleFemaleTotal Newborn and infant101424 1 – 14 years 182240 15 – 29 316091 30 – 49 5752109 50 or over232649 Unknown235 Total141177318 95

96. Core Maths Support Programme Highbridge House 16–18 Duke Street Reading RG1 4RU E-mail: cmsp@educationdevelopmenttrust.com Call: 0118 902 1243