Mean for Continuous Data mathematics .pptx

Math

1. Averages from Grouped Data

2. 0 < t ≤ 10 Jess says: the interval means all the numbers from 0 to 10 Explain what the interval means: Actually they start from 1 to 10 as it doesn’t include 0 Chan says: Who is correct? Explain your answer.

3. Height Frequency 20 ≤ h ≤ 30 30 ≤ h ≤ 40 40 ≤ h ≤ 50 50 ≤ h ≤ 60 60 ≤ h ≤ 70 Height Frequency 20 ≥ h > 30 30 ≥ h > 40 40 ≥ h > 50 50 ≥ h > 60 60 ≥ h > 70 Jess wrote the following tables: Explain what Jess has done WRONG.

4. 25 99 51 48 82 The data lists the height of corn crop (in cm) 95 89 21 73 94 33 71 62 66 102 109 94 42 70 35 101 109 83 47 60 27 104 44 29 38 31 119 90 87 115 54 59 48 47 95 81 76 22 83 92 99 65 52 44 96 85 22 53 56 49 70 89 57 68 32 How can we calculate the mean?

5. 25 99 51 48 82 95 89 21 73 94 33 71 62 66 102 109 94 42 70 35 101 109 83 47 60 27 104 44 29 38 31 119 90 87 115 54 59 48 47 95 81 76 22 83 92 99 65 52 44 96 85 22 53 56 49 70 89 57 68 32 Height Frequency 20 ≤ h < 30 30 ≤ h < 40 40 ≤ h < 50 50 ≤ h < 60 60 ≤ h < 70 70 ≤ h < 80 80 ≤ h < 90 90 ≤ h < 100 100 ≤ h < 110 110 ≤ h < 120 Height Frequency 20 ≤ h < 40 40 ≤ h < 60 60 ≤ h < 80 80 ≤ h < 100 100 ≤ h < 120 Complete both tables. 6 5 7 8 5 5 8 9 5 2

6. 25 99 51 48 82 89 21 73 94 33 71 62 66 102 109 94 42 70 35 101 109 83 47 60 27 104 44 29 38 31 119 90 87 115 54 59 48 47 95 81 76 22 83 92 99 65 52 44 96 85 22 53 56 49 70 89 57 68 32 Height Frequency 20 ≤ h < 30 30 ≤ h < 40 40 ≤ h < 50 50 ≤ h < 60 60 ≤ h < 70 70 ≤ h < 80 80 ≤ h < 90 90 ≤ h < 100 100 ≤ h < 110 110 ≤ h < 120 Height Frequency 20 ≤ h < 40 40 ≤ h < 60 60 ≤ h < 80 80 ≤ h < 100 100 ≤ h < 120 Does it matter which table we use to calculate the mean? Complete both tables. 6 5 7 8 5 5 8 9 5 2 11 15 10 17 7 95

7. Time Frequency (f) 0 - 4 5 - 9 10 - 14 15 - 19 Your Turn: Estimate the mean. 8 10 4 2

8. Time Frequency (f) Mid point (x) fx 0 - 4 5 - 9 10 - 14 15 - 19 Total Your Turn: Estimate the mean. 24 168 Mean =   Mean = = 7   8 10 4 2 8 x 2 = 16 10 x 7 = 70 4 x 12 = 48 2 x 17 = 34 2 12 7 17

9. Length Frequency (f) 100 < m ≤ 200 200 < m ≤ 300 300 < m ≤ 400 400 < m ≤ 500 500 < m ≤ 600 What is the MODAL interval? 14 6 4 6 6 The mode is 6 so the modal interval is 6 Explain why Chan is WRONG.

10. Time Frequency (f) 0 - 4 5 - 9 10 - 14 15 - 19 What is the MODAL interval of each of these tables? 8 10 4 2 Length Frequency (f) 100 < m ≤ 200 200 < m ≤ 300 300 < m ≤ 400 400 < m ≤ 500 500 < m ≤ 600 600 < m ≤ 700 700 < m ≤ 800 800 < m ≤ 900 9 10 9 25 Height Frequency (f) 1.0 < h ≤ 1.5 1.5 < h ≤ 2.0 2.0 < h ≤ 2.5 2.5 < h ≤ 3.0 3.0 < h ≤ 3.5 3.5 < h ≤ 4.0 1 2 1 12 9 8 9 1 1 (a) (b) (c) 5 - 9 400 < m ≤ 500 2.0 < h ≤ 2.5

11. Length Frequency (f) 100 < m ≤ 200 200 < m ≤ 300 300 < m ≤ 400 400 < m ≤ 500 500 < m ≤ 600 What is the MEDIAN interval? 14 6 4 6 6 The median is 4 as its in the middle so the interval is 300<m≤400 Explain why Jess is WRONG.

12. Length Frequency (f) Cumulative Frequency 100 < m ≤ 200 200 < m ≤ 300 300 < m ≤ 400 400 < m ≤ 500 500 < m ≤ 600 What is the MEDIAN interval? 14 6 4 6 6 14 20 24 30 36 n = 36 Median value = 36÷2 = 18 th value 1 st - 14 th value 15 th - 20 th value Median interval = 200 < m ≤ 300

13. Length Frequency (f) 1.0 < h ≤ 1.5 1.5 < h ≤ 2.0 2.0 < h ≤ 2.5 2.5 < h ≤ 3.0 3.0 < h ≤ 3.5 Your Turn: Work out the median interval. 15 11 19 8 2

14. Length Frequency (f) Cumulative Frequency 1.0 < h ≤ 1.5 1.5 < h ≤ 2.0 2.0 < h ≤ 2.5 2.5 < h ≤ 3.0 3.0 < h ≤ 3.5 15 26 45 53 55 n = 55 Median value = 55÷2 = 27.5 th value 1 st - 15 th value 16 th - 26 th value Median interval = 2.0 < m ≤ 2.5 Your Turn: Work out the median interval. 15 11 19 8 2 27 th - 45 th value

15. Alma planted some seeds to test different composts. After a week she measured the heights of thirty seedlings, in millimetres, and recorded her results in the table below. Height of seedling (h) Frequency Mid point fx 0 < h ≤ 5 5 2.5 12.5 5 < h ≤ 10 5 7.5 37.5 10 < h ≤ 15 4 12.5 50 15 < h ≤ 20 9 17.5 157.5 20 < h ≤ 25 7 22.5 157.5 415 For each the following statements decide if they are TRUE or FALSE or not enough information (NEI) The range of the heights was 25. True False The modal interval of the heights was 5. True False There were 5 seeds with a height of 20mm or more. The median interval is 15 < h ≤ 20. True False True False The estimated mean = 415 ÷ 5. True False PROBLEM SOLVING: NEI NEI NEI NEI NEI

16. Alma planted some seeds to test different composts. After a week she measured the heights of thirty seedlings, in millimetres, and recorded her results in the table below. Height of seedling (h) Frequency Mid point fx 0 < h ≤ 5 5 2.5 12.5 5 < h ≤ 10 5 7.5 37.5 10 < h ≤ 15 4 12.5 50 15 < h ≤ 20 9 17.5 157.5 20 < h ≤ 25 7 22.5 157.5 415 For each the following statements decide if they are TRUE or FALSE or not enough information (NEI) The range of the heights was 25. True False The modal interval of the heights was 5. True False There were 5 seeds with a height of 20mm or more. The median interval is 15 < h ≤ 20. True False True False The estimated mean = 415 ÷ 5. True False PROBLEM SOLVING: NEI NEI NEI NEI NEI

17. The frequency tables show the price of bread in different shops. For each the following statements decide if they are TRUE or FALSE or not enough information (NEI) The range of the price of bread was 19p. True False The modal interval price of bread was 10 . True False 3 loaves of bread cost 87p The median interval is 95 < h ≤ 99. NEI True False True False The estimated mean = 1855 ÷ 20. True False YOUR TURN: PROBLEM SOLVING: Price (p) Frequency Mid point fx 80-84 2 82 164 85-89 3 87 261 90-94 5 92 460 95-99 10 97 970 1855 NEI NEI NEI NEI

18. The frequency tables show the price of bread in different shops. For each the following statements decide if they are TRUE or FALSE or not enough information (NEI) The range of the price of bread was 19p. True False The modal interval price of bread was 10 . True False 3 loaves of bread cost 87p The median interval is 95 < h ≤ 99. NEI True False True False The estimated mean = 1855 ÷ 20. True False YOUR TURN: PROBLEM SOLVING: Price (p) Frequency Mid point fx 80-84 2 82 164 85-89 3 87 261 90-94 5 92 460 95-99 10 97 970 1855 NEI NEI NEI NEI

19. a) Fiona works in a call centre. She records the length of each phone call in the list below. b) Anna planted some seeds to test different composts. After a week she measured the heights of twenty seedlings, in millimetres, and recorded her results in the table below. For each of the following estimate the mean Length of call (mins) Frequency Midpoint  fx   0 ≤ t ≤ 5 3     5 < t ≤ 10 15     10 < t ≤ 15 8     15 < t ≤ 20 40     20 < t ≤ 25 20      Total       Height (mm) Frequency Midpoint  fx   ≤ h ≤ 5       5 < h ≤ 10       10 < h ≤ 15       15 < h ≤ 20       20 < h ≤ 25        Total       1.3, 3.2, 5.4, 8.1, 11.7, 15.4, 13.9, 10.5, 2.8, 7.3, 10.0 22.4, 18.7, 24.5, 21.7, 19.8, 16.4, 17.5, 1.9, 15.3 86 570 3 x 2.5 = 7.5 112.5 100 700 2.5 12.5 7.5 17.5 22.5 450 Mean = = 6.63   20 250 3 x 2.5 = 10 30 37.5 105 2.5 12.5 7.5 17.5 22.5 67.5 4 4 3 6 3 Mean = = 12.5  

20. Height (h cm) Frequency     120 < h ≤ 130 2     130 < h ≤ 150 5     150 < h ≤ 160 4     160 < h ≤ 165 10     165 < h ≤ 180 18     180 < h ≤ 200 7             Time (t mins) Frequency     0 ≤ t ≤ 5 19     5 < t ≤ 10 23     10 < t ≤ 20 7     20 < t ≤ 25 2     25 < t ≤ 40 3     40 < t ≤ 60 1             Distance (d metres) Frequency 1.00 < d ≤ 1.20 3 1.20 < d ≤ 1.30 4 1.30 < d ≤ 1.35 18 1.35 < d ≤ 1.50 15 1.50 < d ≤ 1.70 9 1.70 < d ≤ 2.00 6 Age (a years) Frequency 0 ≤ a ≤ 15 14 15 < a ≤ 25 20 25 < a ≤ 35 12 35 < a ≤ 40 18 40 < a ≤ 50 10 50 < a ≤ 60 14 60 < a ≤ 75 12 a) The heights of a class of Year 10 pupils were recorded as follows: b) In a Health Centre, the times patients had to wait was recorded as follows: c) Here are the long jump records for Year 11 pupils. d) Here are the ages of 100 people in a village. For each of the following: a) write the modal interval, b) estimate the mean and c) work out the median interval 46 7630 250 700 620 1625 125 155 140 162.5 172.5 3105 190 1330 Mean = = 165.87   midpoint fx 55 505 47.5 172.5 87.5 50 2.5 12.5 7.5 22.5 32.5 97.5 50 50 midpoint fx Mean = = 9.18   55 79.025 3.3 5 23.85 21.375 1.1 1.325 1.25 1.425 1.6 14.4 1.85 11.1 midpoint fx Mean = = 1.44   100 3570 105 400 360 675 7.5 30 20 37.5 45 450 55 770 midpoint fx 67.5 810 Mean = = 35.7  

21. Extension Questions: Corbett Maths Sally is raising money for charity for a fun run. The table below has been given to her from the website. Sally says the average donation is £10. By calculating the estimated mean, decide if you agree with Sally. Nathan delivers pizzas. The table below shows information about his delivery times. The pizza company has a promotion that if the delivery time Calculate an estimate for the mean delivery time. What percentage of deliveries took over 30 minutes? Nathan’s manager thinks that the promotion should be changed to 40 minutes. c) Do you agree? Explain your answer.

22. Finding the estimate of the Median – Using Interpolation Number of hours Number of girls (f) Cumulative Frequency 0 ≤ x ≤ 5 3 5 < x ≤ 10 7 10 < x ≤ 15 10 15 < x ≤ 20 15 3 10 20 24 n = 24 Median value = 24÷2 = 12 th value 1 st – 3 rd value 4 th - 10 th value Median value lies in this interval 11 th - 20 th value We are looking for the 2 nd value along this interval of frequency 10   Start of interval Class width 11 +2

23. Your Turn: Finding the estimate of the Median – Using Interpolation Age (x) in years Number of people (f) 10 < x ≤ 20 4 20 < x ≤ 30 18 30 < x ≤ 40 8 40 < x ≤ 50 10

24. Your Turn: Finding the estimate of the Median – Using Interpolation Age (x) in years Number of people (f) Cumulative Frequency 10 < x ≤ 20 4 20 < x ≤ 30 18 30 < x ≤ 40 8 40 < x ≤ 50 10 4 22 30 40 n = 40 Median value = 40÷2 = 20 th value 1 st – 4 th value 5 th - 22 th value Median value lies in this interval We are looking for the 16 th value along this interval of frequency 18   Class width 28.9 Start of interval +16

25. a) Fiona works in a call centre. She records the length of each phone call in the list below. b) Bob asked each of 40 friends how many minutes they took to get to work. The table shows some information about his results For each of the following estimate the median (using interpolation). Length of call (mins) Frequency   0 ≤ t ≤ 5 7   5 < t ≤ 10 15   10 < t ≤ 15 8   15 < t ≤ 20 40   20 < t ≤ 25 20   Time taken (m minutes) Frequency < m ≤ 10  3   10 < m ≤ 20  8   20 < m ≤ 30  11   30 < m ≤ 40  9   40 < m ≤ 50  9  

26. Answers: Finding the estimate of the Median – Using Interpolation Length of call (mins) Frequency Cumulative Frequency 0 ≤ t ≤ 5 7 5 < t ≤ 10 15 10 < t ≤ 15 8 15 < t ≤ 20 40 20 < t ≤ 25 20 7 22 30 70 n = 90 Median value = 90÷2 = 45 th value 1 st – 7 th value 8 th - 22 th value Median value lies in this interval We are looking for the 15 th value along this interval of frequency 40   Class width 16.875 90 23 rd - 30 th value 31 st - 70 th value Start of interval a) Fiona works in a call centre. She records the length of each phone call in the list below. +15

27. Answers: Finding the estimate of the Median – Using Interpolation Time taken (m minutes) Frequency Cumulative Frequency < m ≤ 10  3 10 < m ≤ 20  8 20 < m ≤ 30  11 30 < m ≤ 40  9 40 < m ≤ 50  9 3 11 22 31 n = 40 Median value = 40÷2 = 20 th value 1 st – 3 rd value 4 th - 11 th value Median value lies in this interval We are looking for the 9 th value along this interval of frequency 11   Class width 28.18 40 12 th – 22 nd value Start of interval b) Bob asked each of 40 friends how many minutes they took to get to work. The table shows some information about his results +9

28. 25 99 51 48 82 95 89 21 73 94 33 71 62 66 102 109 94 42 70 35 101 109 83 47 60 27 104 44 29 38 31 119 90 87 115 54 59 48 47 95 81 76 22 83 92 99 65 52 44 96 85 22 53 56 49 70 89 57 68 32 25 99 51 48 82 95 89 21 73 94 33 71 62 66 102 109 94 42 70 35 101 109 83 47 60 27 104 44 29 38 31 119 90 87 115 54 59 48 47 95 81 76 22 83 92 99 65 52 44 96 85 22 53 56 49 70 89 57 68 32

29. Height Frequency (f) Mid point (x) fx 20 ≤ h < 30 30 ≤ h < 40 40 ≤ h < 50 50 ≤ h < 60 60 ≤ h < 70 70 ≤ h < 80 80 ≤ h < 90 90 ≤ h < 100 100 ≤ h < 110 110 ≤ h < 120 Total Height Frequency (f) Mid point (x) fx 20 ≤ h < 40 40 ≤ h < 60 60 ≤ h < 80 80 ≤ h < 100 100 ≤ h < 120 Total

30. Time Frequency (f) 0 - 4 5 - 9 10 - 14 15 - 19 8 10 4 2 Your Turn: Calculate the mean. Length Frequency (f) 100 < m ≤ 200 200 < m ≤ 300 300 < m ≤ 400 400 < m ≤ 500 500 < m ≤ 600 14 6 4 6 6 What is the MODAL interval? Length Frequency (f) 1.0 < h ≤ 1.5 1.5 < h ≤ 2.0 2.0 < h ≤ 2.5 2.5 < h ≤ 3.0 3.0 < h ≤ 3.5 Your Turn: Work out the median interval. 15 11 19 8 2

31. Alma planted some seeds to test different composts. After a week she measured the heights of thirty seedlings, in millimetres, and recorded her results in the table below. Height of seedling (h) Frequency Mid point fx 0 < h ≤ 5 5 2.5 12.5 5 < h ≤ 10 5 7.5 37.5 10 < h ≤ 15 4 12.5 50 15 < h ≤ 20 9 17.5 157.5 20 < h ≤ 25 7 22.5 157.5 415 For each the following statements decide if they are TRUE or FALSE or not enough information (NEI) The range of the heights was 25. True False The modal interval of the heights was 5. True False There were 5 seeds with a height of 20mm or more. The median interval is 15 < h ≤ 20. True False True False The estimated mean = 415 ÷ 5. True False PROBLEM SOLVING: NEI NEI NEI NEI NEI Price (p) Frequency Mid point fx 80-84 2 82 164 85-89 3 87 261 90-94 5 92 460 95-99 10 97 970 1855 The range of the price of bread was 19p. The modal interval price of bread was 10 . 3 loaves of bread cost 87p The median interval is 95 < h ≤ 99. The estimated mean = 1855 ÷ 20. The frequency tables show the price of bread in different shops. True False True False True False True False True False NEI NEI NEI NEI NEI

32. a) Fiona works in a call centre. She records the length of each phone call in the list below. b) Anna planted some seeds to test different composts. After a week she measured the heights of twenty seedlings, in millimetres, and recorded her results in the table below. For each of the following estimate the mean Length of call (mins) Frequency Midpoint  fx   0 ≤ t ≤ 5 3     5 < t ≤ 10 15     10 < t ≤ 15 8     15 < t ≤ 20 40     20 < t ≤ 25 20      Total       Height (mm) Frequency Midpoint  fx   ≤ h ≤ 5       5 < h ≤ 10       10 < h ≤ 15       15 < h ≤ 20       20 < h ≤ 25        Total       1.3, 3.2, 5.4, 8.1, 11.7, 15.4, 13.9, 10.5, 2.8, 7.3, 10.0 22.4, 18.7, 24.5, 21.7, 19.8, 16.4, 17.5, 1.9, 15.3

33. Height (h cm) Frequency     120 < h ≤ 130 2     130 < h ≤ 150 5     150 < h ≤ 160 4     160 < h ≤ 165 10     165 < h ≤ 180 18     180 < h ≤ 200 7             Time (t mins) Frequency     ≤ t ≤ 5 19     5 < t ≤ 10 23     10 < t ≤ 20 7     20 < t ≤ 25 2     25 < t ≤ 40 3     40 < t ≤ 60 1             Distance (d metres) Frequency 1.00 < d ≤ 1.20 3 1.20 < d ≤ 1.30 4 1.30 < d ≤ 1.35 18 1.35 < d ≤ 1.50 15 1.50 < d ≤ 1.70 9 1.70 < d ≤ 2.00 6 Age (a years) Frequency ≤ a ≤ 15 14 15 < a ≤ 25 20 25 < a ≤ 35 12 35 < a ≤ 40 18 40 < a ≤ 50 10 50 < a ≤ 60 14 60 < a ≤ 75 12 a) The heights of a class of Year 10 pupils were recorded as follows: b) In a Health Centre, the times patients had to wait was recorded as follows: c) Here are the long jump records for Year 11 pupils. d) Here are the ages of 100 people in a village. For each of the following: a) write the modal interval, b) estimate the mean and c) work out the median interval

34. Extension Questions: Corbett Maths Sally is raising money for charity for a fun run. The table below has been given to her from the website. Sally says the average donation is £10. By calculating the estimated mean, decide if you agree with Sally. Nathan delivers pizzas. The table below shows information about his delivery times. The pizza company has a promotion that if the delivery time Calculate an estimate for the mean delivery time. What percentage of deliveries took over 30 minutes? Nathan’s manager thinks that the promotion should be changed to 40 minutes. c) Do you agree? Explain your answer.

35. Finding the estimate of the Median – Using Interpolation Number of hours Number of girls (f) Cumulative Frequency 0 ≤ x ≤ 5 3 5 < x ≤ 10 7 10 < x ≤ 15 10 15 < x ≤ 20 15 Age (x) in years Number of people (f) 10 < x ≤ 20 4 20 < x ≤ 30 18 30 < x ≤ 40 8 40 < x ≤ 50 10 Example Your turn Finding the estimate of the Median – Using Interpolation Number of hours Number of girls (f) Cumulative Frequency 0 ≤ x ≤ 5 3 5 < x ≤ 10 7 10 < x ≤ 15 10 15 < x ≤ 20 15 Age (x) in years Number of people (f) 10 < x ≤ 20 4 20 < x ≤ 30 18 30 < x ≤ 40 8 40 < x ≤ 50 10 Example Your turn