1. INTERMEDIATE 2 – ADDITIONAL QUESTION BANK UNIT 2 : Graphs, Charts & Tables You have chosen to study: Please choose a question to attempt from the following: 12 34 EXIT Back to Unit 2 Menu 56 Stem & Leaf Dot Plot Cum Freq Table Dot to boxplot Stem to boxplot Piechart

2. (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Reveal answer EXIT Get hint

3. (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end What would you like to do now?

4. (a) Use this information to find the (i) median (ii) lower & upper quartiles (iii) the semi-interquartile range (b)What is the probability that someone chosen at random earns less than £180? The following stem & leaf diagram shows the distribution of wages for employees in a small factory ….. 162369 17111889 182335677 19128 201556n = 25 218 17 4 = £174 GRAPHS, CHARTS, TABLES : Question 1 Go to full solution Go to Comments EXIT Graphs etc Menu median = £183 Q1 = £171 Q3 = £195 = £12 = 2/5 What would you like to do now?

5. Comments Begin Solution Continue Solution Question 1 Menu Back to Home 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 position = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 What would you like to do now? (NOT 3!!!)

6. Comments Begin Solution Continue Solution Question 1 Menu Back to Home 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (i)Median (ii)lower & upper quartiles (iii) the semi-interquartile range 4. Use SIQR = ½ (Q 3 – Q 1 ) / 2 (iii) SIQR = ½(Q3 – Q1) = (£195 - £171)  2 = £12

7. Question 1 5. Use P = no of favourable / no of data 16 2 3 6 9 17 1 1 1 8 8 9 18 2 3 3 5 6 7 7 19 1 2 8 20 1 5 5 6 n = 25 21 8 17 4 = £174 (b)What is the probability that someone chosen at random earns less than £180? No of favourable ( under £180) = 10 No of data = n = 25 (b) Prob(under £180) = 10 / 25 = 2 / 5. Comments Begin Solution Continue Solution Menu Back to Home

8. Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 Median: the middle number in the ordered list. 25 numbers in the list. 1 – 12 13 14 - 25 12 numbers on either side of the medianmedian is the 13th number in order.

9. Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q1Q1 12 numbers before median. 6 numbers either side of Q 1 is midway between the 6th and 7th number.

10. Comments Menu Back to Home Next Comment 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 25 then the median is 13 th value ie median = £183 (ii) Both 6 th & 7 th values are £171 so Q1 = £171 2. There are 12 values before median so Q 1 position = 13 - (12 + 1) / 2 3. There are 12 values after median so Q 3 = 13 + (12 + 1) / 2 19 th is £192 & 20 th is £198 so Q3 = £195 To find the upper and lower quartiles deal with the numbers on either side of the median separately. Q3Q3 12 numbers after median. 6 numbers either side of Q 3 is midway between the 19th and 20th number.

11. Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Go to full solution Go to Comments Reveal answer Get hint EXIT

12. Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Establish lowest & highest values and draw line with scale. Plot a dot for each piece of data and label diagram. For probability use: P = no of favourable / no of data Go to full solution Go to Comments Graphs etc Menu Reveal answer EXIT What would you like to do now?

13. Charts, Graphs & Tables : Question 2 The weights in grams of 20 bags of crisps were as follows 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 a) Illustrate this using a dot plot. b) What type of distribution does this show? c) If a bag is chosen at random what is the probability it will be heavier than the modal weight? Tightly clustered 3/10 EXIT Go to full solution Go to Comments Graphs etc Menu CLICK

14. Question 2 1. Establish lowest & highest values and draw line with scale. 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 Illustrate this using a dot plot. 26283032 (a) Lowest = 28 & highest = 31. Weights in g 2. Plot a dot for each piece of data and label diagram. Comments Begin Solution Continue Solution Menu Back to Home

15. Question 2 3. Make sure you know the possible descriptions of data. 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 26283032 Weights in g What type of distribution does this show? (b) Tightly clustered distribution. Comments Begin Solution Continue Solution Menu Back to Home

16. Question 2 28 29 29 30 31 30 28 30 29 28 29 30 30 28 28 29 29 29 29 28 26283032 Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. If a bag is chosen at random what is the probability it will be heavier than the modal weight? Mode! Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

17. Comments 3. Make sure you know the possible descriptions of data. 26283032 Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

18. Comments 3. Make sure you know the possible descriptions of data. 26283032 Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

19. Comments 3. Make sure you know the possible descriptions of data. 26283032 Weights in g (b) Tightly clustered distribution. Other types of distribution: Menu Back to Home Next Comment

20. Comments Probability = Number of favourable outcomes Number of possible outcomes Weights in g 4. Use P = no of favourable / no of data No of favourable ( bigger than 29) = 6 No of data = n = 20 (c) Prob(W > mode) = 6 / 20 = 3 / 10. Mode! 2626 2828 3030 3232 To calculate simple probabilities: Menu Back to Home Next Comment

21. Charts, Graphs & Tables : Question 3 The results for a class test were 18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? Go to full solution Go to Comments Reveal answer Get hint EXIT Graphs etc Menu

22. Charts, Graphs & Tables : Question 3 The results for a class test were 18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? Establish lowest & highest values and draw table. Complete each row 1 step at a time, calculating running total as you go. For probability use: P = no of favourable / no of data Use median = (n+1) / 2 to establish in which row median lies. EXIT Go to full solution Go to Comments Graphs etc Menu Reveal answer What would you like to do now?

23. Charts, Graphs & Tables : Question 3 The results for a class test were 18 14 16 17 14 16 13 11 13 13 16 14 13 18 15 10 14 17 13 15 15 18 14 17 13 16 10 14 13 17 (a) Construct a cumulative frequency table for this data. (b) What is the median for this data? (c) What is the probability that a pupil selected at random scored under 14? CLICK Median = 14 1/3 EXIT Go to full solution Go to Comments Graphs etc Menu

24. Question 3 1. Establish lowest & highest values and draw a table. (a) Lowest = 10 & highest = 18 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 (a)Construct a cumulative frequency table for this data. Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 2. Complete each row 1 step at a time, calculating running total as you go. Comments Begin Solution Continue Solution Menu Back to Home

25. Question 3 3. Use median = (n+1) / 2 to establish in which row median lies. 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 (b) What is the median for this data? For 30 values median is between 15th & 16th both of which are in row 14. Median Mark = 14 Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

26. Question 3 1814 16 17 14 16 13 11 13 1316 14 13 18 15 10 14 17 1315 15 18 14 17 13 16 10 14 13 17 Mark Frequency Cum Frequency 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 3 19 4 23 4 27 3 30 7 10 (c) What is the probability that a pupil selected at random scored under 14? No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / 3. 4. Use P = no of favourable / no of data Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

27. Comments For 30 values median is between 15th & 16th both of which are in row 14. Median = 14 Mark Freq Cum Freq 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 4 23 4 27 3 30 7 10 3 19 Median: 1 – 15 Q 2 16 - 30 Find the mark at which the cumulative frequency first reaches between 15 th and 16 th number. Median = 14 Menu Back to Home Next Comment

28. Comments Mark Freq Cum Freq 10 11 12 13 14 15 16 17 18 2 1 3 0 3 6 16 4 23 4 27 3 30 7 10 3 19 No of favourable ( under 14) = 10 No of data = n = 30 (c) Prob(mark<14) = 10 / 30 = 1 / 3. Probability = Number of favourable outcomes Number of possible outcomes To calculate simple probabilities: Menu Back to Home Next Comment

29. Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

30. Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Use median position = (n+1) / 2 to find median Q1 is midpoint from start to median Q3 is midpoint from median to end remember bigger SIQR means more variation (spread) in data. EXIT Go to full solution Go to CommentsReveal answer Graphs etc Menu What would you like to do now?

31. Charts, Graphs & Tables : Question 4 The dot plot below shows the number of matches per box in a sample of 23 boxes. 48 50 52 54 56 58 (a)Find the (i) median (ii) lower quartile (iii) upper quartile (b)Construct a boxplot using this data. (c)In a second sample the semi-interquartile range was 2.5. How does this distribution compare to the above sample? Median = 50 So Q1 = 49 So Q3 = 52 the data is distributed more widely than (or not as clustered as) the above data EXIT Menu Full solutionComments CLICK

32. Question 4 48 50 52 54 56 58 (a)Find the (i) median (ii) lower quartile (iii) upper quartile (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 Comments Begin Solution Continue Solution Menu Back to Home

33. Question 4 48 50 52 54 56 58 (b) Construct a boxplot using this data. 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. 48 50 52 54 56 58 (b)Lowest = 48, Q1 = 49, Q2 = 50, Q3 = 52 & Highest = 58. Comments Begin Solution Continue Solution Menu Back to Home

34. Question 4 48 50 52 54 56 58 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c) In a second sample the semi-interquartile range was 2.5. How does this compare? (c)For above sample SIQR = (52 - 49)  2 = 1.5 In a sample where the SIQR is 2.5 the data is distributed more widely than (or not as clustered as) the above data Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

35. Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 23 numbers in the list: 1 - 1112 13 - 23 Q2Q2 11 numbers on either side of the median The median: Menu Back to Home Next Comment

36. Comments (a)(i) Sample size = 23 so median position is 12. ie (23+1)  2 Median = 50 1. Use median = (n+1) / 2 to find median 2. There are 11 values before median so Q 1 position = 12 - (11 + 1) / 2 3. There are 11 values after median so Q 3 position = 12 + (11 + 1) / 2 (ii) Middle of 1 st 11 is position 6. So Q1 = 49 (iii) Middle of 2 nd 11 is position 18. So Q3 = 52 For quartiles: 1 - 5 6 7 - 11 12 Q2Q2 Q1Q1 Q3Q3 13 - 17 18 19 - 23 12 Q2Q2 Now count through the list until you reach the 6 th, 12 th,and 18 th number in the list. Menu Back to Home Next Comment

37. Comments 5. Calculate SIQR then compare remember bigger SIQR means more variation (spread) in data. (c)For above sample SIQR = (52 - 49)  2 = 1.5 In a sample where the SIQR is 2.5 the data is distributed more widely than or not as clustered as the above data The semi-interquartile range is a measure of the range of the “middle” 50%. S.I.R. = (Q 3 - Q 1 ) 1212 Remember: when asked to compare data always consider average and spread. It is a measure of how spread-out and so how “consistent” or “reliable” the data is. Menu Back to Home Next Comment

38. Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 Full solution Comments Reveal answer Get hint EXIT

39. Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 Use median position = (n+1) / 2 to find median position Q1 is midpoint from start to median Q3 is midpoint from median to end When comparing two data sets comment on spread and average Full solution Comments Reveal answer Menu EXIT What now?

40. Charts, Graphs & Tables : Question 5 The stem & leaf diagram below shows the weight distribution of 26 people when they joined a slimming club. (a)Find the median, lower & upper quartiles for this data. (b)Use the data to construct a boxplot. (c)The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 60 70 80 90 100 110 120 median = 87 Q1 = 77 Q3 = 99 Full solution Comments Menu EXIT CLICK

41. Question 5 1. Use median = (n+1) / 2 to find median (a)(i) Since n = 26 then the median is between 13 th & 14 th value ie median = 87 (ii) so Q1 = 77 2. There are 13 values before median so Q 1 position is 6 th value 3. There are 13 values after median so Q 3 position is 20 th position so Q3 = 99 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg (a)Find the median, lower & upper quartiles for this data. Comments Begin Solution Continue Solution Menu Back to Home

42. Question 5 6 0 2 7 1 3 5 7 7 8 2 2 2 5 6 6 8 9 9 4 4 6 9 9 10 5 7 7 11 12 1 1 3 11 4 = 114kg 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. 60 70 80 90 100 110 120 (b) Use the data to construct a boxplot. Comments Begin Solution Continue Solution Menu Back to Home

43. Question 5 5. Compare spread and relevant average. (c) The boxplot below shows the weight distribution for these people after several months. Compare the two & comment on the results. 60 70 80 90 100 110 120 (c)Lightest has put on weight – lowest now 65, heaviest 3 have lost weight – highest now 115, median same but overall spread of weights has decreased as Q3-Q1 was 22 but is now only 15. Comments Begin Solution Continue Solution Menu Back to Home What would you like to do now?

44. Comments 4. Draw number line with scale. Make sure you note highest & lowest as well as Q 1, Q 2, Q 3. (b)Lowest = 60, Q1 = 77, Q2 = 87, Q3 = 99 & Highest = 123. 60 70 80 90 100 110 120 Box Plot : LowestHighestQ1Q1 Q2Q2 Q3Q3 Remember: To draw a boxplot you need a “five-figure summary”: five-figure summary Menu Back to Home Next Comment

45. Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? Go to full solution Go to Comments Reveal answer Get hint Graphs etc Menu EXIT

46. Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? angle 360° = amount 630 EXIT Go to full solution Go to Comments Reveal answer Graphs etc Menu What would you like to do now?

47. Charts, Graphs & Tables : Question 6 The pie chart below shows the breakdown of how a sample of 630 people spent their Saturday nights. clubbing 144° x°x° theatre cinemaWatching TV (a)How many people went clubbing? (b) If 84 people went to the theatre then how big is x°? = 252 = 48° EXIT Go to full solution Go to Comments Graphs etc Menu What would you like to do now?

48. Question 6 clubbin g 144° x°x° theatre cinemaWatching TV How many people went clubbing? 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount 630 144° 360° = amount 630 360 x amount = 144 x 630 amount = 144 x 630  360 = 252 Comments Begin Solution Continue Solution Menu Back to Home

49. Question 6 clubbin g 144° x°x° theatre cinemaWatching TV 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = 84 630 630 x angle = 360° x 84 angle = 360° x 84  630 = 48° (b) If 84 people went to the theatre then how big is x°? Comments Begin Solution Continue Solution Menu Back to Home

50. Comments 1. Set up ratio of angles and sectors and cross multiply. (a) The angle is 144° so ….. angle 360° = amount 630 144° 360° = amount 630 360 x amount = 144 x 630 amount = 144 x 630  360 = 252 Can also be tackled by using proportion: Amount = x 630 144 360 Menu Back to Home Next Comment

51. Comments Can also be tackled by using proportion: 2. Set up ratio of angles and sectors and cross multiply. (b) The amount is 84 so ….. angle 360° = amount 630 angle 360° = 84 630 630 x angle = 360° x 84 angle = 360° x 84  630 = 48° 84 = x 630 x 360 630 x = 84 x 360 x = 84 360 630 x Menu Back to Home Next Comment End of graphs, charts etc.