Statistical charts and diagrams — Paper 4 · IGCSE Mathematics 0580

9 Answer the whole of this question on one sheet of graph paper. The heights (h cm) of 270 students in a school are measured and the results are shown in the table. h Frequency 120 < h 130 15 130 < h 140 24 140 < h 150 36 150 < h 160 45 160 < h 170 50 170 < h 180 43 180 < h 190 37 190 < h 200 20 (a) Write down the modal group. [1] (b) (i) Calculate an estimate of the mean height. [4] (ii) Explain why the answer to part (b)(i) is an estimate. [1] (c) The following table shows the cumulative frequencies for the heights of the students. h Cumulative frequency h 120 0 h 130 p h 140 q h 150 r h 160 120 h 170 170 h 180 213 h 190 250 h 200 270 Write down the values of p, q and r. [2] (d) Using a scale of 1cm to 5 units, draw a horizontal h-axis, starting at h = 120. Using a scale of 1cm to 20 units on the vertical axis, draw a cumulative frequency diagram. [5] (e) Use your diagram to find (i) the median height, [1] (ii) the upper quartile, [1] (iii) the inter-quartile range, [1] (iv) the 60th percentile. [1] (f) All the players in the school’s basketball team are chosen from the 30 tallest students. Use your diagram to find the least possible height of any player in the basketball team. [2]

19 marks

Mark scheme: 9 (a) 160 < h Y 170 B1 (b) (i) Mid values 125, 135, 145, 155, 165, 175, M1 Allow 1 slip 185, 195 (15 x 125 + 24 x 135 + 36 x 145 + 45 M1 Dep on mid values ± 0.5, allow 1 slip in mid- x 155 + 50 x 165 + 43 x 175 + 37 x 185 values (43830) + 20 x 195) ÷ 270 M1 Dep on previous method 162 or better A1 (162.333..) www4 (ii) Mid-values are an estimate of each B1 e.g. exact values not given interval o.e. (c) p = 15, q = 39, r = 75 B2 B1 for 2 correct. If no labels, take in order given (d) Correct scales S1 9 points correctly plotted ft P3ft P2ft for 7 or 8 correct acc. 1 mm P1ft for 5 or 6 correct Curve or line through 9 points C1ft Dep on ‘S’ shape within 1 small square of 2 points (e) (i) 162 to 164 B1 (ii) 176 to 178 B1 (iii) 28 to 30 B1 (iv) 167.5 to 168.5 B1 (f) Uses 240 or 241 on cumul, freq. axis M1 e.g. annotates graph or shows values in working 186.5 to 188 A1 ww2 [19] IGCSE – NOVEMBER 2005 0580/0581 4 IGCSE – NOVEMBER 2005 0580/0581 4 IGCSE – NOVEMBER 2005 0580/0581 4

This question in 0580/41 Oct/Nov 2005

6 Answer the whole of this question on a sheet of graph paper. Kristina asked 200 people how much water they drink in one day. The table shows her results. Amount of water (x litres) Number of people 0 < x 0.5 8 0.5 < x 1 27 1 < x 1.5 45 1.5 < x 2 50 2 < x 2.5 39 2.5 < x 3 21 3 < x 3.5 7 3.5 < x 4 3 (a) Write down the modal interval. [1] (b) Calculate an estimate of the mean. [4] (c) Make a cumulative frequency table for this data. [2] (d) Using a scale of 4 cm to 1 litre of water on the horizontal axis and 1 cm to 10 people on the vertical axis, draw the cumulative frequency graph. [5] (e) Use your cumulative frequency graph to find (i) the median, [1] (ii) the 40th percentile, [1] (iii) the number of people who drink at least 2.6 litres of water. [2] (f) A doctor recommends that a person drinks at least 1.8 litres of water each day. What percentage of these 200 people do not drink enough water? [2]

18 marks

Mark scheme: 6 (a) 1.5 < x ≤ 2 B1 (b) (8×0.25 + 27×0.75 + 45×1.25 +…….. M1 For mid-values (allow two slips) ……….. 3×3.75) M1 For Σfx (allow two slips) dep on first M1, or mid-values ± 0.05 their 345.5 ÷ 200 M1 for ÷ 200 dep on second M1 1.7275, 1.727, 1.728 or 1.73 cso A1 www 4 (c) 8, 35, 80, 130, 169, 190, 197, 200 B2 If B0, allow M1 for clear attempt to add accumulatively (d) axes correct scale S1 Not reversed and must reach 200 vertically, even if not labelled 8 points plotted ft part (c) P3dep dep on at least M1 in (c) (0.5, 8), (1, 35), (1.5, 80), (2, 130), (2.5, 8 points from their values 169), (3, 190), (3.5, 197), For x-values (upper boundary values), (4, 200) points must touch grid line For y-values, even, must touch grid line, odd must be inside square. P2 for 6 or 7 points ft P1 for 4 or 5 points ft curve (or polygon) either correct or C1 Allow 1 mm tolerance through 8 points and correct shape Ignore any bars drawn if they do not compromise the points and graph (e) (i) 1.65-1.75 B1 (ii) 1.5 B1 (iii) 23 – 29 integers only B2 If B0 allow SC1 for non-integer in correct range, or 172 – 177 seen (may be written on graph) (f) 54 – 56.5 B2 SC1 for figures 108 – 113 or 87 – 92 Accept if written on graph www 2 [18] IGCSE – May/June 2007 0580 and 0581 04

This question in 0580/41 May/June 2007

8 Fifty students are timed when running one kilometre. For Examiner's The results are shown in the table. Use Time 4.0 < t Y 4.5 4.5 < t Y 5.0 5.0 < t Y 5.5 5.5 < t Y 6.0 6.0 < t Y 6.5 6.5 < t Y 7.0 (t minutes) Frequency 2 7 8 18 10 5 (a) Write down the modal time interval. Answer(a) min [1] (b) Calculate an estimate of the mean time. Answer(b) min [4] (c) A new frequency table is made from the results shown in the table above. Time 4.0 < t Y 5.5 5.5 < t Y 6.0 6.0 < t Y 7.0 (t minutes) Frequency 18 (i) Complete the table by filling in the two empty boxes. [1] (ii) On the grid below, complete an accurate histogram to show the information in this new For table. Examiner's Use 40 30 Frequency density 20 10 0 t 4 5 6 7 8 Time (minutes) [3] (iii) Find the number of students represented by 1 cm2 on the histogram. Answer(c)(iii) [1]

10 marks

Mark scheme: 8 (a) 5.5 < t Y 6 B1 Condone poor notation (b) 4.25, 4.75, 5.25, 5.75, 6.25, 6.75 M1 At least 5 correct mid-values seen where x is in the correct interval allow (2 × 4.25 + 7 × 4.75 + 8 × 5.25 + 18 × 5.75 M1 ∑fx + 10 × 6.25 + 5 × 6.75) (= 283.5) one further slip M1 Depend on second method ÷ 50 or their ∑f 5.67 www4 A1 After M3 allow 5.7 isw conversion to mins/secs and reference to classes (c) (i) 17, 15 B1 (ii) Rectangular bars of heights 11.3 B1ft ft their 17 divided by 1.5 and 15 B1ft ft their 15 11.3 plot between 11 and 12 include lines and 15 to be touching the 15 line Correct widths of 1.5 and 1 – no gaps B1 (iii) 2.5 cao B1 [10] IGCSE – October/November 2009 0580 04

This question in 0580/41 Oct/Nov 2009

7 200 students were asked how many hours they exercise each week. Examiner's Use The table shows the results. Time (t hours) 0ItY5 5ItY10 10ItY15 15ItY20 20ItY25 25ItY30 30ItY35 35ItY40 Number of 12 15 23 30 40 35 25 20 students (a) Calculate an estimate of the mean. Answer(a) h [4] (b) Use the information in the table above to complete the cumulative frequency table. Time (t hours) t Y=5 t Y=10 t Y=15 t Y=20 t Y=25 t Y=30 t Y=35 t Y=40 Cumulative frequency 12 27 50 80 120 200 [1] For Examiner's Use 200 180 160 140 120frequency 100 80Cumulative 60 40 20 t 0 5 10 15 20 25 30 35 40 Time (t hours) (c) On the grid, draw a cumulative frequency diagram to show the information in the table in part (b). [4] (d) On your cumulative frequency diagram show how to find the lower quartile. [1] (e) Use your cumulative frequency diagram to find (i) the median, Answer(e)(i) [1] (ii) the inter-quartile range, Answer(e)(ii) [1] (iii) the 64th percentile, Answer(e)(iii) [1] (iv) the number of students who exercise for more than 17 hours. Answer(e)(iv) [2]

15 marks

Mark scheme: 7 (a) 12 × 2.5 + 15 × 7.5 + 23 × 12.5 + 30 × M1 mid-values any three soi 17.5 + 40 × 22.5 + 35 × 27.5 + 25 × M1 Use of Σfx dep on x anywhere in each interval 32.5 + 20 × 37.5 (including lower bound) – allow 2 slips or omissions ÷ 200 M1 Depend on second M 21.9 www 4 A1 (b) 155, 180 1 (c) 8 points plotted ft, ignoring (0, 0) P3ft P2ft for 6 or 7 plotted , P1ft for 4 or 5 plotted Reasonable increasing curve or C1ft Condone starting at (5, 12) and ft only if shape polygon through their 8 points correct. (d) Either horizontal or vertical line at 1 least 1 cm long at y = 50 on the curve (e) (i) 22 – 23 1 (ii) 13.5 – 14.5 1 (iii) 25.5 – 26.5 1 (iv) 136 – 140 must be integer 2 SC1 for 60 – 64 seen and must be integer 2 2

This question in 0580/42 May/June 2010

10 (a) For a set of six integers, the mode is 8, the median is 9 and the mean is 10. For Examiner's The smallest integer is greater than 6 and the largest integer is 16. Use Find the two possible sets of six integers. Answer(a) First set , , , , , Second set , , , , , [5] (b) One day Ahmed sells 160 oranges. He records the mass of each orange. The results are shown in the table. Mass (m grams) 50 < m Y 80 80 < m Y 90 90 < m Y 100 100 < m Y 120 120 < m Y 150 Frequency 30 35 40 40 15 (i) Calculate an estimate of the mean mass of the 160 oranges. Answer(b)(i) g [4] (ii) On the grid, complete the histogram to show the information in the table. For Examiner's Use 5 4 3 Frequency density 2 1 0 m 50 60 70 80 90 100 110 120 130 140 150 Mass (grams) [4] Question 11 is printed on the next page.

13 marks

Mark scheme: 10 (a) 7, 8, 8, 10, 11, 16 5 Mark answer spaces only or clearly indicated and 8, 8, 8, 10, 10, 16 lists. Allow numbers in any order but must be lists of 6 integers B4 for either correct list If not B4 then B1 for a series with mode 8 and B1 for a series with median 9 and B1 for a series with sum 60 (b) (i) (30 × 65 + 35 × 85 + 40 × 95 + 4 M1 for mid-values soi (allow 1 error/omission) with x in correct 40 × 110 + 15 × 135) ÷ 160 and M1 for use of ∑fx interval including both boundaries allow one further error/omission and M1 (dependent on second M) for ÷ 160 94.7 (94.68 – 94.69) www 4 (ii) Heights of 4, 2, 0.5 with correct 4 B3 for 2 correct interval widths or B2 for 1 correct or B1 for all three freq. densities correct but no/incorrect graph

This question in 0580/43 Oct/Nov 2010

8 The table below shows the marks scored by a group of students in a test. For Examiner's Use Mark 11 12 13 14 15 16 17 18 Frequency 10 8 16 11 7 8 6 9 (a) Find the mean, median and mode. Answer(a) mean = median = mode = [6] (b) The table below shows the time (t minutes) taken by the students to complete the test. Time (t) 0 I=t Y=10 10 I=t Y=20 20 I=t Y=30 30 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 2 19 16 14 15 9 (i) Cara rearranges this information into a new table. Complete her table. Time (t) 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 9 [2] (ii) Cara wants to draw a histogram to show the information in part (b)(i). Complete the table below to show the interval widths and the frequency densities. 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Interval 10 width Frequency 0.9 density [3] (c) Some of the students were asked how much time they spent revising for the test. For Examiner's 10 students revised for 2.5 hours, 12 students revised for 3 hours and n students revised for Use 4 hours. The mean time that these students spent revising was 3.1 hours. Find n. Show all your working. Answer(c) n = [4]

15 marks

Mark scheme: 8 14.2 3 M1 for Σ fx (10 × 11 + 8 × 12 + 16 × 13 + 11 × 14 + 7 × 15 + 8 × 16 + 6 × 17 + 9 × 18 ) (1065) (allow one error or omission) M1dep for ÷ Σf (10 + 8 + 16 + 11 + 7 + 8 + 6 + 9) (75) (allow one further error or omission) 14 2 M1 for 37th, 37.5th or 38th seen 13 1 (b) (i) 21, 30, 15 2 B1 for 2 correct (ii) 20 20 10 (10) 3 1, 1, 1 for each correct vertical pair 1.05 1.5 1.5 (0.9) (c) 10 × 5.2 + 12 × 3 + 4 n ( = )1.3 M2 M1 for either numerator or denominator seen 10 + 12 + n multiplying across and collecting terms M1 dep on linear numerator and denominator their (68.2 − 25 − 36) = their (4 − 3.1) × n (n =) 8 www 4 A1 IGCSE – May/June 2011 0580 41

This question in 0580/41 May/June 2011

5 For Examiner′s Height (h cm) 150 < h Ğ 160 160 < h Ğ 165 165 < h Ğ 180 180 < h Ğ 190 Use Frequency 5 9 18 10 The table shows information about the heights of a group of 42 students. (a) Using mid-interval values, calculate an estimate of the mean height of the students. Show your working. Answer(a) … cm [3] (b) Write down the interval which contains the lower quartile. Answer(b) … [1] (c) Complete the histogram to show the information in the table. One column has already been drawn for you. 2 Frequency density 1 0 150 155 160 165 170 175 180 185 190 Height (cm) [4] _____________________________________________________________________________________

8 marks

Mark scheme: 5 (a) 171.25 (or 171 or 171.2 or 171.3) 3 M1 for 5 × 155 + 9 × 162.5 + 18 × 172.5 + 10 × 185 [= 7192.5] www and M1 (dep on M1) for their Σfx ÷ 42 (b) 160 < x ≤ 165 oe 1 (c) Blocks with heights of 1.8, 1.2, 1, with 4 B3 for 2 correct blocks correct interval widths and no gaps or B2 for 1 correct block or B1 for 3 correct frequency densities or heights or 3 correct widths IGCSE – May/June 2013 0580 41 Qu. Answer Marks Part marks 15 7

This question in 0580/41 May/June 2013

7 120 students are asked to answer a question. For Examiner′s The time, t seconds, taken by each student to answer the question is measured. Use The frequency table shows the results. Time 0 < t Y 10 10 < t Y 20 20 < t Y 30 30 < t Y 40 40 < t Y 50 50 < t Y 60 Frequency 6 44 40 14 10 6 (a) Calculate an estimate of the mean time. Answer(a) … s [4] (b) (i) Complete the cumulative frequency table. Time t Y 10 t Y 20 t Y 30 t Y 40 t Y 50 t Y 60 Cumulative frequency 6 104 120 [2] (ii) On the grid below, draw a cumulative frequency diagram to show this information. 120 100 80 Cumulative 60 frequency 40 20 t 0 10 20 30 40 50 60 Time (seconds) [3] (iii) Use your cumulative frequency diagram to fi nd the median, the lower quartile and For Examiner′s the 60th percentile. Use Answer(b)(iii) Median … s Lower quartile … s 60th percentile … s [4] (c) The intervals for the times taken are changed. (i) Use the information in the frequency table on the opposite page to complete this new table. Time 0 < t Y 20 20 < t Y 30 30 < t Y 60 Frequency 40 [2] (ii) On the grid below, complete the histogram to show the information in the new table. One column has already been drawn for you. 4 3.5 3 2.5 Frequency density 2 1.5 1 0.5 t 0 10 20 30 40 50 60 Time (seconds) [3] _____________________________________________________________________________________

18 marks

Mark scheme: 7 (a) 24.7 or 24.66 to 24.67 4 M1 for midpoints soi (condone 1 error or omission) (5, 15, 25, 35, 45, 55) and M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) and M1 (dependent on second M) for ∑fx ÷ 120 (b) (i) 50, 90, 114 2 B1 for 2 correct (ii) Correct curve 3 Ignore section to left of t = 10 or ruled polygon B1 for 6 correct horizontal plots and B1FT for 6 correct vertical plots If 0 scored SC1 for 5 out of 6 correct plots and B1FT for curve or polygon through at least 5 of their points dep on an increasing curve/polygon that reaches 120 vertically (iii) 21.5 to 23 B1 15 to 16.5 B1 24 to 26 4 B2 or B1 for 72 or 72.6 seen (c) (i) 50, 30 2 B1 each (ii) Correct histogram 3FT B1 for blocks of widths 0 – 20, 30 – 60 (no gaps) B1FT for block of height 2.5 or their 50 ÷ 20 and B1FT for block of height 1 or their 30 ÷ 30 IGCSE – October/November 2013 0580 41

This question in 0580/41 Oct/Nov 2013

5 (a) 80 students were asked how much time they spent on the internet in one day. For Examiner′s This table shows the results. Use Time (t hours) 0 < t Y 1 1 < t Y 2 2 < t Y 3 3 < t Y 5 5 < t Y 7 7 < t Y 10 Number of students 15 11 10 19 13 12 (i) Calculate an estimate of the mean time spent on the internet by the 80 students. Answer(a)(i) … hours [4] (ii) On the grid, complete the histogram to show this information. 16 14 12 10 Frequency 8 density 6 4 2 t 0 1 2 3 4 5 6 7 8 9 10 Time (hours) [4] 3 For (b) The probability that Chaminda uses the internet on any day is 5 . Examiner′s Use 3 The probability that Niluka uses the internet on any day is 4 . (i) Complete the tree diagram. Chaminda Niluka 3 Uses the 4 internet Uses the 3 internet 5 Does not … use the internet … Uses the internet … Does not use the internet Does not … use the internet [2] (ii) Calculate the probability, that on any day, at least one of the two students uses the internet. Answer(b)(ii) … [3] (iii) Calculate the probability that Chaminda uses the internet on three consecutive days. Answer(b)(iii) … [2] _____________________________________________________________________________________

15 marks

Mark scheme: 5 (a) (i) 3.81 or 3.812 to 3.813 or 4 M1 for midpoints soi (condone 1 error or omission 3h 49min nfww and M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) and M1 (dep on 2nd M1) for ∑fx ÷ 80 (305 ÷ 80) (ii) Correct histogram 4 B1 for each correct block and B1 for correct widths 2 1 3 1 2 1 (b) (i) , , , oe 2 B1 for or both s in correct place 5 4 4 4 5 4 18  9  2 1 (ii) nfww 3 M2 FT for 1 – their × their 20  10  5 4 3 3 3 1 2 3 or × + × their + their × oe 5 4 5 4 5 4 or 2 1 M1 FT for their × their 5 4 3 1 2 3 or × their + their × oe 5 4 5 4 27 3 3 3 (iii) [0.216] 2 M1 for × × 125 5 5 5 2 2 2

This question in 0580/43 Oct/Nov 2013

1 For 8 (a) Rearrange s = ut + 2 at2 to make a the subject. Examiner′s Use Answer(a) a = … [3] (b) The formula v = u + at can be used to calculate the speed, v, of a car. u = 15, a = 2 and t = 8, each correct to the nearest integer. Calculate the upper bound of the speed v. Answer(b) … [3] (c) The diagram shows the speed-time graph for a car travelling between two sets of traffi c lights. For Examiner′s Use 16 Speed (m/s) 0 10 20 25 Time (seconds) (i) Calculate the deceleration of the car for the last 5 seconds of the journey. Answer(c)(i) … m/s2 [1] (ii) Calculate the average speed of the car between the two sets of traffi c lights. Answer(c)(ii) … m/s [4] _____________________________________________________________________________________

11 marks

Mark scheme: 2 (s ut ) 8 (a) 2 oe nfww 3 M1 for a correct rearrangement to isolate the a term t and M1 for a correct multiplication by 2 and M1 for a correct division by t2 (b) 36.75 cao 3 M2 for 15.5 + 2.5 × 8.5 B1 for two of 15.5, 2.5, 8.5 seen 16 (c) (i) or better [3.2] 1 5 (ii) 11.2 4 M2 for ½(25 + 10)16 (= 280) or M1 for appreciation of distance from area and M1 for their 280 ÷ 25 (dep on M1)

This question in 0580/43 Oct/Nov 2013

7 (a) 1.0 0.8 0.6 Frequency density 0.4 0.2 m 0 10 20 30 40 50 60 70 80 90 100 Mass (grams) The histogram shows some information about the masses (m grams) of 39 apples. (i) Show that there are 12 apples in the interval 70 < m Y 100 . Answer(a)(i) [1] (ii) Calculate an estimate of the mean mass of the 39 apples. Answer(a)(ii) … g [5] (b) The mean mass of 20 oranges is 70 g. One orange is eaten. The mean mass of the remaining oranges is 70.5 g. Find the mass of the orange that was eaten. Answer(b) … g [3] __________________________________________________________________________________________

9 marks

Mark scheme: 24 7 (a) (i) (100 – 70) × 0.4 [= 12] or better 1 Accept × 39 oe 78 (ii) 60.9 or 60.89… nfww 5 B1 for 3 or 4 correct extra frequencies 3, 6, 10, 8 soi M1 for at least 4 of mid-interval values 15, 40, 55, 65, 85 soi M1 for Σfx where x is any value in each interval allow their frequencies provided integers and they must be shown [3 × 15 + 6 × 40 + 10 × 55 + 8 × 65 + 12 × 85] [2375] M1 (dependent on second M1) for ÷ 39 or ÷ (3 + 6 + 10 + 8 +12) (b) 60.5 3 M2 for 20 × 70 – 19 × 70.5 oe or M1 for either 20 × 70 or 19 × 70.5 600 600

This question in 0580/42 May/June 2014

6 A company tested 200 light bulbs to fi nd the lifetime, T hours, of each bulb. The results are shown in the table. Lifetime Number (T hours) of bulbs 0 < T Y 1000 10 1000 < T Y 1500 30 1500 < T Y 2000 55 2000 < T Y 2500 72 2500 < T Y 3500 33 (a) Calculate an estimate of the mean lifetime for the 200 light bulbs. Answer(a) … hours [4] (b) (i) Complete the cumulative frequency table. Lifetime (T hours) T Y 1000 T Y 1500 T Y 2000 T Y 2500 T Y 3500 Number of bulbs [2] (ii) On the grid, draw a cumulative frequency diagram to show this information. 200 150 Cumulative frequency 100 50 T 0 500 1000 1500 2000 2500 3000 3500 Lifetime (hours) [3] (iii) The company says that the average lifetime of a bulb is 2200 hours. Estimate the number of bulbs that lasted longer than 2200 hours. Answer(b)(iii) … [2] (c) Robert buys one energy saving bulb and one halogen bulb. 9 The probability that the energy saving bulb lasts longer than 3500 hours is . 10 3 The probability that the halogen bulb lasts longer than 3500 hours is . 5 Work out the probability that exactly one of the bulbs will last longer than 3500 hours. Answer(c) … [4] __________________________________________________________________________________________

15 marks

Mark scheme: 6 (a) 2000 or 1998.75 or 1998.8 or 4 M1 for midpoints soi (condone 1 error or omission) 1999 nfww (500, 1250, 1750, 2250, 3000) and with x in correct interval M1 for use of ∑fx including both boundaries (condone 1 further error or omission) (5000, 37500, 96250, 162000, 99000) and M1 (dep on 2nd M1) for ∑fx ÷ 200 (b) (i) 10, 40, 95, 167, 200 2 B1 for 2 correct (ii) Correct curve or ruled polygon 3 B1FT their (b)(i) for 5 correct heights within 1mm vertically and B1 for 5 points at upper ends of intervals on correct vertical line and B1FT (dep on at least B1) for increasing curve or polygon through 5 points After 0 scored, SC1FT for 4 correct points plotted (iii) 68 to 80 2 M1 for 120 to 132 seen 21 9 2 1 3 (c) oe 4 M3 for × + × oe or better 50 10 5 10 5 or 9 2 1 3 18 3 M2 for × or × or oe or oe 10 5 10 5 50 50 or 1 2 M1 for sight of and 10 5 Qu Answers Mark Part Marks

This question in 0580/41 Oct/Nov 2014

9 The table shows the height, h cm, of 40 children in a class. Height (h cm) 120 < h  130 130 < h  140 140 < h  144 144 < h  150 150 < h  170 Frequency 3 14 4 6 13 (a) Write down the class interval containing the median. Answer(a) … < h  … [1] (b) Calculate an estimate of the mean height. Answer(b) … cm [4] (c) Complete the histogram. 2 1.5 Frequency density 1 0.5 0 h 120 130 140 150 160 170 Height (cm) [4] __________________________________________________________________________________________

9 marks

Mark scheme: 9 (a) 140 < h ≤ 144 1 (b) 144.875 nfww 4 M1 for at least 4 correct mid-values soi M1 for ∑fx where x is in the correct interval, allow one further error/omission M1 dep for ÷ 40 dependent on second method mark (c) 4 correct blocks 4 B3 for 3 correct blocks B2 for 2 correct blocks B1 for 1 correct block or at least 3 correct frequency densities (1.4, 1, 1, 0.65 )

This question in 0580/42 Feb/March 2015

6 The table shows the time, t minutes, that 400 people take to complete a test. Time taken 0  t  10 10  t  24 24  t  30 30  t  40 40  t  60 60  t  70 (t mins) Frequency 10 90 135 85 70 10 (a) (i) Write down the modal time interval. Answer(a)(i) … min [1] (ii) Calculate an estimate of the mean time taken to complete the test. Answer(a)(ii) … min [4] (b) (i) Complete the table of cumulative frequencies. Time taken t  10 t  24 t  30 t  40 t  60 t  70 (t mins) Cumulative 10 100 400 frequency [2] (ii) On the grid opposite, draw a cumulative frequency diagram to show this information. 400 350 300 250 Cumulative frequency 200 150 100 50 t 0 10 20 30 40 50 60 70 Time taken (minutes) [3] (c) Use your graph to estimate (i) the median time, Answer(c)(i) … min [1] (ii) the inter-quartile range, Answer(c)(ii) … min [2] (iii) the 15th percentile, Answer(c)(iii) … min [2] (iv) the number of people who took more than 50 minutes. Answer(c)(iv) … [2]

17 marks

Mark scheme: 6 (a) (i) 24 < t Y 30 1 (ii) 30.9 or 30.875 nfww 4 M1 for midpoints soi (condone 1 error or omission) 5, 17, 27, 35, 50, 65 soi M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) (50, 1530, 3645, 2975, 3500, 650) and M1 (dep on 2nd M1) for ∑fx ÷ 400 (b) (i) [10 100] 235 320 390 [400] 2 B1 for any two correct SC1 for 235, n, n + 70 n > 235 (ii) Correct curve or polygon 3 B1 for correct horizontal placement B1FT for correct vertical placement B1FT dep on at least B1 for reasonable increasing curve or polygon through their 6 points If zero scored SC1 for 5 out of 6 points correctly plotted (c) (i) 27.5 to 29 1 (ii) 12 to 14 2 B1 for 36 to 38 or 24 seen (iii) 18 to 20 2 B1 for 60 seen or marked on grid (iv) 30 to 45 2 B1 for 355 to 370 seen

This question in 0580/41 May/June 2015

7 (a) A group of 50 students estimated the mass, M grams, of sweets in a jar. The results are shown in the table. Mass (M grams) Number of students 0 < M  200 5 200 < M  300 9 300 < M  350 18 350 < M  400 12 400 < M  500 6 (i) Calculate an estimate of the mean. Answer(a)(i) … grams [4] (ii) Complete this histogram to show the information in the table. 0.4 0.3 Frequency 0.2 density 0.1 M 0 100 200 300 400 500 Mass (grams) [3] (b) A group of 50 adults also estimated the mass, M grams, of the sweets in the jar. The histogram below shows information about their estimates. Use the histograms to make two comparisons between the distributions of the estimates of the students and the adults. 0.4 0.3 Frequency 0.2 density 0.1 M 0 100 200 300 400 500 Mass (grams) Answer(b) 1 … … 2 … … [2] __________________________________________________________________________________________

9 marks

Mark scheme: 7 (a) (i) 316 4 M1 for 100, 250, 325, 375, 450 soi M1 for Σfm with m’s in intervals including boundaries [15800] M1 (dep on 2nd M1) for their Σfm ÷ 50 (ii) Three correct blocks with heights 3 B2 for two correct blocks 0.09, 0.36, 0.24 with correct widths or and no gaps B1 for one correct block or three correct frequency densities soi (b) Students have a greater range of B1 estimates oe [On average] adults estimated a B1 greater mass oe

This question in 0580/42 May/June 2015

6 120 students take a mathematics examination. (a) The time taken, m minutes, for each student to answer question 1 is shown in this table. Time (m minutes) 0 < m G 1 1 < m G 2 2 < m G 3 3 < m G 4 4 < m G 5 5 < m G 6 Frequency 72 21 9 11 5 2 Calculate an estimate of the mean time taken. Answer(a) … min [4] (b) (i) Using the table in part (a), complete this cumulative frequency table. Time (m minutes) m G 1 m G 2 m G 3 m G 4 m G 5 m G 6 Cumulative frequency 72 120 [2] (ii) Draw a cumulative frequency diagram to show the time taken. 120 110 100 90 80 70 Cumulative frequency 60 50 40 30 20 10 0 m 1 2 3 4 5 6 Time (minutes) [3] (iii) Use your cumulative frequency diagram to find (a) the median, Answer(b)(iii)(a) … min [1] (b) the inter-quartile range, Answer(b)(iii)(b) … min [2] (c) the 35th percentile. Answer(b)(iii)(c) … min [2] (c) A new frequency table is made from the table shown in part (a). Time (m minutes) 0 < m G 1 1 < m G 3 3 < m G 6 Frequency 72 (i) Complete the table above. [2] (ii) A histogram was drawn and the height of the first block representing the time 0 < m G 1 was 3.6 cm. Calculate the heights of the other two blocks. Answer(c)(ii) … cm and … cm [3] __________________________________________________________________________________________

19 marks

Mark scheme: 6 (a) 1.35 nfww 4 M1 for 0.5, 1.5, 2.5, 3.5, 4.5, 5.5 soi, M1 for Σfm soi by 162 where m is in correct interval including boundaries M1dep for Σfm ÷ 120 or Σfm ÷ Σf dependent on second M1 earned (b) (i) 93, 102, 113, 118 2 SC1FT for 1 error (ii) Correct diagram 3 B1FT for correct vertical plots and B1 for correct horizontal plots and B1FT dep on at least B1 for reasonable increasing curve or polygon through their 6 points If zero scored, SC1FT for 5 out of 6 correct plots (iii) (a) 0.6 to 0.85 1 (b) 1.3 to 1.7 2 B1 for UQ = 1.7 to 1.9 or LQ = 0.2 to 0.4 (c) 0.3 to 0.6 2FT Allow in correct range provided there is no evidence of reading at 35 or FT their reading at 42 B1 for 42 soi (c) (i) 30 and 18 2 B1 for each (ii) 0.75 and 0.3 3FT FT (their 30) ÷ 40 and (their 18) ÷ 60 B2FT for either 0.75 or 0.3 or M1 for their 30 ÷ 2 or ÷ 20 or for their 18 ÷ 3 or ÷ 20

This question in 0580/41 Oct/Nov 2015

6 The table shows information about the masses, m grams, of 160 apples. Mass (m grams) 30 < m  80 80 < m  100 100 < m  120 120 < m  200 Frequency 50 30 40 40 (a) Calculate an estimate of the mean. Answer(a) … g [4] (b) On the grid, complete the histogram to show the information in the frequency table. 2.5 2 1.5 Frequency density 1 0.5 m 0 40 80 120 160 200 Mass (grams) [3] (c) An apple is chosen at random from the 160 apples. Find the probability that its mass is more than 120 g. Answer(c) … [1] (d) Two apples are chosen at random from the 160 apples, without replacement. Find the probability that (i) they both have a mass of more than 120 g, Answer(d)(i) … [2] (ii) one has a mass of more than 120 g and one has a mass of 80 g or less. Answer(d)(ii) … [3] __________________________________________________________________________________________

13 marks

Mark scheme: 6 (a) 101.5625 or 102 or 101.5 to 101.6 4 M1 for 55, 90, 110, 160 soi nfww M1 for Σfm with frequencies and each m in or on a boundary of a correct interval 2750, 2700, 4400, 6400 M1 dep on 2nd M for ÷ 160 (b) Correct histogram drawn with 3 B1 for each correct block correct widths and heights If zero scored, SC1 for correct heights or 1, 1.5 and 2 (no gaps) frequency densities 40 (c) oe 1 160 1560 40 39 (d) (i) oe 2 M1 for × 25440 160 159 4000 40 50 50 40 (ii) oe 3 M2 for × + × oe 25 440 160 159 160 159 or M1 for one of these products soi

This question in 0580/43 Oct/Nov 2015

4 200 people run 10 km. The table shows some information about the times, t minutes, taken to run the 10 km. Time 30 1 t G 40 40 1 t G 45 45 1 t G 50 50 1 t G 55 55 1 t G 60 60 1 t G 80 (t minutes) Frequency 8 22 95 55 14 6 (a) Howard takes 40 minutes to run the 10 km. Calculate his average speed in kilometres per hour. … km/h [2] (b) Calculate an estimate of the mean time. … min [4] (c) Complete the histogram to show the information in the table. 20 18 16 14 12 Frequency density 10 8 6 4 2 0 t 30 40 50 60 70 80 Time (minutes) [4] (d) (i) Use the frequency table opposite to complete the cumulative frequency table. Time t G 40 t G 45 t G 50 t G 55 t G 60 t G 80 (t minutes) Cumulative 8 30 194 200 frequency [1] (ii) Draw a cumulative frequency diagram to show the information in the table above. 200 180 160 140 120 Cumulative frequency 100 80 60 40 20 0 t 30 40 50 60 70 80 Time (minutes) [3] (iii) Use your diagram to find (a) the median, … min [1] (b) the 90th percentile, … min [1] (c) the number of people who took more than 58 minutes to run the 10 km. … [2]

18 marks

Mark scheme: 4 (a) 15 2 M1 for 10 ÷ 40 [× 60] (b) 49.2 nfww 4 M1 for 35, 42.5, 47.5, 52.5, 57.5, 70 soi M1 for Σ fx 8 × 35 + 22 × 42.5 + 95 × 47.5 + 55 × 52.5 + 14 × 57.5 + 6 × 70 M1 dep for their Σ fx ÷ 200 (c) Fully correct histogram 4 B3 for 4 correct blocks or B2 for 2 or 3 correct blocks or B1 for 1 correct block If zero scored, SC1 for correct frequency densities 0.8, 19, 11, 2.8, 0.3 soi (d) (i) 125, 180 1 (ii) Correct diagram 3 B1FT their (d)(i) for 6 correct heights within correct square(including boundaries) or touching correct line if should be on a grid line and B1 for 6 points at upper ends of intervals on correct vertical line and B1FT (dep on at least B1) for increasing curve or polygon through 6 points If zero scored, SC1FT for 5 correct points plotted (iii) (a) 48 to 49 1 (b) 55 1 (c) 8 to 14 2FT B1FT for 186 to 192 seen

This question in 0580/42 Oct/Nov 2016

7 (a) (i) 100 80 60 Cumulative frequency 40 20 0 2 4 6 8 10 12 14 16 18 20 Price in $(thousands) The cumulative frequency diagram shows information about the prices of 100 cars on Website A. Use the information to complete this table. Lower Upper Inter-quartile Median quartile quartile range $ $7600 $ $ [2] (ii) This table shows information about the prices of cars on Website B. Lower Upper Inter-quartile Median quartile quartile range $7600 $10 800 $13 600 $6000 Here are two statements comparing the distributions of the prices of cars on Website A and Website B. For each statement write True or False. Give a reason for each answer, stating clearly which statistic you use to make your decision. (a) The prices of cars on Website A are lower than the prices of cars on Website B. … because … … [1] (b) A greater percentage of cars have a price more than $13 600 on Website A compared to Website B. … because … … [1] (b) The table shows the prices of cars on Website B. Price ($P) Number of cars 0 1 P G 6 000 9 6 000 1 P G 8 000 29 8 000 1 P G 10 000 20 10 000 1 P G 12 000 14 12 000 1 P G 14 000 21 14 000 1 P G 22 000 27 Calculate an estimate of the mean price of the 120 cars. $ … [4] (c) The price of a car is $8760. Bryan pays a deposit of 25% of this price and then 24 equal monthly payments. After 24 months, he will have paid a total of $9948. Calculate the cost of one monthly payment. $ … [3]

11 marks

Mark scheme: 7 (a) (i) 6000 [7600] 10200 4200 2 B1 for 6000 or 10200 If B0 then B1FT for their (UQ – LQ) (ii)(a) True, median price is lower 1 No inclusion of other statistic (ii)(b) False, A’s UQ < 13 600 oe 1FT FT their UQ in (a)(i) (b) 11 025 4 Listed values are in thousands M1 for 3, 7, 9, 11, 13, 18 soi M1 for Σfm [1323] M1 (dep on second M1) for their Σfm ÷ 120 (c) 323.25 nfww 3 M2 for 9948 – 0.25 × 8760 or M1 for 0.25 × 8760 8 (a) Attempt to use 18 – r in M1 Pythagoras’ 144 = r2 – 324 + 18r + 18r – r2 B2 or B1 for 324 – 18r – 18r + r2 oe 468 = 36r oe A1 Correct simplification with no errors  12   132 + 132 − 24 2  (b) [2 ×] sin–1   oe M1 or cos =   or better or  13  2 × 13 × 13    5  [180 – ] 2 × sin–1    13  134.76… A1 Not 67.4 × 2

This question in 0580/43 Oct/Nov 2016

4 The table shows information about the time, t minutes, taken for each of 150 girls to complete an essay. Time (t minutes) 60 1 t G 65 65 1 t G 70 70 1 t G 80 80 1 t G 100 100 < t G 150 Frequency 10 26 34 58 22 (a) Write down the interval that contains the median time. … 1 t G … [1] (b) Calculate an estimate of the mean time. … min [4] (c) Rafay looks at the frequency table. (i) He says that it is not possible to work out the range of the times. Explain why he is correct. … … [1] (ii) He draws a pie chart to show this information. Calculate the sector angle for the interval 65 1 t G 70 minutes. … [2] (d) A girl is chosen at random. Work out the probability that she took more than 100 minutes to complete the essay. … [1] (e) Two girls are chosen at random. Work out the probability that, to complete the essay, (i) they both took 65 minutes or less, … [2] (ii) one took 65 minutes or less and the other took more than 100 minutes. … [3] (f) The information in the frequency table is shown in a histogram. The height of the block for the 60 1 t G 65 interval is 5 cm. Complete the table. Time (t minutes) 60 1 t G 65 65 1 t G 70 70 1 t G 80 80 1 t G 100 100 1 t G 150 Height of block 5 (cm) [3]

17 marks

Mark scheme: 4(a) 80 < t ⩽100 1 4(b) 86 nfww 4 M1 for midpoints soi M1 for use of Σfx with x in correct interval including both boundaries M1 (dep on 2nd M1) for Σfx ÷ 150 4(c)(i) Reference to not knowing the 1 individual values so we do not know the highest or the lowest values 4(c)(ii) 62.4 2 M1 for 26 ÷ 150 or 360 ÷ 150 soi 4(d) 22 1 oe 150 4(e)(i) 90 2 10 9 oe M1 for × 22350 150 149 100 After zero scored, SC1 for answer oe 22500 4(e)(ii) 440 3 10 22 22 10 oe M2 for × + × oe 22350 150 149 150 149 or 10 22 22 10 M1 for × or × oe 150 149 150 149 440 After zero scored, SC1 for answer oe 22500 4(f) 13, 8.5, 7.25, 1.1 3 B2 for 3 correct or B1 for 1 correct or for 3 correct FD.s 5.2, 3.4, 2.9, 0.44 oe

This question in 0580/43 Oct/Nov 2017

9 (a) The table shows the amount of time, T minutes, 120 people each spend in a supermarket one Saturday. Time (T minutes) Number of people 10 1 T G 30 16 30 1 T G 40 18 40 1 T G 45 22 45 1 T G 50 40 50 1 T G 60 21 60 1 T G 70 3 (i) Use the mid-points of the intervals to calculate an estimate of the mean. … min [4] (ii) Complete this histogram to show the information in the table. 8 6 Frequency 4 density 2 0 T 0 10 20 30 40 50 60 70 Time (minutes) [4] (b) This histogram shows the amount of time, T minutes, 120 people each spend in the supermarket one Wednesday. 8 6 Frequency 4density 2 0 T 0 10 20 30 40 50 60 70 Time (minutes) Make a comment comparing the distributions of the times for the two days. … … [1]

9 marks

Mark scheme: 9(a)(i) 42.8 or 42.79 … nfww 4 M1 for mid-values soi M1 for Σfm where m is any value in interval including boundaries M1 (dep on second M1) for their Σfm ÷ 120 9(a)(ii) Blocks of height 1.8 4.4 8 2.1 with 4 B1 for each correct block correct widths If B0, SC1 for correct frequency densities seen 9(b) Valid general comment about 1 e.g. [On average], shoppers spend less time distributions shopping on Wednesday oe

This question in 0580/42 Oct/Nov 2018

9 100 students were each asked how much money, $m, they spent in one week. The frequency table shows the results. Amount ($m) 0 1 m G 5 5 1 m G 10 10 1 m G 20 20 1 m G 30 30 1 m G 50 Frequency 16 38 30 9 7 (a) Calculate an estimate of the mean. $ … [4] (b) Complete the cumulative frequency table below. Amount ($m) m G 5 m G 10 m G 20 m G 30 m G 50 Cumulative 16 100 frequency [2] (c) On the grid, draw the cumulative frequency diagram. 100 80 60 Cumulative frequency 40 20 0 0 10 20 30 40 50 m Amount ($) [3] (d) Use your cumulative frequency diagram to find an estimate for (i) the median, $ … [1] (ii) the interquartile range, $ … [2] (iii) the number of students who spent more than $25. … [2]

14 marks

Mark scheme: 9(a) 12.8[0] 4 M1 for midpoints soi M1 for use of ∑fm with m in correct interval including both boundaries M1 (dep on 2nd M1) for ∑fm ÷ 100 9(b) 54 84 93 2 B1 for 2 correct or 1 error and 2 correct or FT 9(c) correct diagram with all points 3 B1FT their (b) for plots at 5 correct heights correctly plotted B1 for 5 points at upper ends of intervals on correct vertical line B1FT (dep on at least B1) for increasing curve or polygon through 5 points After 0 scored, SC1FT for 4 correct points plotted 9(d)(i) 9 to 9.8 final answer 1 9(d)(ii) 8.5 to 11.5 2 B1 for [UQ =] 15.5 to 17.5 or [LQ =] 6 to 7 seen 9(d)(iii) 10, 11 or 12 2 B1 for 88 to 90 seen or for answer between 10 and 12

This question in 0580/42 May/June 2019

3 (a) The average speeds, in km/h, of cars travelling along a road are recorded. The box-and-whisker plot shows this information. 40 45 50 55 60 65 70 75 80 85 Speed (km/h) Find (i) the lowest speed recorded, … km/h [1] (ii) the median, … km/h [1] (iii) the interquartile range. … km/h [1] (b) Another car takes 18 seconds to travel 400 m along this road. Calculate the average speed of this car in km/h. … km/h [3]

6 marks

Mark scheme: 3(a)(i) 43 1 3(a)(ii) 65 1 3(a)(iii) 13 1 3(b) 80 3 400 60 × 60 M2 for × oe 18 1000 400 Or M1 for 18 60 × 60 or for their speed in m/s × 1000 400 18 or for and soi 1000 60 × 60

This question in 0580/43 Oct/Nov 2020

7 (a) The box-and-whisker plot shows information about the marks scored by some students in a test. 0 10 20 30 40 50 60 70 80 90 100 Mark (i) Write down the median mark. … [1] (ii) Work out the range. … [1] (iii) Jais scored a mark in the test that was higher than the marks scored by 75% of the students. Write down a possible mark for Jais. … [1] (iv) This box-and-whisker plot shows information about the marks scored by the same students in a second test. 0 10 20 30 40 50 60 70 80 90 100 Mark Make one comparison between the distributions of marks in the two tests. … … [1] (b) The table shows information about the height, h cm, of each of 50 plants. Height (h cm) 0 1 h G 20 20 1 h G 30 30 1 h G 34 34 1 h G 40 40 1 h G 60 Frequency 4 9 20 15 2 Calculate an estimate of the mean. … cm [4] (c) Some apples are weighed and the mass, m grams, of each apple is recorded. The table shows the results. Mass (m grams) 100 1 m G 110 110 1 m G 115 115 1 m G 125 125 1 m G 140 Frequency 50 x 44 51 The histogram shows some of the information from the table. 8 6 Frequency density 4 2 0 m 100 110 120 130 140 Mass (grams) (i) Work out the value of x. x = … [1] (ii) Complete the histogram. [2]

11 marks

Mark scheme: 7(a)(i) 70 1 7(a)(ii) 78 1 7(a)(iii) Value in range 86 <V ≤ 90 1 7(a)(iv) One general comment interpreting 1 the median comparison nfww e.g. Students did better on second test oe OR One general comment interpreting IQR/range comparison nfww e.g. Students marks were more consistent on the 2nd test oe 7(b) 31.2 4 M1 for mid-values soi M1 for Σfm where m is any value in interval including boundaries M1 (dep on second M1) for their Σfm ÷ 50 7(c)(i) 38 1 7(c)(ii) Blocks of heights 4.4 and 3.4 with 2 B1 for each correct block correct widths If B0 scored, SC1 for both correct frequency densities soi

This question in 0580/42 Feb/March 2021

1 (a) The list shows 15 midday temperatures, in degrees Celsius, in Suntown. 17 21 21 18 23 22 25 19 21 17 19 18 21 24 23 (i) Complete the stem-and-leaf diagram to show this information. 1 7 2 Key: 1|7 represents 17 °C [2] (ii) Find the median. … °C [1] (iii) Find the upper quartile. … °C [1] (iv) Rahul draws a pie chart to show this information. Calculate the sector angle for the number of days the temperature is 18 °C. … [2] (b) 0 50 100 150 200 Mass (grams) The box-and-whisker plot shows information about the masses, in grams, of some apples. (i) Find the median. … g [1] (ii) Find the range. … g [1] (iii) Find the interquartile range. … g [1] (c) (i) The time, t minutes, spent on homework in one week by each of 200 students is recorded. The table shows the results. Time (t minutes) 40 1 t G 60 60 1 t G 80 80 1 t G 90 90 1 t G 100 100 1 t G 150 Frequency 6 10 70 84 30 Calculate an estimate of the mean. … min [4] (ii) A new table with different class intervals is completed. Time (t minutes) 40 1 t G 90 90 1 t G 150 Frequency 86 114 On a histogram the height of the bar for the 40 1 t G 90 interval is 17.2 cm. Calculate the height of the bar for the 90 1 t G 150 interval. … cm [2]

15 marks

Mark scheme: 1(a)(i) 1 7 7 8 8 9 9 2 1 1 1 1 2 3 3 4 5 2 B1 for one row correctly ordered or for fully correct unordered stem-and-leaf diagram or for a correct diagram with one error or omission 1(a)(ii) 21 1 1(a)(iii) 23 1 1(a)(iv) 48 2 M1 for   2 360 15  or   360 2 15  1(b)(i) 120 1 1(b)(ii) 130 1 1(b)(iii) 60 1 1(c)(i) 93.4 4 M1 for mid-values soi M1 for fx M1 dep on second M for 200   fx 1(c)(ii) 19 2 M1 for 86 50 or 114 60

This question in 0580/41 May/June 2022

2 (a) Daisy records her 50 homework marks. The table shows the results. Homework mark 15 16 17 18 19 20 Frequency 1 3 19 11 10 6 (i) Write down the range. … [1] (ii) Write down the mode. … [1] (iii) Find the median. … [1] (iv) Calculate the mean. … [3] (b) 21 33 20 25 21 34 22 21 20 30 18 The list shows Ed’s scores in 11 tests. (i) Complete the stem-and-leaf diagram to show this information. 1 2 3 Key: 2|5 represents 25 [2] (ii) Find the median. … [1] (iii) Find the interquartile range. … [2]

11 marks

Mark scheme: 2(a)(i) 5 1 2(a)(ii) 17 1 2(a)(iii) 18 1 2(a)(iv) 17.88 3 M2 for (1×15 + 3×16 + 19×17 + 11×18 + 10×19 + 6×20) ÷ 50 oe or M1 for 1×15 + 3×16 + 19×17 + 11×18 + 10×19 + 6×20 oe 2(b)(i) 2 1 8 2 0 0 1 1 1 2 5 B1 for two rows correct or for fully correct unordered stem-and- 3 0 3 4 leaf diagram 2(b)(ii) 21 1 2(b)(iii) 10 nfww 2 B1 for [upper qtile] = 30 or [lower qtile] = 20 soi

This question in 0580/42 Oct/Nov 2023

2 The table shows the number of each type of bird seen in a garden on Monday. Type of bird Frequency Pie chart sector angle Goldfinch 8 96° Jay 6 Starling 11 Robin 5 (a) Find the percentage of the birds that are Starlings. … % [2] (b) (i) In the table, complete the column for the pie chart sector angle. [2] (ii) Complete the pie chart to show the information in the table. Goldfinch [2] (c) On Tuesday, the number of Goldfinches seen in the garden increased by 262.5%. Calculate the number of Goldfinches seen on Tuesday. … [2] (d) One of the most common birds in the world is the Red-Billed Quelea which lives in Sub-Saharan Africa. There are approximately 1500 million of these birds in this area. (i) Write 1500 million in standard form. … [1] (ii) The land area of Sub-Saharan Africa is approximately 21.2 million square kilometres. Work out the average number of these birds per square kilometre. … birds/km2 [2]

11 marks

Mark scheme: 2(a) 2 2 11 36.7 or 36.66 to 36.67 or 36 M1 for [ 100] oe 3 8 + 6 + 11 + 5 2(b)(i) 72, 132 and 60 2 M1 for 360 ÷ (8 + 6 + 11 + 5) oe or 96 ÷ 8 2(b)(ii) Correct pie chart drawn 2 For 2 marks, strict FT their angles for correct pie chart only if angles add up to 360. B1FT for one correct sector 2(c) 29 2  262.5  M1 for 8   1 +  oe  100  or B1 for 21 2(d)(i) 1.5  109 1 2(d)(ii) 70.8 or 70.75… 2 M1 for 1500 [million] ÷ 21.2 [million]

This question in 0580/43 Oct/Nov 2023

8 Guillaume measures the speed of each of 100 cars. The results are shown in the table. Speed (v km/h) 30 1 v G 4 0 40 1 v G 45 45 1 v G 50 50 1 v G 70 Frequency 15 20 35 30 (a) Guillaume draws a pie chart for this data. Calculate the angle for the interval 45 1 v G 50 . … [2] (b) Calculate an estimate of the mean speed. … km/h [4] (c) Complete the histogram to show the data in the table. 8 7 6 5 Frequency density 4 3 2 1 030 40 50 60 70 v Speed (km / h) [3]

9 marks

Mark scheme: 8(a) 126 2 35 M1 for  360 100 8(b) 48.375 4 M1 for mid-values 35, 42.5, 47.5, 60 soi M1 for 15 × 35 + 20 × 42.5 + 35 × 47.5 + 30 × 60 fx M1 dep dep on second M1 100 8(c) Correct histogram with correct widths 3 B1 for each column and heights 1.5, … 7, 1.5 If 0 scored, SC1 for freq. densities 1.5, ...7, 1.5 seen 15 35 30 or , , 10 5 20

This question in 0580/43 Oct/Nov 2024

15 Virat records the height of each of 80 sunflowers. The results are shown in the table. Height (h m) 1.2 1 h # 1 .5 1.5 1 h # 1 .6 1 .6 1 h # 1 .7 1.7 1 h # 1.9 Frequency 12 20 34 14 (a) Calculate an estimate of the mean height. … m [4] (b) Draw a histogram to show the information in the table. 400 300 Frequency density 200 100 0 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 h Height (m) [3]

7 marks

Mark scheme: 15(a) 1.606 25 4 M1 for mid-points soi (1.35, 1.55, 1.65, 1.8 ) M1 for use of fm with m in correct interval including both boundaries M1 dep (dep on 2nd M1) for fm  80 15(b) Correct histogram 3 B2 for 3 correct blocks or B1 for 2 correct blocks If 0 scored, SC1 for 3 correct freq densities soi (40, 200, 340, 70)

This question in 0580/42 Feb/March 2025

19 The table shows some information about the heights of 200 plants. Cumulative Height (h cm) frequency h G 30 6 h G 40 42 h G 50 76 h G 60 170 h G 70 190 h G 80 200 (a) Draw a cumulative frequency diagram to show this information. 200 160 120 Cumulative frequency 80 40 0 20 30 40 50 60 70 80 h Height (cm) [3] (b) Find an estimate of the 75th percentile. … cm [1]

4 marks

Mark scheme: 19(a) A correct cumulative frequency 3 B1 for correct horizontal placement for 6 plots curve through [(20, 0)], (30, 6) (40, B1 for correct vertical placement for 6 plots 42), (50, 76), (60, 170), (70, 190) B1FT dep on at least B1 for reasonable and (80, 200) increasing curve or polygon through their 6 points If 0 scored, SC1FT for 5 out of 6 points correctly plotted 19(b) Reading at 150 FT their increasing 1 curve

This question in 0580/43 May/June 2025

24 The histogram shows information about the distance some people travel to work. 4 3 Frequency density 2 1 0 0 10 20 30 40 50 d Distance (km) (a) Use the histogram to complete the table. Distance (d km) Number of people 0 1 d G 5 5 1 d G 20 20 1 d G 50 [3] (b) The people who travel a distance greater than 35 km to work have a car allowance. Calculate an estimate of the number of these people who have a car allowance. … [1]

4 marks

Mark scheme: 24(a) 14 3 B1 for each correct answer 37 24 24(b) 6 1

This question in 0580/43 May/June 2025