Interpreting statistical data — Paper 4 · IGCSE Mathematics 0580

7 The speeds (v kilometres/hour) of 150 cars passing a 50 km/h speed limit sign are recorded. A cumulative frequency curve to show the results is drawn below. 150 140 130 120 110 100 90 80Cumulative frequency 70 60 50 40 30 20 10 0 30 35 40 45 50 55 60 Speed (v kilometres / hour) (a) Use the graph to find (i) the median speed, [1] (ii) the inter-quartile range of the speeds, [2] (iii) the number of cars travelling with speeds of more than 50 km/h. [2] (b) A frequency table showing the speeds of the cars is Speed (v km/h) 30<v 35 35<v 40 40<v 45 45<v 50 50<v 55 55<v 60 Frequency 10 17 33 42 n 16 (i) Find the value of n. [1] (ii) Calculate an estimate of the mean speed. [4] (c) Answer this part of this question on a sheet of graph paper. Another frequency table for the same speeds is Speed (v km/h) 30<v 40 40<v 55 55<v 60 Frequency 27 107 16 Draw an accurate histogram to show this information. Use 2 cm to represent 5 units on the speed axis and 1 cm to represent 1 unit on the frequency density axis (so that 1 cm2 represents 2.5 cars). [5]

15 marks

Mark scheme: 7 (a)(i) Median 46.5 B1 (ii) IQR 9.5 www B2 SC1 for 42 or 51.5 seen (iii) 48 B2 SC1 for 102 seen (b)(i) n = 32 B1 (ii) Midpts 32.5, 37.5, 42.5, 47.5, 52.5, 57.5 M1 At least 5 correct s.o.i. 10x32.5 + 17x37.5 + 33x42.5 + 42x47.5 M1* Dep on first M1 or midpoints ±0.5 + their 32x52.5 + 16x57.5 [6960] Allow 1 more slip ∑fx / 150 M1 Dep on 2nd M1* 46.4 A1 (c) Horizontal Scale correct S1 Implied by correct use. Ignore vertical scale 3 correct widths on their scale (f.t ) W1√ no gaps For each block of correct width 2.7 cm H1 For scale error double or half, award H1, H1, H1 7.1(3) or 7.2 cm H1 for correct f.t heights 3.2 cm H1 After H0, SC1 for 3 correct frequency densities written or for heights 2.7cm, 7.1cm and 3.2cm drawn on doubled/ halved horizontal scale. 15 2 2 √

This question in 0580/41 May/June 2005

9 Diagram 1 Diagram 2 Diagram 3 The first three diagrams in a sequence are shown above. The diagrams are made up of dots and lines. Each line is one centimetre long. (a) Make a sketch of the next diagram in the sequence. [1] (b) The table below shows some information about the diagrams. Diagram 1 2 3 4 --------- n Area 1 4 9 16 --------- x Number of dots 4 9 16 p --------- y Number of one centimetre lines 4 12 24 q --------- z (i) Write down the values of p and q. [2] (ii) Write down each of x, y and z in terms of n. [4] (c) The total number of one centimetre lines in the first n diagrams is given by the expression 2 n3 + fn2 + gn. 3 (i) Use n = 1 in this expression to show that f + g =10 . [1] 3 32 (ii) Use n = 2 in this expression to show that 4f + 2g = . [2] 3 (iii) Find the values of f and g. [3] (iv) Find the total number of one centimetre lines in the first 10 diagrams. [1]

14 marks

Mark scheme: 9 (a) Sketch of 4 by 4 diagram B1 (b) (i) 25, 40 B1,B1 (ii) n2 B1 (n + 1)2 oe B1 (n + 1)2 + n2 – 1 or 2n2 + 2n) or B2 Any one of these oe isw and if B0 2n(n + 1) oe allow SC1 for their (n + 1)2 + their (n2) – 1 or an expression containing 2n2, as the highest order term, soi (c) (i) 2 + f + g = 4 B1 3 (ii) 2 × 2 3 + f × 2 2 + g × 2 oe M1 ie for substituting 2 3 32 E1 No errors Allow 10, 2 3 10., 10.7, … 4 f + 2 g = 3 (iii) 2 f + 2 g = 20 M1 for correctly setting up for elimination 3 32 of one variable 4 f + 2 g = 3 ( f = ) (,2 g = ) 4 oe cao A1A1 www 3 accept 6 3 for 2 3 (iv) 880 cao B1 [14]

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

3 80 boys each had their mass, m kilograms, recorded. For The cumulative frequency diagram shows the results. Examiner's Use 80 60 Cumulative 40 frequency 20 0 m 30 40 50 60 70 80 90 Mass (kg) (a) Find (i) the median, Answer(a)(i) kg [1] (ii) the lower quartile, Answer(a)(ii) kg [1] (iii) the interquartile range. Answer(a)(iii) kg [1] (b) How many boys had a mass greater than 60kg? Answer(b) [2] (c) (i) Use the cumulative frequency graph to complete this frequency table. For Examiner's Use Mass, m Frequency 30 I m Y 40 8 40 I m Y 50 50 I m Y 60 14 60 I m Y 70 22 70 I m Y 80 80 I m Y 90 10 [2] (ii) Calculate an estimate of the mean mass. Answer(c)(ii) kg [4]

11 marks

Mark scheme: 3 (a) (i) 63 to 63.5 1 (ii) 50 to 50.5 1 (iii) 21.5 to 22.5 1 (b) 46 2 B1 for 34 seen (could be on graph) (c) (i) 12, 14 1, 1 (ii) {35 × 8 + 45 × their 12 + 55 × 14 + 65 × 22 + 75 × their 14 + 85 × 10} M3 M1 for mid-values soi (allow 1 error/omit) with x in correct ÷ their 80 (or 80) and M1 for use of ∑fx boundary including both ends (at least 4 products) (4920 seen implies M2) and M1 depend on 2nd M for dividing by their 80 (or 80) (not 54 or less) 61.5 cao A1 www4

This question in 0580/42 Oct/Nov 2010

6 For Examiner's 200 Use 180 160 140 120 Cumulative 100 frequency 80 60 40 20 m 0 1 2 3 4 5 6 7 8 9 10 Mass (kilograms) The masses of 200 parcels are recorded. The results are shown in the cumulative frequency diagram above. (a) Find (i) the median, Answer(a)(i) kg [1] (ii) the lower quartile, Answer(a)(ii) kg [1] (iii) the inter-quartile range, Answer(a)(iii) kg [1] (iv) the number of parcels with a mass greater than 3.5 kg. Answer(a)(iv) [2] (b) (i) Use the information from the cumulative frequency diagram to complete the grouped For frequency table. Examiner's Use Mass (m) kg 0 I m Y 4 4 I m Y 6 6 I m Y 7 7 I m Y 10 Frequency 36 50 [2] (ii) Use the grouped frequency table to calculate an estimate of the mean. Answer(b)(ii) kg [4] (iii) Complete the frequency density table and use it to complete the histogram. Mass (m) kg 0 I m Y 4 4 I m Y 6 6 I m Y 7 7 I m Y 10 Frequency 9 16.7 density 40 35 30 25 Frequency 20 density 15 10 5 m 0 1 2 3 4 5 6 7 8 9 10 Mass (kilograms) [4]

15 marks

Mark scheme: 6 (a) (i) 5.8 1 (ii) 4.6 to 4.65 1 (iii) 2.35 to 2.5 1 (iv) 172 or 171 2 SC1 for 28 or 29 (b) (i) 72 to 76, 38 to 42 2 Must be integers. B1 either. (ii) Their correct Σfx ÷ 200 4 M1 for 3 or 4 correct mid-values seen 2, 5, 6.5, 8.5 M1 for Σfx, ft their frequencies and x anywhere in interval, including boundaries 36 × 2 + (72 to 76) × 5 + (38 to 42) × 6.5 + 50 × 8.5 M1 for ÷ 200 or their 200 (dependent on second M1) (74, 40 give 1127 then 5.635 (or 5.64 or 5.63)) Other pairs of frequencies from (b)(i) must have a sum of 114 to gain the A mark. (iii) p ÷ 2, q, where p, q are from (b)(i) 2ft B1 either ft (ft their table) Histogram with two new columns of correct width B1 Two correct heights 2ft B1 ft (ft their freq. densities) IGCSE – May/June 2011 0580 43 7 (a) Correct tree diagram. 5 B1 for labels flower and not flower First pair B1 for 107 and 103 B1 for next three branches after flowers B1 for clear labels for colours B1 for 23 , 14 and 121 in correct places If three branches at ends of both branches of first pair, lose final B, unless probabilities of 0 indicated. (b) 33 3 M2 for 1 – 107 × 14 (M1 for 107 × 14 or 40 o.e. (0.825) cao 7 1 10 × (1− 4 ) ) oe or M2 for 103 + 107 × 23 + 107 × their 121 or 103 + 107 × 34 oe (c) 7 cao 2 M1 for 120 × 107 × their 121 8 (a) Arc centre D, radius 6 cm 1 (b) (i) Perp bisector of AB, with two pairs 2 At least 3 cm from AB. SC1 accurate without of arcs arcs or accurate arcs (but no choice) (ii) Bisector of angle B, with arcs 2 At least 5 cm from B. SC1 accurate without arcs or accurate arcs (but no choice) (c) (i) Q at intersection of loci 1 Dependent on at least both SC1’s (ii) 2.7 cm to 2.9 cm cao 1 Dependent on (c)(i) (d) Region inside arc, to left of perp bisector 1 Dependent on at least both SC1’s in (b) and below angle bisector

This question in 0580/43 May/June 2011

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

4 The cumulative frequency diagram shows information about the time taken, t minutes, by 60 students to complete a test. 60 50 40 Cumulative frequency 30 20 10 t 0 10 20 30 40 50 60 70 80 90 100 Time taken (minutes) (a) Find (i) the median, … min [1] (ii) the inter-quartile range, … min [2] (iii) the 40th percentile, … min [2] (iv) the number of students who took more than 80 minutes to complete the test. … [2] (b) Use the cumulative frequency diagram to complete the frequency table below. Time taken 0  t  40 40  t  60 60  t  70 70  t  80 80  t  90 90  t  100(t minutes) Frequency 8 4 [3] (c) On the grid below, complete the histogram to show the information in the table in part (b). 3 2 Frequency density 1 t 0 10 20 30 40 50 60 70 80 90 100 Time taken (minutes) [4]

14 marks

Mark scheme: 4 (a) (i) 64 1 (ii) 16 to 16.5 2 M1 for UQ = 71 to 71.5 or LQ =55 (iii) 62 2 B1 for 24 indicated (iv) 6 2 B1 for 54 seen (b) [8] 12 23 11 [ 4] 2 3 B2 for 1 incorrect reading FT others B1 for 2 correct (c) Blocks of height 0.6 2.3 1.1 0.4 4FT FT their (b) for heights with correct widths B1FT for each correct block If B0, SC1 for blocks of widths 20, 10, 10, 10 or for their correct frequency densities 6000

This question in 0580/42 Feb/March 2016

3 (a) 200 students estimate the volume, V m3, of a classroom. The cumulative frequency diagram shows their results. 200 180 160 140 120 Cumulative frequency 100 80 60 40 20 V 0 50 100 150 200 250 300 350 400 450 500 Volume (m3) Find (i) the median, … m3 [1] (ii) the lower quartile, … m3 [1] (iii) the inter-quartile range, … m3 [1] (iv) the number of students who estimate that the volume is greater than 300 m3. … [2] (b) The 200 students also estimate the total area, A m2, of the windows in the classroom. The results are shown in the table. Area (A m2) 20 1 A G 60 60 1 A G 100 100 1 A G 150 150 1 A G 250 Frequency 32 64 80 24 (i) Calculate an estimate of the mean. Show all your working. … m2 [4] (ii) Complete the histogram to show the information in the table. 2 Frequency density 1 A 0 50 100 150 200 250 Area (m2) [4] (iii) Two of the 200 students are chosen at random. Find the probability that they both estimate that the area is greater than 100 m2. … [2]

15 marks

Mark scheme: 3 (a) (i) 400 1 (ii) 350 1 (iii) 70 1 (iv) 170 2 B1 for 30 seen (b) (i) Mid-values 40, 80, 125, 200 soi M1 Σ fx with correct frequencies and x’s in M1 correct intervals or on boundaries of correct intervals M1(dep) Dependent on second M1 ÷ 200 A1 SC2 for correct answer without working 106 nfww (ii) Correct histogram 4 B1 for correct widths and B1 for each rectangle of correct height at 0.8, 1.6, 1.6 (up to B3) After 0 scored, SC1 for 3 correct frequency densities seen 10712 104 103 (iii) oe isw 2 M1 for × oe 39800 200 199

This question in 0580/41 May/June 2016

2 (a) 200 students record the time, t minutes, for their journey from home to school. The cumulative frequency diagram shows the results. Cumulative frequency 200 180 160 140 120 100 80 60 40 20 t 0 5 10 15 20 25 30 35 40 Time (minutes) Find (i) the median, … min [1] (ii) the lower quartile, … min [1] (iii) the inter-quartile range, … min [1] (iv) the 15th percentile, … min [1] (v) the number of students whose journey time was more than 30 minutes. … [2] (b) The 200 students record the time, t minutes, for their journey from school to home. The frequency table shows the results. Time (t minutes) 0 1 t G 10 10 1 t G 15 15 1 t G 20 20 1 t G 30 30 1 t G 60 Frequency 48 48 60 26 18 (i) Calculate an estimate of the mean. … min [4] (ii) On the grid, complete the histogram to show the information in the frequency table. 12 11 10 9 8 Frequency density 7 6 5 4 3 2 1 t 0 10 20 30 40 50 60 Time (minutes) [4]

14 marks

Mark scheme: 2 (a) (i) 15 to 15.2 1 (ii) 10.8 to 11 1 (iii) 9 to 9.2 1FT FT 20 – their (a)(ii) (iv) 10 1 (v) 24 2 B1 for 176 written (b) (i) 16.75 nfww 4 isw attempted time conversion after correct answer M1 for 5, 12.5, 17.5, 25, 45 soi M1 for Σ fx M1 dep for their Σ fx ÷ 200 (ii) Fully correct histogram 4 B1 for each correct block If zero scored, SC1 for frequency densities of 9.6, 12, 2.6 and 0.6 seen

This question in 0580/41 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

6 (a) There are 100 students in group A. The teacher records the distance, d metres, each student runs in one minute. The results are shown in the cumulative frequency diagram. 100 90 80 70 60 Cumulative 50 frequency 40 30 20 10 0 d 100 200 300 400 Distance (metres) Find (i) the median, … m [1] (ii) the upper quartile, … m [1] (iii) the inter-quartile range, … m [1] (iv) the number of students who run more than 350 m. … [2] (b) There are 100 students in group B. The teacher records the distance, d metres, each of these students runs in one minute. The results are shown in the frequency table. Distance 100 1 d G 200 200 1 d G 250 250 1 d G 280 280 1 d G 32 0 320 1 d G 400(d metres) Number of 20 22 30 16 12students (i) Calculate an estimate of the mean distance for group B. … m [4] (ii) Complete the histogram to show the information in the frequency table. 1 0.8 0.6 Frequency density 0.4 0.2 0 d 100 200 300 400 Distance (metres) [4] (c) For the 100 students in group B, the median is 258 m. Complete the statement. On average, the students in group A run … than the students in group B. [1]

14 marks

Mark scheme: 6(a)(i) 280 1 6(a)(ii) 320 1 6(a)(iii) 90 1 6(a)(iv) 10 2 M1 for 90 written 6(b)(i) 250.2 nfww cao 4 M1 for at least 4 correct mid-values M1 for Σfx M1 dep on second M1 for Σfx ÷ 100 6(b)(ii) Correct completion of 4 B1 for each correct block histogram If zero scored, then SC1 for correct frequency densities seen 6(c) [22 m] further oe 1

This question in 0580/42 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

11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The sequence of diagrams above is made up of small lines and dots. (a) Complete the table. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of 4 10 18 28 small lines Number of 4 8 13 19 dots [4] (b) For Diagram n find an expression, in terms of n, for the number of small lines. … [2] (c) Diagram r has 10 300 small lines. Find the value of r. r = … [2] (d) The number of dots in Diagram n is an 2 + bn + 1. Find the value of a and the value of b. a = … b = … [2]

10 marks

Mark scheme: 11(a) 40 54 4 B1 for each 26 34 11(b) 2 2 B1 for a quadratic expression n + 3n or n (n + 3) oe or for 2nd common difference 2 (at least 2 shown) or for 2 correct equations seen or for subtracting n2 11(c) 100 2 M1 for their (b) = 10300 seen 11(d) 1 2 B1 for each [a = ] oe or M1 for one correct equation 2 or for 2nd difference = 1 soi (at least 2 shown) and 5 [b =] oe 2

This question in 0580/42 May/June 2019

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