Edexcel A-level Maths · Topic S2
Practice questions covering data presentation and interpretation from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Data presentation and interpretation, each with its mark scheme and worked method. Reveal the answer when you have had a go.
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(4)
(Total for Question 1 is 4 marks)
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(2)
(Total for Question 2 is 2 marks)
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(3)
(Total for Question 3 is 3 marks)
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(2)
(Total for Question 4 is 2 marks)
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(3)
(Total for Question 5 is 3 marks)
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(4)
(Total for Question 6 is 4 marks)
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Printable question pack — S2 Data presentation and interpretationAll 40 questions across 4 specification points — 34 more than the 6 above — with the full mark scheme. Guest accessible · browser print.10 questions shown as printed, 8 reference-only · newest series first
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Edexcel 9MA0 · June 2024 · Paper 31 · Q2 · 6 marks
6 marks
Examiner report
Whilst most students showed they had a good grasp of the concepts of correlation and regression some failed to give sufficient detail in their answers. In part (a) the requirement to “interpret” meant, as usual, that we needed a fully contextual response. Some simply stated “negative correlation” or gave the value of the gradient as – 1.28 whereas a full interpretation of the rate expected to see mention of the height decreasing by 1.28 m(etres) for each second of flight. In part (b) most students interpreted this situation as a one-tailed test and stated the hypotheses correctly in terms of but a few equated this equal to – 0.510 rather than 0 for the null hypothesis. Almost all the students gave a critical value that matched their alternative hypothesis, and an encouraging number gave a correct contextual conclusion to the test mentioning both height and time. Some however lost the final accuracy mark for having contradictory or incorrect statements such as “reject H0” followed by a correct contextual conclusion such as “there is insufficient evidence of a negative correlation between h and t”. Although we did not require a comparison of the test statistic with the critical value if one was seen it had to be correct and some students lost the final mark for statements such as – 0.510 < – 0.5494 or for comparing 0.510 with the significance level of 0.05 In part (c) most students realised that the given equation was not an appropriate model but did not give a strong enough explanation as to why this was so. Those who noted that the scatter diagram showed a curved set of data scored the mark whilst those who suggested that the points seemed to form a (negative) quadratic probably also gave themselves lead in to part (d). Of those who did not score the mark in part (c) some simply discussed correlation rather than stating that the line had negative gradient throughout whereas the scatter diagram had a positive gradient initially and then a negative gradient. Others just compared one or two points rather than concentrating on the overall shape of the scatter diagram and some simply stated “non-linear” without developing their answer and drawing out a difference between the scatter diagram and Amar’s linear model. Part (d) proved challenging for most students and many simply moved on to the next question. There were many convoluted attempts involving calculus or multiplying out the bracket or substituting a value such as (1, 28). Students should be encouraged to consider the number of marks available as a guide to the amount of work required to answer the question. Those who had identified the scatter diagram as representing a parabola and saw the connection with the quadratic nature of Jane’s model could simply estimate the equation of the axis of symmetry and usually chose a suitable value of k.
Amar measures a bird's height h metres at time t seconds for 10 values of t. The regression line is h = 38.6 - 1.28t and the product moment correlation coefficient is -0.510. The scatter diagram shows points at approximately (1, 28), (1.5, 34), (2, 36), (2.5, 38), (3.5, 40), (4.5, 39), (5, 37), (6.5, 33), (7, 27) and (8.5, 20), so the height peaks at about t = 3.5 seconds before falling.
Edexcel 9MA0 · June 2024 · Paper 31 · Q3 · 6 marks
6 marks
Examiner report
It is encouraging to see most students now attempting the question on the large data set and a growing familiarity with it. In part (a) many wrote down rainfall or windspeed but other quantitative variables such as hours of sunshine or maximum wind gust were sometimes seen but these are not variables available for overseas locations. Some students gave answers like “trace” or “outlier” which have appeared in previous papers but had no relevance to this question. Part (b) was a straightforward calculation and was accessible to all the students. Almost everyone scored the mark for the mean and most had a correct expression for the standard deviation, but some used their rounded value of 15.2 for the mean and this meant that their final answer for the standard deviation was not accurate to 3 significant figures. A few students omitted the square root sign or forgot to square the mean. The final part of the question required students to know that Perth on the large data set was in the southern hemisphere and therefore the warmer months would be during our autumn and winter. They were told that the mean air temperature for the month in question was 19.4 and needed to observe that this was larger than the overall mean of 15.2 for the 6 months of the large data set. Using their knowledge that the large data set only covers the months from May to October they could then infer that the month in question was October. Many were able to identify that the month to be identified was a warmer month (and scored the first mark) but then suggested an answer such as November or December which of course are not available in the large data set. It was sometimes difficult to know whether a candidate was using “summer” to simply refer to the months from June to August or referring to the warmest months of the year in Perth, often an example month was given which helped clarify what the student meant.
Ming is studying the large data set for Perth in 2015. He intended to analyse Daily Mean Air Temperature but selected the wrong spreadsheet variable, obtaining mean 5.3 and standard deviation 12.4.
Edexcel 9MA0 · June 2023 · Paper 31 · Q3 · 7 marks
7 marks
Examiner report
Once again the large data set aspects of this question caused problems for some students but most scored some marks here. In part (a) a few responses simply quoted standard textbook answers about cleaning data by removing outliers and others were distracted by the class intervals for the data. Many though did realise that the “tr” values needed attention but a number didn’t seem to realise that trace entries had a value between 0 and 0.05 and therefore removing them would have a big effect on the calculations Ben wanted to carry out. Those that did appreciate the value of the “tr” usually suggested a suitable numerical value (usually 0) to replace them with and scored both marks. Part (b) was, unsurprisingly, answered very well with many scoring full marks. A few lost marks for truncating rather than rounding their answers for example 2.11 in (i) whilst others were convinced that the “tr” values should have been removed and had a denominator of 155 despite the question telling them that all 184 values were used. It is still surprising to see a small number of students using a denominator of 9 or 10 (the number of classes in the table) and a slightly larger proportion who couldn’t use the standard deviation formula, given in the formula booklet, correctly. It was encouraging to see a good number of students demonstrating some familiarity with the large data set and scoring some marks in part (c). In (i) we were simply looking for an awareness that because the large data set only had data for May to October this would not be representative of the whole population. Some demonstrated a lack of understanding of the principle of using a sample to make inferences by stating that 184 days was not a sufficiently large sample to represent the weather for the whole year. In this case the problem was not the sample size but the fact that the sample was not representative. In (ii) even some of those who had failed to mention May to October in (i) were able to score here by stating that because the missing months were in winter and they expected more rainfall then, the actual mean would be higher. The great storm of October 1987 featured in many incorrect answers to part (c).
Ben studies Daily Total Rainfall x mm in Leeming for 1987 using all the relevant large-data-set entries. His grouped table is: x = 0, 0.1-0.5, 0.6-1.0, 1.1-1.9, 2.0-4.0, 4.1-6.9, 7.0-12.0, 12.1-20.9, 21.0-32.0, tr; frequencies = 55, 18, 18, 21, 17, 9, 9, 6, 2, 29 respectively. Here tr denotes a trace amount. Using all 184 values, Ben estimates sum(x) = 390 and sum(x^2) = 4336.
Edexcel 9MA0 · June 2023 · Paper 31 · Q6 · 14 marks
14 marks
Examiner report
This question also proved to be quite challenging. In part (a) most students could use the histogram to calculate the number of patients spending between 10 and 30 hours in hospital but many failed to convert this to a probability by dividing by 90. Some students tried using small squares, but errors were often made and a divisor of 1600 not 1800 was often seen. Part (b) though was not answered so well. Some did explain that the histogram was not symmetric or was skewed and so a normal distribution wouldn’t be suitable. Some focussed on the continuous nature of the variable and thought it was therefore suitable whilst others thought that T was discrete and therefore a normal wasn’t suitable. The question wanted reference to the histogram and there was no expectation or requirement to find probabilities from the normal distribution nor compare the mean of the data with 14.9 If further calculations were expected these would be asked for and reflected in the mark allocation. Many students were able to use integration by parts correctly to establish the result in part (c). Some started the wrong way around (using u = e – x) and soon realised they were not heading for the given answer and there was some muddling over the signs when applying the limits. Part (d) was not answered so well with only a minority of students realising that the area under the curve should equal 90. Those who did realise this were often able to apply the answer to part (c) though some used n = 40 rather than n = 4. The first part of (e) was often answered well and some realised that the answer to part (c) could be used to answer (e)(ii) but some tried using n = 10 and 30 and others stopped after find an estimate for the number of patients as 53.1 and failed to complete the question to find the required probability. In part (f) only a small number of candidates realised that the fact that some patients might stay longer than 40 hours was a limitation of the model. Others did not give any context in their answers or discussed the shape of the curve without considering the stated upper limit.
A medical researcher studies the number of hours T that a patient stays in hospital after an operation, using a random sample of 90 patients. A histogram has the following class intervals and frequency densities: 0-4: 2.5; 4-7: 3.0; 7-12: 4.2; 12-16: 4.0; 16-20: 3.5; 20-40: 1.0. The sample mean is 14.9 hours and the standard deviation is 9.3 hours. Tomas suggests T ~ N(14.9, 9.3^2). Xiang suggests modelling the frequency polygon by y = kx e^(-x) for 0 <= x <= 4, where x is measured in tens of hours and k is constant.
Edexcel 9MA0 · June 2022 · Paper 31 · Q3 · 7 marks
7 marks
Examiner report
Part (a) tested knowledge of the data set. Some students were able to identify tr or trace as the required answer, but many displayed their lack of knowledge of the data set, with n/a being a typical wrong answer, or 0 or 0.025. Many students gave no answer of any sort and there were a few nonsensical or confused answers such as explaining how the rainfall might be collected (a particular favourite being ‘with a bucket and ruler’) or how it might be recorded (in an excel spreadsheet). The vast majority of students were able to find the mean correctly in Part (b). Mistakes if any in (i) tended to be rounding the answer to 5.6 without previously showing a more accurate answer, hence losing the single mark. A few students showed the calculation without evaluating it. Disturbingly there was a tiny minority who seemed to have formulae involving square root signs etc. It is disappointing to see evidence of students studying statistics at advanced level who still do not know how to calculate a mean. There were more errors (ii) although well answered by the majority. Errors tended to be missing the square root in their expression for the standard deviation giving the variance as the final answer, with students believing this to be the standard deviation; subtracting the mean rather than the square of the mean premature approximation of the mean leading to inaccurate final answers and giving the answer to 2sf or an even lesser degree of accuracy. A small number of students gave an answer with no working and were then at the mercy of having obtained a sufficiently accurate answer or risk losing these relatively easy marks. The calculation and use of 𝑆𝑆𝑥𝑥𝑥𝑥 was seen occasionally. Part (c) proved to be more problematic. Many students lacked knowledge of the LDS and did not know the relative geographical locations of Camborne and Leuchars within the UK. As such, a significant number of students made the incorrect assumption that Leuchars is in the South and Camborne is in the North. However, when students were able to locate the relative positions of the two places correctly, they were generally able to draw the correct conclusion, resulting from comparing the mean rainfall between the two places. The main errors included • not making reference to the relative positions of the two locations in their answer e.g. stating that Camborne was in the South with no reference to the relative position of Leuchars, • not making any statements about either of the geographical locations at all and only compared means, with responses such as: no, as the mean is less in Leuchars than in Camborne. • not drawing the correct conclusion after correctly comparing locations and means • not stating any conclusion. • incorrectly comparing standard deviations rather than means. There was a wide variety of responses to part (d). Successful students often stated one or more of the conditions of the Binomial Distribution, with reasons why this distribution was not suitable in this case. e.g. • the probability needs to be independent but the probability of rainfall is not independent as it often depends on the weather from the day before. • the probability of rain needs to be constant but the probability was taken from May to October and summer is June to September and therefore the probability is likely to be higher than in the summer when there is less rain Errors included • referring to independence without context • commenting on the great storm of 1987 or global warming • stating that the data is out of date • the sample size is too small • some values are trace values and/or rainfall is influenced by other factors.
Dian uses the large data set to investigate Daily Total Rainfall r mm for Camborne. For the 31 days of August 2015, n = 31, sum(r) = 174.9 and sum(r^2) = 3523.283.
Edexcel 9MA0 · June 2022 · Paper 31 · Q6 · 9 marks
9 marks
Examiner report
In part (a) a large number of students simply wrote ‘negative correlation’, with no context. The word Interpret in the question hints that context is required. Of the students that did try to truly interpret the correlation in a contextual statement, the vast majority were successful. Whilst many students realised that one- tailed hypotheses were needed in part (b) few used ρ for the parameter. Most students had ± the correct CV but the final A1 was lost by many students, as they compared the positive value 0.3887 to the negative test statistic. This showed these students were unfamiliar with the meaning of the CV/CR and how it related to their test statistic. Part (d) differentiated well between students with basic knowledge and those with deeper understanding. The majority of students scored 5/5 on this part. Almost all used the LH approach on the MS. This was usually well done, though a good number made errors with the laws of indices and maintained a + between the two terms after making the equation into powers of 10. The students knew their log rules, but many did only so to a basic degree, and came unstuck during their working. Sometimes this was to such an extent that they lost the marks. A surprisingly high number of students used natural logarithms and hence the exponential function in this question showing a less secure grip of this topic.
Anna records resting heart rate h beats per minute and weekly exercise time m minutes for a random sample of 19 students. The scatter diagram shows a clear curved downward pattern: h generally falls from roughly 70-76 at low m towards roughly 62-65 by m = 150-400, with the curve becoming flatter as m increases. She codes x = log_10(m) and y = log_10(h). The product moment correlation coefficient between x and y is -0.897, and the regression line of y on x is y = -0.05x + 1.92.
Edexcel 9MA0 · October 2021 · Paper 31 · Q2 · 6 marks
6 marks
Examiner report
Most scored the mark in part (a) by identifying that the correlation was negative. When we ask students to “describe” we are simply requiring a non-contextual description. We will ask them to “interpret” when we require a contextual description. In this case we allowed students to describe the correlation in terms of the variables so answers such as “as x increases y decreases” were accepted. Some thought the correlation was strong and others weak but at this stage all we wanted was the negative description. Part (b) caused some problems with some stating that the scatter diagram showed negative correlation but failing to say whether or not this was compatible with Marc’s suggestion. Some students felt that there was no correlation shown by the scatter diagram and did not reconsider this even after correctly answering part (d). Most answered part (c) correctly though a number “lost” the minus sign between their calculator and writing down the answer on the page. Usually there were sufficient figures given for us to award the mark when the minus sign was included. In part (d) most attempted the hypotheses and the majority used ρ. Some didn’t have “= 0” for their null hypothesis with alternative values or inequalities being used and of course some used r or simply wrote the hypotheses in words. We allowed (+) for the critical value provided the actual value was compatible with their alternative hypothesis and many achieved this mark but few scored the final mark as the conclusion was not related to the context of the question.
Marc records, for a random sample of 16 students, x = the number of letters in the student's last name and y = the number of letters in the student's first name. The scatter diagram plots the points (3,7), (6,7), (8,4), (7,4), (5,6), (3,8), (11,5), (3,5), (4,8), (5,4), (4,7), (9,4), (7,5), (10,5), (6,6), (6,3), on axes 0 <= x <= 12 and 0 <= y <= 10.
Edexcel 9MA0 · October 2021 · Paper 31 · Q3 · 8 marks
8 marks
Examiner report
Although this question involved familiarisation with the large data set nearly 70% scored 3 or more marks here. Many students knew the units used for Daily Mean Pressure (or wrote “Pascals” which was allowed). The coding caused problems for many in part (b) finding y was easy enough but then adding the 1010 seemed to elude a great number of the candidates. Another frequently seen error was to add 1010 to Σy leading to the common wrong answer of 40.8. Part (c) was the most successful part with many correct answers seen; sometimes the effect of coding was clearly stated but often, I suspect, it was simply ignored and the correct answer obtained anyway. Occasionally students felt they should add the 1010 on after reaching the correct value of 12.1 and this just lost the first method mark for this part. The final part proved challenging. There was evidence that students knew the relative locations of the 3 sites and often they deduced that it was high pressure and therefore the winds circulated clockwise. The major problem though was that students did not seem to appreciate that Cardinal Wind Direction in the large data set gives the direction from which the wind comes. This led to an incorrect allocation of directions in the table, some students were clearly confused by this and felt that the circulation must be anticlockwise and then a correct allocation of wind directions could, of course, be deduced but this lost the first mark. A small percentage of students though did achieve full marks for this question.
Stav studies the large data set for September 2015. He codes Daily Mean Pressure x using y = x - 1010. For all 30 days at Hurn, sum(y) = 214 and sum(y^2) = 5912. On 26/09/2015, Daily Mean Pressure was 1029 at Heathrow, 1028 at Hurn and 1028 at Leuchars. The three Cardinal Wind Directions were W, NE and E, in random order. Assume the three locations were under a single region of pressure. In the UK, winds circulate clockwise around high pressure and anticlockwise around low pressure.
Edexcel 9MA0 · October 2020 · Paper 31 · Q2 · 7 marks
7 marks
Examiner report
Very few students were unable to identify the negative correlation in part (a) with many attempting to describe the strength as well (split fairly evenly between weak and strong) though this was not required here. In part (b) it was clear that more students were familiar with the large data set. Many identified “rainfall” as the variable on the y-axis and they usually gave the correct units although some thought it should be mm3. The most common incorrect answer here was “cloud cover” but, of course, this variable is not available on the large data set for Perth. Part (c) was a standard hypothesis test for correlation. It was encouraging to see a good number using ρ for the hypotheses but fewer realised that the test was two-tailed. The phrase “…believes there is a correlation between…” shows that a two tailed test should be used, however many were swayed by the negative value of r and opted for a one tailed test. It was good to see that most chose the critical value from the tables that was compatible with their alternative hypothesis and this meant that they could also access the final mark provided their comment was in context, “support for Stav’s belief” was perhaps the simplest way of including the context. There were some good answers to part (d) but many others were unable to pull together the various parts of the question to make a correct inference. Some stated that the humidity was high and other assumed this but interpreted the negative value of r correctly and were then able to state that they would expect the number of hours of sunshine to be low. Some gave a suggested value for the number of hours of sunshine but this needed to be realistic and values more than 5 were not accepted. A few students knew (from the large data set) that humidity greater than 95% implies foggy conditions and argued correctly that this would suggest a low amount of sunshine.
A random sample of 15 days is taken from the large data set for Perth in June and July 1987. A scatter diagram compares two variables. The x-axis is Daily Mean Temperature in degC, running from 0 to 20. The y-axis is unlabelled and has a scale break near the origin. The 15 plotted points show a clear negative trend, with the higher y-values mainly at x about 11 to 15 and the lower y-values mainly at x about 15 to 19; two observations lie on the horizontal axis. No numerical y-scale is shown.
Edexcel 9MA0 · October 2020 · Paper 31 · Q3 · 10 marks
10 marks
Examiner report
Part (a) should have been an easy start to the question but a surprisingly large number of students gave an answer of 33 by failing to consider the outliers to determine the range. Part (b) though was answered very well and nearly all students answered (c) correctly. The presence of the formula for standard deviation in the formula booklet meant that a good number answered part (d) correctly although occasionally some forgot to take the square root of failed to square the mean. In (e) most gave a suitable calculation to support their statement that Taruni would identify only 1 outlier. Some candidates used the median of 20 rather than the mean of 22.5 though. The last two parts were more challenging but it was encouraging to see a good number of attempts. Many were able to deduce that because the mean did not change then a + b = 45 but fewer were able to establish that, because the median increases, both values had to be greater than 20. Some did state these two conditions but sadly did not choose suitable values (a = 25 and b = 20 was a common error) for their final answer. In (g) many students stated that the new values would be “close to the mean” but this was not precise enough to ensure that the standard deviation would decrease. Some of the very strongest candidates did realise that because both values were within 1 standard deviation of the mean then they could be sure that the new standard deviation would be smaller. This appreciation that standard deviation represents “average distance from the mean” is clearly not well known.
Each of 27 people is timed while completing a puzzle. A box-and-whisker plot for time x minutes has minimum 7, lower quartile 14, median 20, upper quartile 25 and maximum 68. Its upper whisker ends at 40, and separate points are plotted at 46 and 68. For the 27 times, sum(x) = 607.5 and sum(x^2) = 17623.25.
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Everything here stays free. If data presentation and interpretation is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.