Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Statistics worksheet — GCSE Higher
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- (c) 69 — Method: multiply the mean by the number of values to find the total, then subtract the total of the known values. Working: the total of all five scores is 68 × 5 = 340. The total of the four known scores is 55 + 62 + 74 + 80 = 271. The fifth score is 340 − 271 = 69. Subtracting the other way round, 271 − 340 = −69, gives the right size answer with the wrong sign. Guessing that the missing score simply equals the mean, 68, ignores that the four known scores are not themselves centred on 68. Multiplying the mean by 4 instead of 5, 68 × 4 = 272, then 272 − 271 = 1, undercounts how many scores there are. Always multiply the mean by the TOTAL number of values before subtracting.
- (d) 210 — 350 − 150 = 200. 200 ÷ 10 = 20, so the gradient is 20. Using the point (5, 150): 20 × 5 = 100, so 150 − 100 = 50 is the intercept, giving the line y = 20x + 50. At x = 8: 20 × 8 = 160, and 160 + 50 = 210, so the estimated number of visitors is 210. Choosing 160 stops after 20 × 8 = 160 and forgets to add the intercept of 50. Choosing 250 comes from averaging the two given y-values: 150 + 350 = 500, and 500 ÷ 2 = 250, instead of using the line's equation. Choosing 240 assumes the visitors are directly proportional to the hours of sunshine using the first point, 150 × 8 ÷ 5 = 240, which ignores that the line does not pass through the origin.
- (a) 15 — Year 11 has 50 − 28 = 22 pupils in total. Of the 22 pupils who walk in total, 15 are in Year 10, so 22 − 15 = 7 Year 11 pupils walk. Subtracting that from the Year 11 total gives 22 − 7 = 15 Year 11 pupils who are driven. Choosing 28 takes the whole school's driven total, 50 − 22 = 28, and treats it as if it were Year 11's alone, without separating the year groups. Choosing 7 correctly finds how many Year 11 pupils walk but stops there, giving that figure instead of the number who are driven. Choosing 35 comes from 50 − 15, subtracting the Year 10 walkers from the whole school total rather than working within Year 11.
- (a) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (a) A vertical line chart (discrete numerical data) — The number of pets is discrete numerical data — whole-number values such as 0, 1, 2, 3 or 4 — recorded for one variable, so a vertical line chart is the chart specified for this kind of data. A bar chart is used for categorical data, such as favourite colour, not numerical values counted like this. A pie chart shows proportions of a whole and does not show the frequency of each separate value. A scatter graph compares two different variables against each other, and only one variable, the number of pets, is recorded here.
- (d) 1100 — Method: to combine two samples of different sizes, add the faulty counts together and add the sample sizes together before scaling up, rather than treating the two samples separately. Working: the combined sample found 34 + 21 = 55 scratched cases out of 100 + 50 = 150 cases checked, a proportion of 55 ÷ 150. Applying that proportion to the week's production of 3,000 gives an estimate of 55 ÷ 150 × 3000 = 1100 scratched cases. Averaging the two shifts' proportions instead of combining their totals, (34 ÷ 100 + 21 ÷ 50) ÷ 2 = 0.38, gives 0.38 × 3000 = 1140 — this treats the two samples as equally weighted even though Shift A checked twice as many cases as Shift B. Using only Shift A's sample, 34 ÷ 100 × 3000 = 1020, ignores Shift B's cases completely. Using only Shift B's sample, 21 ÷ 50 × 3000 = 1260, ignores Shift A's cases completely. When two samples are different sizes, combine their totals before finding the proportion — do not average the two proportions, and do not use only one shift's sample.
- (d) 60 — Method: on a histogram the frequency of a class is its frequency density × its class width, and the frequencies of all the classes add up to the total, so turn each labelled bar into a frequency and subtract their total from 250. Working: 10 ≤ age < 20 has width 20 − 10 = 10, so its frequency is 4.5 × 10 = 45; 20 ≤ age < 35 has width 35 − 20 = 15, so its frequency is 6 × 15 = 90; 50 ≤ age < 70 has width 70 − 50 = 20, so its frequency is 2.75 × 20 = 55. Those three come to 45 + 90 + 55 = 190, and the total is 250, so the missing frequency is 250 − 190 = 60. Answer: the class 35 ≤ age < 50 has 60 members. Watch what you do with the total and the three frequencies you have found: giving the total, 250, as the answer forgets that three bars have already accounted for some of the members; giving 190, the total of the other three classes, reports how many members are not in this class rather than how many are; and leaving one of the three out of the subtraction, for example 45 + 90 = 135 and 250 − 135 = 115, still owes the class at 50 ≤ age < 70 its 55 members.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (b) Q was faster on average and more consistent — Method: compare the medians for the average and the interquartile ranges for the spread, remembering that a shorter time is faster and a smaller interquartile range means more consistent. Working: the median for class Q is 35 seconds against 38 seconds for class P, so class Q was faster on average; the interquartile range for class P is 46 − 24 = 22 seconds and for class Q it is 44 − 30 = 14 seconds, so class Q's times are more tightly grouped. Answer: class Q was faster on average and more consistent. The distractors: calling Q slower comes from comparing the lower quartiles, 30 against 24, as though a quartile were the average; calling Q less consistent comes from using the gap between the median and the upper quartile as the spread, 44 − 35 = 9 against 46 − 38 = 8, instead of the full interquartile range; the statement that Q was both slower and less consistent comes from making both of those mistakes together.
- (a) No, the size of the fire affects both of the quantities — Method: correlation says that two quantities change together; a claim that one of them produces the other is a further claim, and it needs evidence that a scatter graph on its own cannot give. Working: the graph does show strong positive correlation, so more engines did go with greater damage. But neither quantity was set by the researchers: both were decided by how large the fire was. A large blaze brings many appliances and also destroys a great deal, while a small one brings few and destroys little, so a third quantity is driving both of the recorded ones. Answer: no, because the size of the fire affects both of the quantities. The distractors: saying the correlation is negative contradicts the graph, which shows the two quantities rising together, and reaching the right verdict from a false reading of the data is not the reason the mark is for; saying that strong positive correlation shows one quantity causes the other is the assumption the question exists to test, and no strength of correlation can establish cause; saying the points lie close to the line of best fit describes how strong the correlation is, and strength and cause are different matters entirely.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (c) Class X has the higher median and the wider spread — Method: compare the two box plots statistic by statistic — median for location, and the interquartile range for spread — checking the true value of each rather than assuming a pattern. Working: Class X's median is 60 and Class Y's is 58, so Class X's median is the higher one. Class X's interquartile range is 70 − 45 = 25 and Class Y's is 65 − 50 = 15 (and the ranges follow the same order: 95 − 20 = 75 against 80 − 35 = 45), so Class X also has the wider spread. Answer: Class X has both the higher median and the wider spread. Watch that each half of a compound statement is checked separately: claiming Class Y has the higher median and the wider spread gets both comparisons backwards; claiming Class X has the higher median but the narrower spread keeps the median right while reading the spread the wrong way round; and claiming Class Y has the higher median but the narrower spread swaps the median comparison while getting the spread right.
- (b) Town B — higher median and smaller IQR — Method: 'higher and more consistent' needs two comparisons — the median for typical price, and the interquartile range for spread, with a smaller interquartile range meaning more consistent. Working: Town B's median, £235,000, is higher than Town A's, £220,000. Town A's interquartile range is 310 − 180 = 130 and Town B's is 260 − 175 = 85, so Town B's interquartile range is the smaller of the two. Answer: Town B has both the higher median and the smaller interquartile range, so it is the town with higher, more consistent prices. Watch which combination of median and interquartile range each statement claims, and for which town: claiming Town A has the higher median and the smaller interquartile range gets both comparisons wrong, since Town B leads on both; claiming Town A has the higher median (still wrong) but the larger interquartile range at least reads the spread correctly, without it rescuing the false median claim; and claiming Town B has the higher median (correct) but the larger interquartile range misreads the spread — Town B's interquartile range is the smaller of the two, not the larger.
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