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.
Answer key: Statistics worksheet — GCSE Higher
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- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
- (d) 25% — First find the number of fruit cakes: 80 − 34 − 26 = 20. Then write this as a percentage of the total: 20 ÷ 80 × 100 = 25%. Giving 20% comes from reporting the count of fruit cakes, 20, directly as a percentage, without dividing by the total of 80 first. Giving 32.5% computes the percentage of chocolate cakes instead of fruit cakes: 26 ÷ 80 × 100 = 32.5%. Giving 42.5% computes the percentage of sponge cakes instead of fruit cakes: 34 ÷ 80 × 100 = 42.5%.
- (c) 156 cm — Method: to combine two groups' means, multiply each group's mean by its own number of pupils, add the two totals together, then divide by the total number of pupils in both groups. Working: 20 × 150 = 3,000 cm for the boys and 10 × 168 = 1,680 cm for the girls, giving a combined total of 3,000 + 1,680 = 4,680 cm. Dividing by all 30 pupils gives 4,680 ÷ 30 = 156 cm. Giving 159 cm averages the two means, (150 + 168) ÷ 2, treating the two groups as if they had the same number of pupils, when there are twice as many boys as girls. Giving 4,680 cm finds the correct combined total height but stops there, forgetting the final division by the 30 pupils. Giving 234 cm divides the combined total by 20, the number of boys only, forgetting that the total also includes the 10 girls. Always weight each mean by its own group size, and always divide by the TOTAL number of pupils in both groups combined.
- (b) A person's shoe size and their favourite colour — A person's shoe size is not linked to which colour they prefer, so these two show no correlation. The other three pairs are all genuinely correlated: distance travelled and fuel used rise together, which is positive correlation; hours of revision and test score generally rise together, which is also positive correlation; and as outdoor temperature rises, fewer woolly hats are sold, which is negative correlation. Negative correlation is still a real relationship between two variables — it is not the same thing as no relationship at all, so the temperature and hats pair is not the answer to this question.
- (c) Because every pupil has an equal chance of being picked — Method: whether a sample represents its population is decided by the selection method, not by the size of the sample, so ask whether the method gives every member of the population the same chance of being chosen. Working: the names are drawn at random from a list of all 10,000 pupils, so each pupil has the same chance, 500 out of 10,000, of being drawn, and no group of pupils is more likely to appear than any other; that is what keeps bias out of the sample. Answer: because every pupil has an equal chance of being picked. The distractors: the reply about 5% treats the sampling fraction as the test of fairness, but a badly chosen 5% is still biased and a well chosen 1% is not; the reply about 500 being large enough makes size the test instead, which is the same mistake in another form, since a large sample drawn from one school would still misrepresent the city; the reply about the most willing pupils describes self-selection, which hands the choice of who is in the sample to the pupils who feel most strongly about the question.
- (c) 42 — Method: find the target cumulative frequency, 90% of the total, locate the class it falls in from the plotted points, then interpolate: lower boundary, plus the extra distance needed into the class divided by the class's frequency, times its width. Working: 90% of 320 is 0.9 × 320 = 288. The plotted points show a cumulative frequency of 280 at d = 40 and 320 at d = 50, so the class 40 ≤ d < 50 has frequency 320 − 280 = 40 and width 50 − 40 = 10, and 288 falls inside it. The extra distance needed into the class is 288 − 280 = 8, and 8 ÷ 40 × 10 = 2, so the diameter is 40 + 2 = 42. Answer: the estimated diameter is 42 mm. Watch which point and which class the interpolation actually uses: reading off d = 40, the plotted point just below the target, instead of interpolating the extra 8 ball bearings into the next 10 mm, stops one step short of the true answer; finding the diameter below which only 10% lie instead of 90% gives a target of 0.1 × 320 = 32, which falls in the class 10 ≤ d < 20 — the extra distance into that class is 32 − 30 = 2, and 2 ÷ 60 × 10 = 0.3, so this route gives 10 + 0.3 = 10.3, the bottom decile rather than the top 90%; and interpolating within the class 30 ≤ d < 40 instead of 40 ≤ d < 50, as though 288 had not yet reached a cumulative frequency of 280, treats the extra distance as 288 − 190 = 98, and 98 ÷ 90 × 10 = 10.9, giving 30 + 10.9 = 40.9, one class too early.
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (c) Route 2, because its interquartile range is smaller — Method: for a journey where turning up on time matters, what matters is not the typical (median) time but how predictable it is — a smaller interquartile range means the middle half of journeys cluster closer together. Working: Route 1's median, 34 minutes, is in fact lower than Route 2's, 41 minutes, so Route 1 is faster on average; but Route 1's interquartile range, 22 minutes, is far larger than Route 2's, 6 minutes, so Route 1's times are much less predictable. Answer: Priya should choose Route 2, because its interquartile range is smaller, even though it is slower on average. Watch which statistic answers the question actually asked: Route 1 does not have the smaller interquartile range, Route 2 does, so picking Route 1 for that reason misreads the table; Route 1's median genuinely is the lower one, but a lower median answers 'which is faster', not 'which is more reliable'; and Route 2's median is not the lower one, so that claim about Route 2 is simply false.
- (d) 16 — Method: on a box plot the interquartile range is the width of the box itself, upper quartile take away lower quartile. Working: the upper quartile is 38 and the lower quartile is 22, so 38 − 22 = 16. Answer: the interquartile range is 16 years. Watch which part of the box plot you are reading: the whole line from whisker to whisker gives the range, 61 − 15 = 46; the left half of the box alone gives median take away lower quartile, 29 − 22 = 7; and the right half of the box alone gives upper quartile take away median, 38 − 29 = 9 — neither half is the interquartile range on its own.
- (b) 6 — Method: frequency density = frequency ÷ class width. Working: the class 12 ≤ h < 18 has width 18 − 12 = 6, so frequency density = 36 ÷ 6 = 6. Answer: the frequency density is 6 seedlings per cm. Watch which numbers you use: taking the lower bound, 12, as the width instead of 18 − 12 = 6 gives 36 ÷ 12 = 3; dividing the total number of seedlings, 90, rather than this class's frequency, 36, by the width gives 90 ÷ 6 = 15, a density that belongs to no single class; and multiplying instead of dividing gives 36 × 6 = 216, far too large a density for so narrow a class.
- (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (a) £16,000 — Method: first find the quarter with the highest sales figure, then subtract quarter 1's sales from it — remembering that every figure is given in THOUSANDS of pounds. Working: the highest sales figure is quarter 2, at £34,000 (34 thousand pounds). The increase from quarter 1 is £34,000 − £18,000 = £16,000. Giving £34,000 reads off the highest sales figure on its own, without subtracting quarter 1's sales — that is the highest quarter's total, not the increase. Giving £12,000 uses quarter 3's sales, 30, the SECOND-highest figure, instead of quarter 2's 34, the actual highest — 30 − 18 = 12, but quarter 3 is not the quarter with the highest sales. Giving £16 gets the subtraction right, 34 − 18 = 16, but forgets that every figure in the question is in thousands of pounds, so the increase is £16,000, not £16. Always identify the correct quarter FIRST, and always check the units the numbers are given in before writing your final answer.
- (d) 300 pupils — Method: an estimate for a whole population is made by finding the proportion in the sample and applying that same proportion to the population. Working: in the sample 30 of the 50 pupils prefer sport, a proportion of 30 ÷ 50 = 0.6, and applying that proportion to the school gives 0.6 × 500 = 300 pupils. Answer: 300 pupils, and it is only an estimate, because a different random sample of 50 would give a slightly different figure. The distractors: 200 pupils comes from scaling up the 20 pupils in the sample who did not prefer sport, 20 × 10, which answers the opposite question; 150 pupils comes from reading 30 out of 50 as 30% and taking 30% of 500; 60 pupils comes from working out the proportion correctly as 60% and then writing the 60 down as a number of pupils instead of applying it to the 500.
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