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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- (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) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (d) Route 1, as its times vary by 6 minutes rather than 20 — Method: work out an average and a measure of spread for each route, then decide which matters to a commuter who must arrive on time every day. Working: for Route 1, 22 + 23 + 24 + 24 + 25 + 25 + 26 + 26 + 27 + 28 = 250 and 250 ÷ 10 = 25, so the mean is 25 minutes, and the range is 28 − 22 = 6 minutes. For Route 2, 18 + 19 + 20 + 20 + 21 + 22 + 26 + 30 + 36 + 38 = 250 and 250 ÷ 10 = 25, so the mean is also 25 minutes, but the range is 38 − 18 = 20 minutes. The means give no reason to prefer either route; the spreads do, because a commuter who must never be late has to allow for the worst day, which is 28 minutes on Route 1 and 38 minutes on Route 2. Answer: Route 1, as its times vary by 6 minutes rather than 20. The distractors: saying Route 2 has the lower mean assumes that its quicker-looking early times must pull the average down, when both routes total 250 minutes over the ten days; choosing Route 2 for its fastest journey of 18 minutes judges a route by its best day, and the commuter has to survive its worst; saying either route will do uses the equal means and ignores the spread altogether, which is the one thing that separates the two routes.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
- (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) 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.
- (b) 60 — Method: for n ordered values, the upper quartile sits at position 3(n + 1) ÷ 4, counting from the smallest. Working: n + 1 = 11 + 1 = 12; 3 × 12 = 36 and 36 ÷ 4 = 9, so the upper quartile is the 9th value in the list 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65, which is 60. Answer: the upper quartile is 60 seconds. Watch which quartile you find: counting to the 6th value gives the median, 52, not the upper quartile; using 3 × 11 = 33 and 33 ÷ 4 = 8.25 without adding 1 to n first, then rounding down, reaches the 8th value, 58, not the 9th; and counting to the 3rd value uses the lower quartile's position, 45, the wrong end of the list.
- (c) Branch A waits longer, and Branch A is more consistent — Method: compare the two branches using a measure of location (the median) for who waits longer, and a measure of spread (the interquartile range) for who is more consistent — a smaller interquartile range means more consistent. Working: Branch A's median, 12 minutes, is higher than Branch B's, 9 minutes, so Branch A's customers wait longer on average. Branch A's interquartile range, 5 minutes, is smaller than Branch B's, 11 minutes, so Branch A's waiting times vary less. Answer: Branch A waits longer, and Branch A is also the more consistent of the two. Watch that each half of the comparison uses the right statistic and reads it correctly: swapping both readings gives Branch B the longer wait and the greater consistency, when neither is true; keeping the median comparison right but reading a larger interquartile range as 'more consistent' has the direction of spread backwards; and swapping only the median comparison keeps the correct branch for consistency but gives the wrong branch the longer wait.
- (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) 1.2 pupils per mark — Method: on a histogram whose class intervals are not all the same width the height of a bar is not the frequency but the frequency density, found by dividing the frequency of the class by the width of that class, so that the area of the bar represents the frequency. Working: the class 40 ≤ m < 60 holds 24 pupils, and its width is 60 − 40 = 20 marks, so the frequency density is 24 ÷ 20 = 1.2. Answer: 1.2 pupils per mark. The distractors: 24 pupils per mark comes from plotting the frequency itself as the height, which is only correct when every class has the same width; 2.4 pupils per mark comes from dividing by 10, the width of the narrowest class, instead of by the width of this class; 0.2 pupils per mark comes from dividing by the 120 pupils in the year group, which gives the proportion of pupils in the class and not a frequency density.
- (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.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (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.
- (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.
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