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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- (c) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (d) Only internet users reach the website; others are excluded. — Method: a sample is biased when it systematically leaves out part of the population, or systematically over-represents another part. Working: anyone without internet access, or who does not visit the council's website, has NO chance of being included — the sample is drawn only from internet-using residents, which is not the whole town. Saying too many people might respond because the survey is free confuses bias with sample size — bias is about who CAN be reached, not how many respond. Saying people might lie describes a different problem, response honesty, not who was sampled in the first place. Saying online surveys cannot be anonymous is not a reason connected to bias at all. A sample is biased when part of the population has no chance of being included, whatever the reason for that.
- (c) The data show a link only; a third factor may affect both — Method: a study of this kind measures two quantities and reports how they change together; deciding that one of them produces the other is a further claim, and it needs evidence that the measurements alone cannot give. Working: the study shows that more coffee goes with better concentration, which is a positive correlation; but a third factor that was never measured, such as how motivated someone is, could raise both the coffee drinking and the concentration, and the concentration could equally be what leads to the extra coffee. Answer: the data show a link only, because a third factor may be affecting both quantities, so no claim about cause can be made. The distractors: calling the conclusion safe because the correlation is positive treats the direction of a correlation as proof of cause, which no direction can give; calling it wrong because the correlation is negative misreads the direction of the relationship, since the study reports both quantities rising together; saying the two quantities are not linked denies the correlation the study actually found, when what fails is only the claim about cause.
- (c) 70 — Method: estimate the median from the cumulative frequency table by interpolation: find its position, n ÷ 2, locate the class it falls in, then add the fraction of the way through that class (adjusted for the cumulative frequency reached before it) to the class's lower boundary. Working: there are 180 riders, so the median is at position 180 ÷ 2 = 90. Before the class 60 ≤ d < 80 the cumulative frequency is 60, and by the end of it, 120, so the 90th rider falls in this class; its frequency is 120 − 60 = 60 and its width is 80 − 60 = 20. The extra distance needed into the class is 90 − 60 = 30, and 30 ÷ 60 × 20 = 10, so the median is 60 + 10 = 70. Answer: the estimated median distance is 70 km. Watch which numbers the interpolation actually uses: reading off just the class's lower boundary, 60, ignores how far into the class the 90th rider falls; using the target position, 90, as the extra distance instead of subtracting the 60 riders already counted before the class gives 90 ÷ 60 × 20 = 30, so 60 + 30 = 90, overshooting by treating the whole position as if none of it had already been counted; and using the total number of riders, 180, instead of half of it as the target position lands in the very last class, giving an estimate of 130 km — further than any rider is known to have ridden by that point in the table.
- (c) 145 g — Method: find the position of the median from the total frequency, locate the class that contains it, then use linear interpolation inside that class, assuming the apples in it are spread evenly. Working: the median is the 100 ÷ 2 = 50th apple; the running totals are 10, then 10 + 30 = 40, then 40 + 40 = 80, so the 50th apple lies in the class 140 ≤ m < 160; it is the 50 − 40 = 10th of the 40 apples in that class, and the class is 20 g wide, so the median is 140 + (10 ÷ 40) × 20 = 140 + 5 = 145. Answer: an estimated median of 145 g. The distractors: 150 g comes from giving the midpoint of the class that contains the median instead of interpolating inside it; 140 g comes from stopping at the lower boundary of that class, which locates the class but not the value; 155 g comes from measuring the 5 g step down from the upper boundary, 160 − 5, instead of up from the lower boundary.
- (d) 648 kg — Method: to estimate a total from a sample, multiply the sample's mean by the number of items in the whole population, then check the units the question asks for. Working: 32.4 g × 20,000 = 648,000 g. Converting to kilograms, 648,000 ÷ 1,000 = 648 kg. This is only an estimate, not an exact total, because it assumes every one of the 20,000 packets has exactly the sample mean mass, when in reality individual packets vary above and below it. Giving 1.62 kg multiplies the mean by 50, the SAMPLE size, instead of by 20,000, the number of packets actually made that day — this finds the total mass of the 50 sampled packets, not the day's production. Giving 32.4 kg treats the sample mean itself, in grams, as if it already were the day's total mass in kilograms, skipping the scaling up altogether. Giving 648,000 kg correctly scales the mean up to the whole day's production but never converts the answer from grams to kilograms, leaving it 1,000 times too large. Always scale a sample's mean up by the SIZE OF THE WHOLE POPULATION, and always finish by checking the units the question asks for.
- (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.
- (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.
- (b) 1 — Method: for data in a frequency table, find the position of the median using (n + 1) ÷ 2, then read off the value at that position from the cumulative frequencies. Working: there are 19 pupils, so the median is the 10th value. The cumulative frequencies are 7 (up to 0 pets), 10 (up to 1 pet), 14 (up to 2 pets) and 19 (up to 3 pets). The 10th value falls at the end of the '1 pet' group, so the median is 1 pet. Giving 0 pets is the mode — the category with the highest frequency, 7 — not the median. Giving 3, the highest number of pets minus the lowest, finds the range, a different statistic entirely. Giving 19 states the total number of pupils, not a number of pets at all. Find the middle POSITION first, then read off the value it belongs to — do not confuse it with the mode, the range or the total.
- (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.
- (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) 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.
- (d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
- (c) 46 — Method: the cumulative frequency table gives the number of runners below each time; to find the number at or above a time, subtract that cumulative frequency from the total. Working: the cumulative frequency for t < 40 is 74, so 120 runners in total take away the 74 who finished in under 40 minutes: 120 − 74 = 46. Answer: 46 runners took 40 minutes or longer. Watch which boundary and which subtraction you use: reading off t < 50 instead of t < 40 and subtracting, 120 − 110 = 10, answers a different question, '50 minutes or longer'; giving 74 itself as the answer reports how many finished below 40 minutes, the opposite of what was asked; and subtracting the two nearby cumulative frequencies, 110 − 74 = 36, finds how many took between 40 and 50 minutes, not everyone from 40 minutes upward.
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