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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- (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.
- (c) Club members probably eat differently from most people — Method: a sample is biased when the group it is drawn from differs from the population in the very thing the survey is measuring, so compare the subgroup with the population on that quantity. Working: the survey measures eating habits, and people who join a sports club take more exercise than average and are known to eat differently from the town as a whole, so their replies pull the results away from the true picture for the town however many of them are asked. Answer: club members probably eat differently from most people. The distractors: the reply about the number of members treats bias as a question of size, but a large biased sample is still biased; the reply that the members were picked at random is false, since the council picked a club rather than picking residents, and it confuses bias with non-response; the reply that everyone asked lives in the town notes something true of the members but draws the false conclusion that the sample therefore covers the town, when a sample must reflect a population and not merely be taken from inside it.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (b) All 2,000 members of the sports centre. — Method: in a survey, the population is the whole group the survey is trying to find out about, and the sample is the smaller group actually asked. Working: this survey wants to know what the sports centre's members think, so the population is every one of the 2,000 members — whether or not they were personally asked. Saying the population is the 100 members who were asked names the sample, not the population; the sample is drawn FROM the population, so it is smaller than it, not the same as it. Saying the population is everybody who lives in Ipswich widens the group far beyond who the survey is actually about — plenty of Ipswich residents are not members of the sports centre at all, so they are outside this survey altogether. Saying the population is the members who say they are unhappy confuses the population with a result of the survey: whether a member turns out to be happy or unhappy is something the survey finds out, not part of the definition of who is being studied. The population is always the whole group the question is about, before any sampling or any results come in.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (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.
- (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.
- (b) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (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.
- (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.
- (c) Neither causes the other; sunshine links both. — Both ice cream sales and sunburn cases tend to rise on hot, sunny days, so the amount of sunshine is a third factor linked to both — neither variable causes the other. Saying ice cream sales cause the sunburn assumes a causal link in one direction that the correlation alone cannot establish. Saying sunburn cases cause the ice cream sales assumes the reverse causal link, which is no more justified. Saying a strong correlation always means causation is the general error this question is testing: correlation, however strong, does not by itself prove that one variable causes the other.
- (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.
- (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.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (d) 25 — Method: the median is estimated at position n ÷ 2 in the cumulative frequency table, then interpolated across the class it falls in: lower boundary, plus the fraction of the way through the class, times the class width. Working: there are 80 sacks, so the median sits at position 80 ÷ 2 = 40. Before the class 20 ≤ m < 30 the cumulative frequency is 22, and by the end of it, it is 58, so this class holds the 40th sack; its frequency is 58 − 22 = 36 and its width is 30 − 20 = 10. The extra distance needed into the class is 40 − 22 = 18, and 18 ÷ 36 × 10 = 5, so the median is 20 + 5 = 25. Answer: the estimated median mass is 25 kg. Watch which numbers the interpolation uses: reading off just the lower boundary of the median class, 20, ignores how far into that class the 40th sack actually falls; treating n ÷ 2 = 40 itself as the median mass mistakes a position in the list for a mass in kilograms; and using the target position, 40, as the extra distance into the class instead of subtracting the sacks already counted changes the calculation to 20 + 40 ÷ 36 × 10. That comes to 20 + 11.1 = 31.1, overshooting the class because it never subtracts the 22 sacks already counted before it.
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