Printable · GCSE Foundation · ages 14-16
Statistics worksheet — GCSE Foundation
Fifteen questions across the statistics statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Statistics worksheet — GCSE Foundation
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- 1.A gardener plants 600 daffodil bulbs. In March she digs up 20 of them, chosen at random, and finds that 17 have flowered. Write down what she can conclude about the 600 bulbs.
- 2.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
- 3.A company makes 50,000 light bulbs a day and wants to check how long they last before they fail. Testing a bulb to find out how long it lasts destroys it. Give a reason why the company should test a sample of bulbs rather than every bulb it makes.
- 4.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 5.A random sample of 50 pupils at a school were asked whether they prefer sport to music. 30 of the 50 pupils said they prefer sport. The school has 500 pupils altogether. Work out an estimate for the number of the 500 pupils who prefer sport.
- 6.On a scatter graph of the age of a car, in years, and its value, in pounds, the points fall from left to right. Write down the type of correlation shown.
- 7.There are 10,000 pupils in a city. A researcher picks a sample of 500 of them by drawing names at random from a list of every pupil in the city. Give the reason why this method gives a representative sample.
- 8.A netball team scored 50, 83 and 68 points in three matches. Work out the mean number of points scored.
- 9.The masses of eight school bags, in kilograms, are 3, 4, 4, 5, 6, 7, 8 and 11. Work out the median mass.
- 10.A shop records the colour of every car that uses its car park one Tuesday. Its frequency table reads: black 44, silver 37, blue 26, red 18, white 12. Write down the frequency of black cars.
- 11.A scatter graph plots the shoe size and the spelling test score of 25 pupils. The points are scattered with no pattern across the graph. Write down the type of correlation shown.
- 12.A vertical line chart shows the number of goals scored by a football team in each of its 20 matches: 0 goals in 4 matches, 1 goal in 7 matches, 2 goals in 6 matches and 3 goals in 3 matches. Write down the modal number of goals.
- 13.A two-way table records whether each of the 30 pupils in a class passed maths and whether they passed science. 18 pupils passed maths, 12 pupils passed science and 8 pupils passed both. Work out how many pupils passed at least one of the two subjects.
- 14.A stem-and-leaf diagram, described in words, shows the ages of 9 people at a family party. The stem is the tens digit: stem 1 has leaves 4 and 8; stem 2 has leaves 0, 3, 5 and 9; stem 3 has leaves 1 and 6; stem 4 has leaf 2. Work out the median age.
- 15.The age, in years, and the score in a reaction test are recorded for eight members of a sports club: (14, 92), (18, 88), (23, 85), (27, 80), (31, 78), (36, 74), (42, 70), (49, 65). The eight pairs are plotted on a scatter graph. Describe the correlation between age and score.
Answer key
- (d) About 510 of the 600 bulbs are likely to have flowered — Method: the proportion found in a random sample is used as an estimate of the proportion in the whole population, and the conclusion is stated as an estimate, never as a fact about every member. Working: 17 of the 20 bulbs dug up had flowered, so the sample proportion is 17 ÷ 20 = 0.85, and applying that proportion to the whole planting gives 0.85 × 600 = 510 bulbs. A different random sample of 20 would very probably give a slightly different figure, so 510 is an estimate. Answer: about 510 of the 600 bulbs are likely to have flowered. The distractors: saying exactly 510 have flowered takes an estimate from a sample of 20 as a count of all 600, which no sample can deliver; saying exactly 17 of the 600 have flowered reports the sample count as though it were the population count, leaving the other 580 bulbs out of the answer altogether; saying about 20 have flowered uses the size of the sample as the estimate, when 20 is the number of bulbs she dug up rather than a number that flowered.
- (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.
- (d) Testing destroys bulbs, so testing all leaves none to sell. — Method: testing every item in a population instead of a sample is a census — sensible only when testing does not use up or destroy what is being tested. Working: here, testing a bulb to find its lifespan destroys it, so testing all 50,000 bulbs would leave nothing left to sell — a sample lets the company estimate the typical lifespan without destroying its whole stock. Extra electricity used in testing is not the real reason a census is avoided here — it is the destruction of the product that matters. Saying a sample is always more accurate than a full census is the wrong way round: a census, if it could be carried out, gives the exact figure for the whole population — it is testing being destructive, not a lack of accuracy, that rules it out here. There is no law against testing every item a company makes — nothing in the question suggests that. When testing destroys the item being tested, sampling is necessary, not just convenient.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (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.
- (a) Negative correlation — As the age of the car increases, the points fall towards a lower value, so the value decreases as the age increases. This falling pattern is a negative correlation. A positive correlation would show the points rising together instead. No correlation would apply only if the points showed no pattern at all, and correlation is not the same as causation — strong causation is not a type of correlation.
- (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) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
- (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.
- (d) 44 — Method: the frequency of a category in a frequency table is the number of times that category was counted, and it is read from the row for that category. Working: the rows of the table pair each colour with its count, and the row for black is paired with the count 44, so the frequency of black cars is 44. Answer: 44 cars — a frequency is a count of cars, not a colour and not a percentage. The distractors: 37 comes from reading the count paired with silver, that is from reading the wrong row of the table; 137 comes from adding every count in the table, 44 + 37 + 26 + 18 + 12, which gives the total number of cars rather than the frequency of one colour; 5 comes from counting how many different colours the table lists instead of how many cars were black.
- (a) No correlation — Shoe size has no real relationship with spelling ability, and the points here are scattered with no rising or falling trend, so this is no correlation. A positive correlation would show the points rising together, and a negative correlation would show them falling as one increases; neither pattern is present here. Strong correlation is not correct either, since strength only applies once a positive or negative trend exists, and there isn't one.
- (b) 1 — The four frequencies are 4, 7, 6 and 3 matches, and the largest of these is 7, which corresponds to 1 goal, so the modal number of goals is 1. Choosing 7 confuses the frequency, how many matches, with the number of goals itself. Choosing 2 uses the second-largest frequency, 6 matches, instead of the largest. Choosing 3 uses the smallest frequency, which belongs to the fewest matches, not the most.
- (a) 22 — Method: the two subject totals overlap, because every pupil who passed both subjects has been counted once in the maths total and once again in the science total; adding the totals therefore counts those pupils twice, and the overlap has to be taken off once. Working: 18 + 12 = 30, and the 8 pupils who passed both have been counted twice in that 30, so the number who passed at least one subject is 30 − 8 = 22. Answer: 22 pupils, a count of pupils, and it is less than the 30 in the class, which leaves 8 pupils who passed neither. The distractors: 30 comes from adding the two subject totals and never removing the overlap, so it counts the 8 pupils twice; 14 comes from taking the 8 away twice, 18 + 12 − 8 − 8, removing an overlap that was only counted twice once too often; 18 comes from writing down the larger of the two subject totals on its own, which leaves out every pupil who passed science but not maths.
- (d) 25 — In order, the nine ages are 14, 18, 20, 23, 25, 29, 31, 36 and 42, and with 9 values the median is the 5th one, which is 25. Choosing 23 takes the 4th value instead of the 5th. Choosing 29 takes the 6th value instead of the 5th. Choosing 26 comes from averaging the 4th and 6th values, 23 + 29 = 52, and 52 ÷ 2 = 26, a method that is only needed when there is an even number of values.
- (c) Strong negative correlation — Method: correlation is described by two things — the direction the points take as the graph is read from left to right, and how closely the points lie to a single straight line. Working: reading the pairs in order of age, the ages rise 14, 18, 23, 27, 31, 36, 42, 49 while the scores fall 92, 88, 85, 80, 78, 74, 70, 65; the score falls at every single step, with no reversal anywhere, so the points fall from left to right and lie close to a straight line. Answer: strong negative correlation — negative for the falling direction, strong because every point follows the pattern. The distractors: strong positive correlation comes from noticing a clear pattern and calling any clear pattern positive, without checking the direction; weak negative correlation comes from reading the direction correctly but judging points that do not lie exactly on a straight line to be only loosely related, when these eight fall without a single exception; no correlation comes from reading a falling trend as though it showed no relationship at all, when a falling trend is itself a relationship.
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