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 quality inspector weighs a random sample of 50 packets of crisps from one day's production and finds their mean mass is 32.4 g. The factory makes 20,000 packets that day. Work out an estimate for the total mass, in kg, of all the packets made that day.
- 2.A survey of 200 car owners in Bristol records the colour of each car: silver 74, black 52, blue 40 and red 34. Write down which average should be used to describe a typical car in this survey, and give a reason for your answer.
- 3.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.
- 4.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.
- 5.A bar chart shows the highest temperature recorded on each of five days. The temperatures were Monday 20 °C, Tuesday 22 °C, Wednesday 18 °C, Thursday 25 °C and Friday 23 °C. Write down the day on which the temperature was highest.
- 6.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
- 7.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
- 8.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.
- 9.A school has 1,000 pupils. A student wants to estimate how many of them walk to school, so she asks 8 pupils in her own class. She says her sample is large enough to give a reliable estimate for the whole school. Is she right? Give a reason for your answer.
- 10.A timetable shows the departure times of a bus from Leeds to Harrogate: 08:15, 08:50, 09:25, 10:05, 10:40. Chloe arrives at the bus stop at 09:30. Work out how many minutes she has to wait for the next bus.
- 11.A scatter graph shows the height, x cm, and the mass, y kg, of 20 pupils in Year 10. The heights on the graph run from 150 cm to 180 cm, and the line of best fit is y = 0.9x − 85. Nadia puts x = 90 into this equation to estimate the mass of a two-year-old child who is 90 cm tall. Is her estimate reliable? Give a reason for your answer.y = 0.9x − 85
- 12.Five pupils spent these numbers of minutes on their homework: 50, 65, 55, 90, 60. Work out the median time.
- 13.A café in York counts the number of customers in each of the nine hours it is open on one day: 4, 4, 4, 11, 13, 15, 18, 22 and 25. The owner says that a typical hour has about 4 customers, because 4 is the mode. Is the owner right? Give a reason for your answer.
- 14.A pie chart shows the favourite sport of the 40 pupils in a class: football 50%, netball 25%, tennis 15% and swimming 10%. Work out how many pupils chose football.
- 15.A council in Leeds wants to know what local people think about letting shops stay open later in the evening. It rings landline telephone numbers between 10 am and 2 pm on a Tuesday. Write down which group is most likely to be under-represented in the sample, and give a reason for your answer.
Answer key
- (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.
- (d) The mode, because colours cannot be added or ordered — Method: an average can only be used on data that supports the operation it needs. A mean needs the values to be added and divided, a median needs them to be placed in order, and a range needs one value to be taken away from another; a mode needs only counting, so it is the average available when the data are categories rather than numbers. Working: the data collected here are colours, silver, black, blue and red. The numbers 74, 52, 40 and 34 count the cars of each colour, they do not measure them, and 74 + 52 + 40 + 34 = 200 simply returns the size of the survey. No colour can be added to another, and there is no order that puts blue before red, so of the four averages only the one found by counting survives. Answer: the mode, because colours cannot be added or ordered, and the mode is silver. The distractors: the mean is said to use all 200 colours, and a mean of the four frequencies, 200 ÷ 4 = 50, is a number of cars rather than a colour, so it describes nothing about a typical car; the median is said to put the colours in order, but ordering the frequencies 34, 40, 52, 74 orders the counts, not the colours, and gives 46, again a number of cars; the range is not an average at all, and 74 − 34 = 40 measures the gap between the commonest and rarest counts, which is a measure of spread.
- (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.
- (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) Thursday — Method: on a bar chart the tallest bar belongs to the greatest value, so compare the five temperatures and then read off the day that the greatest one belongs to. Working: the temperatures are 20 °C, 22 °C, 18 °C, 25 °C and 23 °C; in order of size these are 18, 20, 22, 23 and 25, so the greatest temperature is 25 °C, and the day recorded with 25 °C is Thursday. Answer: Thursday — the answer is a day, not a temperature. The distractors: Friday comes from stopping at the final temperature listed instead of comparing all five; Wednesday comes from picking out the shortest bar, 18 °C, and so answering for the lowest temperature rather than the highest; Monday comes from writing down the first day in the chart without comparing any of the temperatures at all.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (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) 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.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (a) 35 minutes — Chloe arrives at 09:30, so she needs the first departure time that is at or after 09:30. The bus at 09:25 has already left five minutes before she arrives, so it is missed, not caught; the next one on the timetable is 10:05. From 09:30 to 10:00 is 30 minutes, then a further 5 minutes to 10:05, so the wait is 35 minutes. Giving 40 minutes measures the gap from the missed 09:25 bus to the 10:05 bus, counting from the bus Chloe missed instead of from the time she arrived. Giving 5 minutes uses the missed 09:25 bus, working out 09:30 − 09:25 the wrong way round. Giving 45 minutes comes from treating the times as decimals — 10.05 − 9.30 = 0.75, then × 60 = 45 — instead of working in hours and minutes, where 05 minutes past the hour is not 0.05 of an hour.
- (c) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
- (c) 60 minutes — Method: the median is the middle value once the data have been put in order of size, so the list must be sorted before any position is read. Working: in order the times are 50, 55, 60, 65, 90 minutes; there are 5 values, so the middle position is the third and the time sitting there is 60 minutes. Answer: 60 minutes. The distractors: 64 minutes comes from working out the mean, 320 ÷ 5, instead of the median; 70 minutes comes from taking the time halfway between the shortest and the longest, (50 + 90) ÷ 2; 40 minutes comes from working out the range, 90 − 50, which measures spread rather than centre.
- (c) No, the mode here is the lowest value of the nine — Method: an average is meant to stand for the data as a whole, so test any proposed average by asking how many values it sits near. Working: the value 4 appears three times and every other count appears once, so 4 is indeed the mode. But those three hours are the quiet ones at the start of the day, and the other six counts run from 11 up to 25; putting the nine counts in order, the middle one is the fifth, which is 13. So the mode sits at the very bottom of the data, with six of the nine hours far above it. Answer: no, because the mode here is the lowest value of the nine, so it describes the quiet opening hours rather than a typical hour. The distractors: saying the mode can only be used when no value repeats reverses the definition, since a mode exists only because a value does repeat; saying the mode is the value that occurs most often is a correct definition, but being the commonest value does not make a value typical when it lies at one end of the data; saying the mode is the best average for any list of numbers ignores the fact that mean, median and mode each describe a population well in different circumstances.
- (b) 20 — Method: each sector of a pie chart stands for a percentage of the whole set of data, so the number of pupils in a sector is that percentage of the total number of pupils. Working: the football sector is 50% and the whole pie chart stands for all 40 pupils, so the number of pupils is 50% of 40 = 0.5 × 40 = 20. Answer: 20 pupils, which is a count of pupils rather than a percentage. The distractors: 10 comes from taking the netball sector, 25% of 40, instead of the football sector; 80 comes from doubling the 40 pupils instead of halving them, which is what multiplying by 50% would do the wrong way round; 40 comes from writing down the total number of pupils in the class and never applying the percentage at all.
- (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.
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