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.
Statistics worksheet — GCSE Foundation
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- 1.A study found that people who drink more coffee tend to concentrate better at work. A coffee company says that this shows that drinking coffee improves concentration. Give the reason why this conclusion cannot be drawn.
- 2.A bar chart shows the number of books borrowed from a school library on each of five days. The bar heights are: Monday 12, Tuesday 18, Wednesday 9, Thursday 15 and Friday 21. Work out the total number of books borrowed over the five days.
- 3.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.
- 4.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.
- 5.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 6.A school posts a questionnaire about school dinners to the families of all 200 pupils. Only 30 families send theirs back, and 27 of those 30 say they are unhappy with school dinners. Give a reason why this result may not represent all 200 families.
- 7.A pie chart is divided into 10 equal sectors. 6 of those sectors stand for the people who chose purple. Altogether 120 people were asked. Work out how many of them chose purple.
- 8.A factory makes 4,000 light bulbs a day. In a random sample of 80 of one day's bulbs, 3 were faulty. Work out an estimate for the number of faulty bulbs the factory makes in a day.
- 9.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
- 10.A school has 1,500 pupils. The head teacher takes a random sample of 150 of them from the school register and asks how long they spend on homework. Rory says the sample is too small for the result to mean anything. Is Rory right? Give a reason for your answer.
- 11.On a scatter graph of the arm span and the height of some pupils, the points rise from left to right. Write down the type of correlation this shows.
- 12.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.
- 13.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.
- 14.The rainfall, in millimetres, was recorded in Cambridge on six days: 12, 5, 9, 15, 3 and 11. Work out the range of the rainfall.
- 15.A bar chart is drawn for a set of categorical data. Write down what the height of each bar represents.
Answer key
- (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.
- (d) 75 — Method: to find a total from a bar chart, add together the height of every bar. Working: 12 + 18 + 9 + 15 + 21 = 75 books. Leaving out Wednesday's bar by mistake, 12 + 18 + 15 + 21 = 66, misses one day out of the total. Giving 21 states only Friday's bar, the tallest one, not the total of all five days. Dividing the total by the number of days, 75 ÷ 5 = 15, finds the mean number of books per day, not the total borrowed. Add up every single bar — do not stop at the biggest one, and do not divide once you have added them all.
- (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) 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) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
- (d) Only families with strong feelings bothered to reply. — Method: a survey has non-response bias when only some of the people asked actually reply, and those who do are not a typical cross-section of everyone who was asked. Working: only 30 of the 200 families sent back their questionnaire, and 27 of those 30 — the great majority — said they were unhappy. Families who feel strongly about an issue, particularly those with a complaint, are far more likely to make the effort to reply than families who are simply satisfied and see no need to say anything, so the 30 replies over-represent unhappy families. Saying the families who replied were picked at random by the school gets the sampling the wrong way round: nobody picked them — they picked themselves by deciding to reply, and that is precisely why they are not a typical cross-section of all 200. Saying postal surveys always have low response rates restates that the response was low without explaining why a low response rate, on its own, makes a result unrepresentative — it is the reason FOR the low response, not the low response itself, that causes the bias here. Saying that the 27 unhappy replies show most families are unhappy is exactly the mistake the question is warning against: it treats the loudest 30 replies as if they stood for the other 170 who never sent theirs back. A low response rate is a warning sign only because the people who bother to reply are rarely typical of everyone who was asked.
- (c) 72 — Method: when a pie chart is divided into equal sectors, each sector stands for the same share of the people asked, so the fraction of the sectors that are shaded is also the fraction of the people. Working: 6 sectors out of 10 are purple, which is the fraction 6/10 of the whole pie chart; one tenth of the 120 people is 120 ÷ 10 = 12 people, so six tenths is 6 × 12 = 72 people. Answer: 72 people, a count of people rather than a number of sectors. The distractors: 48 comes from working with the 4 sectors that are not purple, 4 × 12, and so answering for the wrong part of the chart; 60 comes from turning the fraction 6/10 into 60% and then writing the 60 down as though it were a number of people; 6 comes from writing down the number of purple sectors instead of the number of people those sectors stand for.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (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.
- (a) No, 150 pupils are a tenth of the school, chosen at random — Method: judge a sample on two things, whether every member of the population had the same chance of being chosen, and whether the sample is large enough to carry a pattern. Working: the 150 pupils were drawn from the register of every pupil in the school, so no year group or set is shut out and no pupil chooses to take part; and 150 ÷ 1,500 = 0.1, so one pupil in ten has been asked. A random sample of that share is ample for an estimate of how long the school's pupils spend on homework. Answer: no, because 150 pupils are a tenth of the school and were chosen at random. The distractors: saying a random sample always gives the exact school figure reaches the same verdict for a reason that is false, since a second random sample of 150 would give a slightly different mean; saying 150 pupils cannot be picked at random from 1,500 treats randomness as something only a whole population can have, when drawing names from the register is exactly how a random sample is taken; saying that only asking all 1,500 could show anything rejects sampling altogether, which would leave no way to study any population too large to count.
- (b) Positive correlation — Method: the type of correlation is named from the direction the points take as the scatter graph is read from left to right. Working: the points rise from left to right, so as the arm span read on the horizontal axis increases, the height read on the vertical axis increases as well; two quantities that increase together show positive correlation. Answer: positive correlation. The distractors: negative correlation comes from naming the direction the wrong way round, since a negative correlation needs the points to fall as the graph is read from left to right; no correlation comes from treating points that are spread out rather than sitting exactly on a line as though they showed no relationship; direct proportion comes from confusing a rising trend with proportion, which would additionally need the line through the points to pass through the origin and would mean doubling one quantity doubles the other.
- (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.
- (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.
- (a) 12 mm — Method: the range is the highest value minus the lowest value. Working: the highest rainfall is 15 mm and the lowest is 3 mm, so the range is 15 − 3 = 12 mm. Giving 15 mm alone states the highest value, not the range. Giving 3 mm alone states the lowest value, not the range. Sorting the six values, 3, 5, 9, 11, 12 and 15, and averaging the middle two, (9 + 11) ÷ 2 = 10 mm, finds the median, a completely different statistic. The range always needs BOTH the highest and the lowest value — never just one of them.
- (c) The frequency of that category — Method: a bar chart for categorical data has one bar for each category, and the vertical scale on which the bars are measured is a count. Working: a bar drawn twice as tall as another tells you that twice as many items of data fell into its category, so the height measures how many items of data belong to that one category, which is exactly what a frequency is. Answer: the height of each bar is the frequency of that category — a count of items of data. The distractors: the number of different categories comes from reading the vertical scale as though it counted the bars, which is shown along the horizontal axis instead; the total of all the data comes from treating one bar as though it stood for the whole data set rather than for one category; the mean of all the data comes from confusing a bar chart with a measure of average, which no single bar can show.
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