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.The lowest temperature in a garden in Sheffield was recorded on each of seven days: 3 °C, 5 °C, 2 °C, 7 °C, 4 °C, 6 °C and 3 °C. Write down what the range of these seven temperatures describes.
- 2.A scatter graph shows the midday temperature, x °C, and the number of ice creams sold at a seaside kiosk, y. The line of best fit is y = 3x − 20. Give a reason why the y-intercept of this line of best fit is not a sensible estimate of the number of ice creams sold.y = 3x − 20
- 3.A scatter graph plots the number of years, x, that 20 employees have worked at a company against their salary, y. All the plotted points lie between x = 1 and x = 15. Write down the word used to describe an estimate for y made using a value of x that lies between 1 and 15.
- 4.A vet records the weights, in kilograms, of 7 dogs, in the order she weighs them: 18, 12, 50, 10, 20, 14, 16. Work out the median weight.
- 5.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.
- 6.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.
- 7.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.
- 8.Work out the median of these five numbers: 2, 4, 7, 12, 26
- 9.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.
- 10.Write down what a scatter graph is used to show.
- 11.A pupil writes this question for a school survey: “Do you agree that learning matters and that we should be set more homework?” Write down what is wrong with the survey question.
- 12.A sports centre in Ipswich has 2,000 members. It wants to know what its members think of its opening hours, so it asks a random sample of 100 of them. Write down what the population is in this survey.
- 13.A frequency polygon for the mass, in kg, of 40 parcels at a delivery depot is drawn by plotting one point at the midpoint of each class, joined by straight lines: (5, 6), (15, 10), (25, 16), (35, 6), (45, 2). Every class has a width of 10 kg. Write down the modal class.
- 14.Ben and Chloe each sat five maths tests. Ben's marks were 62, 64, 65, 66 and 68. Chloe's marks were 40, 52, 65, 78 and 90. Both pupils have a mean mark of 65. Their teacher says the mean on its own does not describe the two sets of marks well. Give a reason why the teacher is right.
- 15.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
Answer key
- (d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (b) Interpolation — The salary is being estimated for a value of x between 1 and 15, which is inside the range of x-values that were actually plotted, so this is interpolation. Extrapolation would apply if the estimate used a value of x below 1 or above 15, outside the plotted range. Correlation describes the relationship between the two variables, not the reliability of an estimate, and causation describes one variable actually causing a change in the other, which is a different idea altogether — neither is the word being asked for here.
- (c) 16 kg — Method: sort the seven weights before finding the middle value. Working: in order, the weights are 10, 12, 14, 16, 18, 20 and 50 kg. There are 7 values, so the median is the 4th one: 16 kg. Reading off the 4th weight in the order the vet recorded them, 10 kg, skips the sorting step and is not the median. Working out the mean, 140 ÷ 7 = 20 kg, finds a different average altogether. Working out the range, 50 − 10 = 40 kg, finds the spread, not the middle value. Always sort your data first — the median lives in the ordered list, not the collection order.
- (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.
- (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.
- (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) 7 — Method: the median is the middle value when the data are written in order of size, and with an odd number of values there is exactly one middle value. Working: the numbers are already in order, 2, 4, 7, 12, 26, and there are 5 of them, so the middle position is the third and the value sitting there is 7. Answer: 7, with two values below it and two above it. The distractors: 10.2 comes from working out the mean, 51 ÷ 5, instead of the median; 14 comes from taking the value halfway between the smallest and the largest, (2 + 26) ÷ 2; 24 comes from working out the range, 26 − 2, which measures spread rather than centre.
- (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.
- (a) The relationship between two variables — Method: what a diagram shows is decided by what has to be known before a single mark can be plotted on it. Working: every point on a scatter graph is plotted from a pair of measurements taken from the same person or object, one read on the horizontal axis and one on the vertical axis; having two measurements for each point is what makes it possible to look for a pattern between them, and the pattern between two variables is what the graph displays. Answer: a scatter graph shows the relationship between two variables. The distractors: the frequency of each single value is what a bar chart or a vertical line chart shows, and it needs only one list of values; how a total is shared between categories is what a pie chart shows; how one quantity changes over time is what a time series line graph shows, in which one of the two axes is always time.
- (d) It asks two things at once and invites agreement — Method: a survey question is faulty when a reply to it cannot be read as evidence about one single thing, so check how many claims it contains and whether its wording pushes the reader one way. Working: the question joins two separate claims, that learning matters and that more homework should be set, so a reply of yes could mean either of them or both and cannot be counted as evidence about homework; the opening words “Do you agree” also invite agreement instead of leaving the reader free to say no. Answer: it asks two things at once and invites agreement. The distractors: the reply calling it too short mistakes length for clarity, when the fault is that too much has been packed in rather than too little; the reply about long words is false, since every word in the question is an everyday one and the fault lies in what is being asked rather than in the vocabulary used to ask it; the reply that the question is fine takes a yes or no answer as proof that the question works, which is exactly what a double question defeats.
- (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) 20 kg ≤ mass < 30 kg — The modal class is the class with the highest frequency. Reading the plotted points, the frequencies are 6, 10, 16, 6 and 2, so the highest frequency is 16, plotted at the midpoint 25. A class of width 10 centred on 25 runs from 25 − 5 = 20 to 25 + 5 = 30, so the modal class is 20 kg ≤ mass < 30 kg. Writing '25 kg' gives only the midpoint, not the class — the modal class is an interval, not a single value. '10 kg ≤ mass < 20 kg' is the class before the peak, centred on 15, which has frequency 10, not the highest. '30 kg ≤ mass < 40 kg' is the class after the peak, centred on 35, which has frequency 6, not the highest.
- (a) Chloe's marks are far more spread out than Ben's — Method: a mean reports where a set of values sits, and two sets can sit in the same place while behaving quite differently, so a measure of spread has to be worked out as well. Working: Ben's marks add to 62 + 64 + 65 + 66 + 68 = 325 and 325 ÷ 5 = 65; Chloe's add to 40 + 52 + 65 + 78 + 90 = 325 and 325 ÷ 5 = 65, so the two means agree, as the question says. The ranges do not: Ben's is 68 − 62 = 6 marks, while Chloe's is 90 − 40 = 50 marks. Ben's five marks all sit within 3 marks of 65; Chloe's lowest is 25 marks below it and her highest 25 marks above it. Answer: Chloe's marks are far more spread out than Ben's, which is exactly what the mean cannot show. The distractors: saying Ben's marks are more spread out comes from subtracting in the order the values are written, 62 − 68 = −6 against 40 − 90 = −50, and then reading −6 as the larger spread; saying Chloe scored far more marks in total assumes a wider set of marks must add to more, when both totals are 325; saying the two sets vary by the same amount assumes that equal means force equal spread, when the two ranges are 6 and 50.
- (c) The median, £160,000, as one very high price lifts the mean — Method: find both averages, then choose the one that sits closer to the bulk of the data. Working: in order the prices are 140,000, 150,000, 160,000, 170,000 and 580,000, so the median is the third of the five, £160,000. For the mean, 140,000 + 150,000 + 160,000 + 170,000 + 580,000 = 1,200,000 and 1,200,000 ÷ 5 = 240,000, so the mean is £240,000. Four of the five houses sold for £170,000 or less, so a reader told that a typical price is £240,000 would expect to pay at least £70,000 more than any of those four cost. Answer: the median, £160,000, as one very high price lifts the mean. The distractors: £580,000 is the middle value of the list as it is printed, which is the median only when the values have first been put in order; £240,000 is the mean, chosen on the ground that a median ignores three of the five prices, but a median uses all five to find which one is central and is then untroubled by how extreme the outer values are; £155,000 comes from deleting the £580,000 house and taking the mean of what is left, since 140,000 + 150,000 + 160,000 + 170,000 = 620,000 and 620,000 ÷ 4 = 155,000, but a real sale may not be thrown away merely for being large.
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