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.
Answer key: Statistics worksheet — GCSE Foundation
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- (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.
- (a) 3/10 — There are 60 − 9 = 51 right-handed pupils. Of the 24 pupils who prefer Art, 6 are left-handed, so 24 − 6 = 18 are right-handed and prefer Art. The probability that a randomly chosen pupil is right-handed and prefers Art is 18/60, which simplifies to 3/10. Giving 2/5 is 24/60 simplified — the probability of preferring Art, ignoring the right-handed condition entirely. Giving 17/20 is 51/60 simplified — the probability of being right-handed, ignoring the Art condition entirely. Giving 1/10 is 6/60 simplified — the probability of being left-handed and preferring Art, the wrong hand condition.
- (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.
- (d) 20 — 0.4x + 1 = 9, so 0.4x = 9 − 1 = 8, and 8 ÷ 0.4 = 20, so 20 sessions are needed. Choosing 23 divides 9 by 0.4 without first subtracting the 1: 9 ÷ 0.4 = 22.5, rounded up to 23. Choosing 25 subtracts the wrong way, adding the 1 instead of taking it away: 9 + 1 = 10, and 10 ÷ 0.4 = 25. Choosing 2 misplaces the decimal point in the gradient, dividing by 4 instead of by 0.4: 8 ÷ 4 = 2.
- (d) 24 — Method: for a sample in proportion to the population, apply the same fraction that each group makes up of the whole population to the size of the sample. Working: women make up 300 out of the 500 members, a fraction of 300 ÷ 500 = 0.6. Applying that fraction to the sample of 40 gives 0.6 × 40 = 24 women. Splitting the sample evenly, 40 ÷ 2 = 20, ignores that the club has more women than men and treats the two groups as equal in size, which they are not. Misreading the sample size as 50 instead of 40, then applying the 3:2 ratio of women to men, 3 ÷ 5 × 50 = 30, uses the right ratio but the wrong sample total. Working out the number of MEN instead of women, 200 ÷ 500 × 40 = 16, answers a different question — how many men, not how many women, belong in the sample. Always apply each group's own share of the population to the sample size, and check which group the question is actually asking about.
- (c) 10 — Method: work out the mean that can be found straight away, then use total = mean × number of values on the group of three to find the missing number. Working: the mean of 15 and 25 is (15 + 25) ÷ 2 = 40 ÷ 2 = 20, so the group of three must also have a mean of 20; three numbers with a mean of 20 have a total of 20 × 3 = 60, and 20 + 30 = 50 of that total is already accounted for, so x = 60 − 50 = 10. Answer: 10, and checking, (10 + 20 + 30) ÷ 3 = 20. The distractors: 20 comes from working out the mean the two groups share and writing that down as x; −10 comes from dividing the group of three by 2 instead of by 3, which gives x + 50 = 40; 70 comes from reading the total 15 + 25 = 40 as the mean of the pair, which sets the target total at 120 and leaves x = 70.
- (a) The temperature is rising steadily — Method: the trend of a line graph is the overall direction of the readings as time goes on, found by comparing each reading with the one before it. Working: from 10 °C to 14 °C is a rise of 4 °C, and the same comparison from 14 °C to 18 °C, from 18 °C to 22 °C, from 22 °C to 26 °C and from 26 °C to 30 °C gives a rise of 4 °C every time; every reading is greater than the one before it and none of them falls. Answer: the temperature is rising steadily — steadily because the rise is the same size each hour. The distractors: falling steadily comes from reading the six values from right to left, which reverses the direction of time; stays the same comes from noticing that the step of 4 °C is the same each hour and describing the step as constant instead of the temperature; rises and then falls comes from assuming that a line graph has to turn at some point rather than reading the values that are actually given.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (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) A vertical line chart (discrete numerical data) — The number of pets is discrete numerical data — whole-number values such as 0, 1, 2, 3 or 4 — recorded for one variable, so a vertical line chart is the chart specified for this kind of data. A bar chart is used for categorical data, such as favourite colour, not numerical values counted like this. A pie chart shows proportions of a whole and does not show the frequency of each separate value. A scatter graph compares two different variables against each other, and only one variable, the number of pets, is recorded here.
- (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.
- (c) Yes — with an estimate of 320, above the 250 limit. — Method: scale the sample proportion up to the whole batch to get an estimate, then compare that estimate with the 250 limit to reach a decision. Working: in the sample, 4 out of 50 boards are faulty, a proportion of 4 ÷ 50 = 0.08. Applying that proportion to the batch of 4,000 gives an estimate of 0.08 × 4000 = 320 faulty boards. Since 320 is more than 250, the factory should scrap the batch. Inverting the proportion, 50 ÷ 4 = 12.5, and treating that as a percentage of the batch, 12.5% × 4000 = 500, still gives 'yes' but from the wrong fraction, so it overstates the estimate. Comparing the raw number of faulty boards found in the sample, 4, directly with the 250 limit skips the scaling up to the batch altogether, and 4 is nowhere near 250, so that route wrongly says 'no'. Dividing the batch by the sample size, 4000 ÷ 50 = 80, finds how many samples of 50 fit into the batch but stops before multiplying by the 4 faulty boards found, so it also wrongly says 'no'. Always find the proportion in the sample first, scale it up to the whole batch, and only then compare the estimate with the limit given.
- (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) 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.
- (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.
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