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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Answer key: Statistics worksheet — GCSE Foundation
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- (b) Ethan, 10 seconds — Method: over the same distance the fastest runner is the one who takes the least time, so the smallest time in the table is found first and the name is then read from the same row. Working: the four times are 12 seconds, 15 seconds, 10 seconds and 14 seconds; in order of size these are 10, 12, 14 and 15, so the least time is 10 seconds, and the row holding 10 seconds is the row for Ethan. Answer: Ethan, 10 seconds — the time is in seconds, and a smaller time means a faster runner. The distractors: Grace with 15 seconds comes from taking the largest number in the table to mean the fastest runner, which reverses the relationship between time and speed over a fixed distance; Oliver with 12 seconds comes from writing down the first row of the table without comparing the four times; Ethan with 15 seconds comes from identifying the right runner but then reading the time from a different row of the table.
- (c) 156 cm — Method: to combine two groups' means, multiply each group's mean by its own number of pupils, add the two totals together, then divide by the total number of pupils in both groups. Working: 20 × 150 = 3,000 cm for the boys and 10 × 168 = 1,680 cm for the girls, giving a combined total of 3,000 + 1,680 = 4,680 cm. Dividing by all 30 pupils gives 4,680 ÷ 30 = 156 cm. Giving 159 cm averages the two means, (150 + 168) ÷ 2, treating the two groups as if they had the same number of pupils, when there are twice as many boys as girls. Giving 4,680 cm finds the correct combined total height but stops there, forgetting the final division by the 30 pupils. Giving 234 cm divides the combined total by 20, the number of boys only, forgetting that the total also includes the 10 girls. Always weight each mean by its own group size, and always divide by the TOTAL number of pupils in both groups combined.
- (a) Priya — A line of best fit should be drawn so that the plotted points are roughly balanced above and below it. Work out the difference between the two counts for each pupil: Amir 10 − 2 = 8; Kofi 10 − 2 = 8 (10 below and 2 above); Leah 12 − 0 = 12; Priya 6 − 6 = 0. Priya's line has the smallest difference, an exact balance of 6 above and 6 below, so her line is drawn correctly. Amir's line has 10 of the 12 points above it, so it is drawn too low. Kofi's line has 10 of the 12 points below it, so it is drawn too high. Leah's line has every single point below it, so it is not a line of best fit at all.
- (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.
- (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
- (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) 80 — Method: the range is a measure of spread and is found by subtracting the smallest value from the largest. Working: the largest number sold is 100 and the smallest is 20, so the range is 100 − 20 = 80. Answer: 80. The distractors: 100 comes from writing down the largest value and never subtracting the smallest; 60 comes from working out the median, the middle value of 20, 40, 60, 70, 100, instead of the range; 58 comes from working out the mean, 290 ÷ 5, which measures centre rather than spread.
- (b) At 0 guests, the model predicts 2 m of table — The y-intercept of a line of best fit y = mx + c is the value of y when x = 0. Here y = 0.5 × 0 + 2 = 2, so the line predicts a table length of 2 m when there are 0 guests. The 2 m does not grow as more guests arrive — that role belongs to the gradient, 0.5 — so an option saying each extra guest adds 2 m has swapped the two numbers around. The 2 is a length in metres, not a number of guests, so an option requiring 2 guests before set-up has misread its units. And the table length does change with x, since it is 0.5x + 2 and not a fixed value, so an option claiming the table is always 2 m ignores the 0.5x term completely.
- (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.
- (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.
- (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.
- (d) 1/2 — Method: a percentage is turned into a fraction by writing it over 100 and then cancelling the fraction down to its simplest form. Working: the reading sector is 50% of the pie chart, so as a fraction it is 50/100; dividing the numerator and the denominator by 50 gives 1/2. Answer: 1/2 of the pie chart — a fraction of the chart, not a number of pupils. The distractors: 1/4 comes from reading the share of the sport sector, 25%, instead of the share of the reading sector; 1/3 comes from counting the three sectors and assuming that three sectors must each take a third of the chart, which is only true when the sectors are equal; 3/4 comes from adding the reading and sport sectors together, 50% + 25% = 75%, instead of taking the reading sector on its own.
- (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) 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.
- (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.
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