Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher 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 Higher
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- (d) 18 minutes — Method: the lower quartile is the 80 ÷ 4 = 20th value and the upper quartile is the 3 × 80 ÷ 4 = 60th value; locate each inside its class by linear interpolation, then subtract. Working: the 20th value lies between the running totals 8 and 28, so it is in the class 10 ≤ t < 20, which holds 20 journeys across 10 minutes, and it is the 20 − 8 = 12th of them, giving 10 + (12 ÷ 20) × 10 = 16 minutes; the 60th value lies between the running totals 52 and 72, so it is in the class 30 ≤ t < 40, which also holds 20 journeys across 10 minutes, and it is the 60 − 52 = 8th of them, giving 30 + (8 ÷ 20) × 10 = 34 minutes; subtracting, 34 − 16 = 18. Answer: an estimated interquartile range of 18 minutes. The distractors: 20 minutes comes from taking the lower boundaries of the two quartile classes, 30 − 10, which locates the classes but never the values inside them; 40 minutes comes from subtracting the two positions, 60 − 20, instead of the two times; 22 minutes comes from interpolating downwards from each upper boundary rather than upwards from each lower boundary, giving 20 − 6 = 14 and 40 − 4 = 36.
- (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (c) 72° — The angle is 12 ÷ 60 × 360 = 72°. Choosing 150° divides football's frequency of 25 instead of badminton's 12: 25 ÷ 60 × 360 = 150. Choosing 20° finds badminton as a percentage of the members, 12 ÷ 60 × 100 = 20, rather than an angle in degrees. Choosing 90° uses 48, the total of the other three activities, as the total instead of the full 60 members: 12 ÷ 48 × 360 = 90.
- (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.
- (b) £11.00 — 1.5 × 6 = 9, and 9 + 2 = 11, so the estimated cost is £11.00. Choosing £9.00 stops after 1.5 × 6 = 9 and forgets to add the £2. Choosing £12.00 adds the mass and the constant first and then multiplies: 6 + 2 = 8, and 8 × 1.5 = 12.00. Choosing £13.50 swaps the gradient and the intercept, using y = 2x + 1.5 instead: 2 × 6 = 12, and 12 + 1.5 = 13.50.
- (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.
- (a) 49 — Method: add the frequencies of the classes that lie wholly below 35 seconds, then use linear interpolation for the class that 35 cuts through, assuming the calls in that class are spread evenly across it. Working: below 30 seconds there are 15 + 24 = 39 calls; the value 35 lies in the class 30 ≤ t < 50, which is 20 seconds wide and holds 40 calls, and 35 is 35 − 30 = 5 seconds into it, so the estimated share is (5 ÷ 20) × 40 = 10 calls; the estimate is 39 + 10 = 49. Answer: about 49 calls met the target. The distractors: 79 comes from adding the whole of the class 30 ≤ t < 50, 39 + 40, and so counting calls of up to 50 seconds as being under 35; 39 comes from stopping at the class boundary 30 and ignoring the part class altogether; 69 comes from measuring the part of the class from 35 up to 50 instead of from 30 up to 35, giving (15 ÷ 20) × 40 = 30 and then 39 + 30.
- (a) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (b) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (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) 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.
- (b) £1,000 — Wages take up 150° out of 360°, so the amount spent on wages is 150 ÷ 360 × 2400 = £1,000. Choosing £600 uses the repairs angle, 90°, instead of the wages angle: 90 ÷ 360 × 2400 = 600. Choosing £3,600 treats the angle in degrees as if it were a percentage, 150 ÷ 100 × 2400 = 3600, instead of dividing by 360°. Choosing £800 uses the angle for the 'other costs' sector, 360 − 90 − 150 = 120°, instead of the wages sector: 120 ÷ 360 × 2400 = 800.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (c) 40 minutes — Method: for grouped data, estimate the mean using the midpoint of each class — multiply each midpoint by its frequency, add the results, then divide by the total frequency. Working: the midpoints are 10, 30, 50 and 70 minutes. 10 × 5 = 50. 30 × 10 = 300. 50 × 10 = 500. 70 × 5 = 350. Σfx = 50 + 300 + 500 + 350 = 1200. Σf = 5 + 10 + 10 + 5 = 30. Estimated mean = 1200 ÷ 30 = 40 minutes. Using the upper boundary of each class instead of the midpoint — 20 × 5 = 100, 40 × 10 = 400, 60 × 10 = 600, 80 × 5 = 400 — gives a total of 1500 and an estimate of 1500 ÷ 30 = 50 minutes, too high because a boundary is not the middle of the class. Averaging the frequencies themselves, 5, 10, 10 and 5, ignores the times altogether and gives 7.5. Stopping after Σfx = 1200 without dividing by the total frequency gives a number far too large to be a time in minutes. Always find the midpoint of each class before multiplying by the frequency, and always divide by Σf at the end.
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