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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- (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) 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.
- (d) 25% — First find the number of fruit cakes: 80 − 34 − 26 = 20. Then write this as a percentage of the total: 20 ÷ 80 × 100 = 25%. Giving 20% comes from reporting the count of fruit cakes, 20, directly as a percentage, without dividing by the total of 80 first. Giving 32.5% computes the percentage of chocolate cakes instead of fruit cakes: 26 ÷ 80 × 100 = 32.5%. Giving 42.5% computes the percentage of sponge cakes instead of fruit cakes: 34 ÷ 80 × 100 = 42.5%.
- (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.
- (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.
- (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.
- (d) The modal class, as the class with most pupils is shown — Method: a grouped frequency table records how many values fall into each class, but not the values themselves, so any average that needs the individual times can only be estimated from it. Working: the four frequencies are 8, 12, 6 and 4, and 8 + 12 + 6 + 4 = 30, so every pupil is counted. The largest frequency is 12, which belongs to the class 10 < t ≤ 20, and that class can be written down exactly, because finding it needs nothing but the counts the table already gives. Answer: the modal class, as the class with most pupils is shown. The distractors: the mean is said to use all 30 times, but the table does not hold them; the usual method replaces each class by its midpoint, 5, 15, 25 and 35, which gives an estimate of the mean and not its true value; the median is said to be shown, but the table locates only the class holding the 15th and 16th times, which is 10 < t ≤ 20, without saying what either time was; the range is said to be shown, but 0 and 40 are the boundaries of the first and last classes, not the fastest and slowest times actually recorded.
- (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.
- (c) It is not representative, as she picked her own friends — Method: judge a sample by asking whether the pupils in it were chosen in a way that gives the whole school a fair chance of being heard. Working: Isla's friends are a group she formed herself, and friends tend to share tastes, so their favourite programme is likely to match hers rather than the school's, and pupils in other year groups and other friendship groups had no chance at all of being asked; the fault lies in how the pupils were selected, not in how many of them there were. Answer: it is not representative, as she picked her own friends. The distractors: the reply blaming the size claims the pupils were picked at random, which is false here, and it is the common mistake of thinking a biased sample can be cured by making it bigger; the reply calling the sample too large is false in the other direction, as a survey is never spoilt by collecting more replies; the reply that the sample is fine treats attending the school as enough, which would make any group of pupils in the building a fair sample.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (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.
- (b) 120, unreliable — x = 25 is outside 1 to 12 — The line of best fit is y = 4x + 20. 4 × 25 = 100, and 100 + 20 = 120, so the estimate is y = 120. But x = 25 lies far outside the plotted range of 1 to 12 years, so this is an extrapolation, and the estimate is not reliable. Reaching 100 instead of 120 comes from 4 × 25 = 100 with the intercept of 20 left out — still correctly flagged as unreliable, but the wrong value. Calling the estimate reliable simply because it was calculated correctly, giving 120, wrongly assumes that a correct calculation is automatically trustworthy, ignoring that x = 25 lies far beyond the data actually collected. Reaching 68, from 4 × 12 = 48 and 48 + 20 = 68, substitutes x = 12, the top of the plotted range, instead of the assistant's actual x = 25, and wrongly calls that reliable because 12 lies inside the range.
- (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.
- (b) A line graph — Sales recorded at the end of each of the twelve months are time series data, and a line graph is the chart built to show how a value changes over time, with the points usually joined in order. A pie chart is for showing categorical data as shares of a whole, not a trend over time. A pictogram shows a frequency for separate categories using symbols, not a continuous trend. A scatter graph is for comparing two different variables against each other, not one variable over time.
- (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.
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