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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- (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) 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.
- (d) Both the mean and the range increase. — The original mean is 150 + 152 + 155 + 158 + 160 = 775, and 775 ÷ 5 = 155 cm; the original range is 160 − 150 = 10 cm. Including the new height of 170 cm gives a new total of 775 + 170 = 945, and 945 ÷ 6 = 157.5 cm, which is higher than 155 cm, and a new range of 170 − 150 = 20 cm, which is higher than 10 cm, so both the mean and the range increase. Saying the range stays the same ignores that 170 cm is a new, higher maximum than the old 160 cm. Saying the mean stays the same ignores that 170 cm is above the original mean of 155 cm, which pulls the average up. Saying both decrease is the opposite of what happens here.
- (a) 23 minutes — Method: the equation of a line of best fit converts a value of x into a predicted value of y, so substitute the known number of pages for x and evaluate. Working: x is the number of pages, so put x = 10 into y = 2x + 3. Multiplication is carried out before addition, so 2 × 10 + 3 = 23. Answer: 23 minutes, and it is a prediction of the trend rather than a promise about any one chapter. The distractors: 20 minutes comes from working out 2 × 10 and stopping there, leaving out the 3 that the line adds; 26 minutes comes from reading the equation as y = 2(x + 3), adding first and then doubling, so 2 × 13 = 26; 13 minutes comes from adding 10 and 3 and never using the gradient at all, which treats the 2 as though it were not there.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (b) Tea — Method: the mode is the category with the highest frequency. Working: tea has a frequency of 8, which is higher than coffee's 7, hot chocolate's 4 and juice's 1. Mode = tea. Coffee has the second-highest frequency, 7, not the highest, so it is not the mode. Hot chocolate, with a frequency of 4, and juice, with a frequency of 1, both have lower frequencies still. Always compare every frequency in the table before naming the mode — the highest number wins, however close the next one is.
- (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.
- (a) 15 — Year 11 has 50 − 28 = 22 pupils in total. Of the 22 pupils who walk in total, 15 are in Year 10, so 22 − 15 = 7 Year 11 pupils walk. Subtracting that from the Year 11 total gives 22 − 7 = 15 Year 11 pupils who are driven. Choosing 28 takes the whole school's driven total, 50 − 22 = 28, and treats it as if it were Year 11's alone, without separating the year groups. Choosing 7 correctly finds how many Year 11 pupils walk but stops there, giving that figure instead of the number who are driven. Choosing 35 comes from 50 − 15, subtracting the Year 10 walkers from the whole school total rather than working within Year 11.
- (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.
- (a) The median, as the one very large wage does not move it — Method: an average describes a population well when it sits close to most of the values, so compare what each average does when one value lies far from the rest. Working: in order the wages are 420, 440, 460, 480 and 1,500, so the median is the third of the five, £460. The mean uses every wage: 420 + 440 + 460 + 480 + 1,500 = 3,300 and 3,300 ÷ 5 = 660, so the mean is £660. Four of the five people earn less than £660, and the nearest of those four wages is £180 below it, so £660 describes nobody at the garage; £460 sits inside the group of four similar wages. Answer: the median, as the one very large wage does not move it, while that same wage drags the mean £200 above the median. The distractors: saying the median is always larger than the mean is an invented rule, and here the median £460 is smaller than the mean £660; saying the mean is the only average that uses all five wages is true as far as it goes, but using a value and being dragged by it are the same thing when that value is £1,500; saying £660 lies between the smallest and largest wage is true of every mean ever calculated, so it proves nothing about whether this one is typical.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (d) 300 pupils — Method: an estimate for a whole population is made by finding the proportion in the sample and applying that same proportion to the population. Working: in the sample 30 of the 50 pupils prefer sport, a proportion of 30 ÷ 50 = 0.6, and applying that proportion to the school gives 0.6 × 500 = 300 pupils. Answer: 300 pupils, and it is only an estimate, because a different random sample of 50 would give a slightly different figure. The distractors: 200 pupils comes from scaling up the 20 pupils in the sample who did not prefer sport, 20 × 10, which answers the opposite question; 150 pupils comes from reading 30 out of 50 as 30% and taking 30% of 500; 60 pupils comes from working out the proportion correctly as 60% and then writing the 60 down as a number of pupils instead of applying it to the 500.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
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