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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Statistics worksheet — GCSE Foundation
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- 1.The masses of eight school bags, in kilograms, are 3, 4, 4, 5, 6, 7, 8 and 11. Work out the median mass.
- 2.A school posts a questionnaire about school dinners to the families of all 200 pupils. Only 30 families send theirs back, and 27 of those 30 say they are unhappy with school dinners. Give a reason why this result may not represent all 200 families.
- 3.Five friends have heights, in cm, of 150, 152, 155, 158 and 160. A sixth friend, with a height of 170 cm, joins the group. Write down what happens to the mean and the range of the heights once this sixth friend is included.
- 4.A factory makes 4,000 light bulbs a day. In a random sample of 80 of one day's bulbs, 3 were faulty. Work out an estimate for the number of faulty bulbs the factory makes in a day.
- 5.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
- 6.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 7.A shop recorded the number of books it sold on five days: 100, 40, 70, 20, 60. Work out the range of the numbers of books sold.
- 8.A pictogram shows the number of cars sold by a garage in Plymouth each month, where each whole car symbol represents 8 cars sold and a half symbol represents 4 cars. March shows 3 whole symbols and one half symbol. Work out how many cars were sold in March.
- 9.The number of pets owned by each of 19 pupils in a class is recorded: 0 pets — 7 pupils, 1 pet — 3 pupils, 2 pets — 4 pupils, 3 pets — 5 pupils. Work out the median number of pets.
- 10.Write down the statement that correctly describes the difference between correlation and causation.
- 11.A sports shop in Cardiff sold 40 pairs of football boots last month: 4 pairs of size 6, 5 pairs of size 7, 8 pairs of size 8, 13 pairs of size 9 and 10 pairs of size 10. The manager will order 40 pairs for next month and wants as many pairs as possible to be in a size customers will buy. Work out the mean size and the modal size, and write down which of the two he should use.
- 12.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
- 13.A café in York counts the number of customers in each of the nine hours it is open on one day: 4, 4, 4, 11, 13, 15, 18, 22 and 25. The owner says that a typical hour has about 4 customers, because 4 is the mode. Is the owner right? Give a reason for your answer.
- 14.A study found that people who drink more coffee tend to concentrate better at work. A coffee company says that this shows that drinking coffee improves concentration. Give the reason why this conclusion cannot be drawn.
- 15.A pie chart is divided into 10 equal sectors. 6 of those sectors stand for the people who chose purple. Altogether 120 people were asked. Work out how many of them chose purple.
Answer key
- (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (d) Only families with strong feelings bothered to reply. — Method: a survey has non-response bias when only some of the people asked actually reply, and those who do are not a typical cross-section of everyone who was asked. Working: only 30 of the 200 families sent back their questionnaire, and 27 of those 30 — the great majority — said they were unhappy. Families who feel strongly about an issue, particularly those with a complaint, are far more likely to make the effort to reply than families who are simply satisfied and see no need to say anything, so the 30 replies over-represent unhappy families. Saying the families who replied were picked at random by the school gets the sampling the wrong way round: nobody picked them — they picked themselves by deciding to reply, and that is precisely why they are not a typical cross-section of all 200. Saying postal surveys always have low response rates restates that the response was low without explaining why a low response rate, on its own, makes a result unrepresentative — it is the reason FOR the low response, not the low response itself, that causes the bias here. Saying that the 27 unhappy replies show most families are unhappy is exactly the mistake the question is warning against: it treats the loudest 30 replies as if they stood for the other 170 who never sent theirs back. A low response rate is a warning sign only because the people who bother to reply are rarely typical of everyone who was asked.
- (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) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (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.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (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.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (b) 1 — Method: for data in a frequency table, find the position of the median using (n + 1) ÷ 2, then read off the value at that position from the cumulative frequencies. Working: there are 19 pupils, so the median is the 10th value. The cumulative frequencies are 7 (up to 0 pets), 10 (up to 1 pet), 14 (up to 2 pets) and 19 (up to 3 pets). The 10th value falls at the end of the '1 pet' group, so the median is 1 pet. Giving 0 pets is the mode — the category with the highest frequency, 7 — not the median. Giving 3, the highest number of pets minus the lowest, finds the range, a different statistic entirely. Giving 19 states the total number of pupils, not a number of pets at all. Find the middle POSITION first, then read off the value it belongs to — do not confuse it with the mode, the range or the total.
- (d) Correlation is a link; causation is one causing the other — Method: the two words describe different claims — one is about a pattern in the data, the other is about what produced that pattern. Working: correlation says only that two quantities tend to change together, which is something a scatter graph can display; causation says that a change in one quantity actually brings about the change in the other, which needs evidence a scatter graph cannot supply, because a third quantity may be driving both. Answer: correlation is a link between the quantities, while causation is one quantity causing the change in another. The distractors: the statement giving causation as the link and correlation as the cause simply swaps the two words over; the statement that the words mean the same thing is the classic error of reading a correlation as proof of cause; the statement that a scatter graph shows causation but not correlation reverses what a scatter graph can do, since the pattern it displays is exactly the correlation.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (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.
- (c) No, the mode here is the lowest value of the nine — Method: an average is meant to stand for the data as a whole, so test any proposed average by asking how many values it sits near. Working: the value 4 appears three times and every other count appears once, so 4 is indeed the mode. But those three hours are the quiet ones at the start of the day, and the other six counts run from 11 up to 25; putting the nine counts in order, the middle one is the fifth, which is 13. So the mode sits at the very bottom of the data, with six of the nine hours far above it. Answer: no, because the mode here is the lowest value of the nine, so it describes the quiet opening hours rather than a typical hour. The distractors: saying the mode can only be used when no value repeats reverses the definition, since a mode exists only because a value does repeat; saying the mode is the value that occurs most often is a correct definition, but being the commonest value does not make a value typical when it lies at one end of the data; saying the mode is the best average for any list of numbers ignores the fact that mean, median and mode each describe a population well in different circumstances.
- (c) The data show a link only; a third factor may affect both — Method: a study of this kind measures two quantities and reports how they change together; deciding that one of them produces the other is a further claim, and it needs evidence that the measurements alone cannot give. Working: the study shows that more coffee goes with better concentration, which is a positive correlation; but a third factor that was never measured, such as how motivated someone is, could raise both the coffee drinking and the concentration, and the concentration could equally be what leads to the extra coffee. Answer: the data show a link only, because a third factor may be affecting both quantities, so no claim about cause can be made. The distractors: calling the conclusion safe because the correlation is positive treats the direction of a correlation as proof of cause, which no direction can give; calling it wrong because the correlation is negative misreads the direction of the relationship, since the study reports both quantities rising together; saying the two quantities are not linked denies the correlation the study actually found, when what fails is only the claim about cause.
- (c) 72 — Method: when a pie chart is divided into equal sectors, each sector stands for the same share of the people asked, so the fraction of the sectors that are shaded is also the fraction of the people. Working: 6 sectors out of 10 are purple, which is the fraction 6/10 of the whole pie chart; one tenth of the 120 people is 120 ÷ 10 = 12 people, so six tenths is 6 × 12 = 72 people. Answer: 72 people, a count of people rather than a number of sectors. The distractors: 48 comes from working with the 4 sectors that are not purple, 4 × 12, and so answering for the wrong part of the chart; 60 comes from turning the fraction 6/10 into 60% and then writing the 60 down as though it were a number of people; 6 comes from writing down the number of purple sectors instead of the number of people those sectors stand for.
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