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.
Statistics worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- 1.The numbers 2, 4, 6, 8, 10 and 12 have a mean of 7 and a median of 7. The value 100 is now added to the list. Which measure is changed more by adding 100, and why?
- 2.A charity shop in Bath holds 2,000 books. Volunteer A checks a random sample of 50 books and finds 35 paperbacks. Volunteer B checks a different random sample of 50 books and finds 31 paperbacks. Work out the estimate each sample gives for the whole stock, and write down what the shop should do next.
- 3.There are 10,000 pupils in a city. A researcher picks a sample of 500 of them by drawing names at random from a list of every pupil in the city. Give the reason why this method gives a representative sample.
- 4.A scatter graph shows the number of years of experience, x, of 18 sales assistants and their monthly sales, y hundred pounds. The plotted points run from x = 1 to x = 12 years, and the line of best fit is y = 4x + 20. A new assistant has 25 years of experience. Use the line of best fit to estimate a value of y for this assistant, and decide whether the estimate would be reliable.y = 4x + 20
- 5.The marks scored by four pupils in a quiz were 8, 8, 8, 8. Work out the mean, the median and the mode of these marks.
- 6.The daily high temperatures, in °C, in Leeds over five days were 14, 16, 18, 19 and 23. In York over the same five days they were 15, 17, 17, 18 and 18. Write a sentence comparing Leeds and York using both the median and the range.
- 7.A charity shop in Leicester made £2,400 in one month. A pie chart shows how this was spent: repairs took up an angle of 90°, wages took up an angle of 150°, and the rest was other costs. Work out how much was spent on wages.
- 8.A sports centre in Ipswich has 2,000 members. It wants to know what its members think of its opening hours, so it asks a random sample of 100 of them. Write down what the population is in this survey.
- 9.Ben and Chloe each sat five maths tests. Ben's marks were 62, 64, 65, 66 and 68. Chloe's marks were 40, 52, 65, 78 and 90. Both pupils have a mean mark of 65. Their teacher says the mean on its own does not describe the two sets of marks well. Give a reason why the teacher is right.
- 10.A random sample of 50 pupils at a school were asked whether they prefer sport to music. 30 of the 50 pupils said they prefer sport. The school has 500 pupils altogether. Work out an estimate for the number of the 500 pupils who prefer sport.
- 11.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.
- 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 quality inspector weighs a random sample of 50 packets of crisps from one day's production and finds their mean mass is 32.4 g. The factory makes 20,000 packets that day. Work out an estimate for the total mass, in kg, of all the packets made that day.
- 14.A garden centre's sales, in thousands of pounds, at the end of each quarter last year were: quarter 1 — 18, quarter 2 — 34, quarter 3 — 30, quarter 4 — 22. Work out the increase in sales from quarter 1 to the quarter with the highest sales.
- 15.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
Answer key
- (c) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (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) 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.
- (c) mean = 8, median = 8, mode = 8 — Method: work out each measure separately — the mean is the total divided by how many values there are, the median is the middle value once the data are in order, and the mode is the value that occurs most often. Working: the total is 8 + 8 + 8 + 8 = 32 and there are 4 marks, so the mean is 32 ÷ 4 = 8; in order the marks read 8, 8, 8, 8, and the mean of the middle pair is (8 + 8) ÷ 2 = 8; the value 8 occurs 4 times and no other value occurs at all, so the mode is 8. Answer: mean = 8, median = 8, mode = 8 — when every value in a data set is the same, all three measures of central tendency take that value. The distractors: a mean of 32 comes from stopping at the total and never dividing by 4; a mode of 4 comes from writing down how many times 8 occurs instead of the value that occurs; a mean of 2 comes from dividing a single value, 8, by the 4 marks instead of dividing the total by 4.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
- (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) All 2,000 members of the sports centre. — Method: in a survey, the population is the whole group the survey is trying to find out about, and the sample is the smaller group actually asked. Working: this survey wants to know what the sports centre's members think, so the population is every one of the 2,000 members — whether or not they were personally asked. Saying the population is the 100 members who were asked names the sample, not the population; the sample is drawn FROM the population, so it is smaller than it, not the same as it. Saying the population is everybody who lives in Ipswich widens the group far beyond who the survey is actually about — plenty of Ipswich residents are not members of the sports centre at all, so they are outside this survey altogether. Saying the population is the members who say they are unhappy confuses the population with a result of the survey: whether a member turns out to be happy or unhappy is something the survey finds out, not part of the definition of who is being studied. The population is always the whole group the question is about, before any sampling or any results come in.
- (a) Chloe's marks are far more spread out than Ben's — Method: a mean reports where a set of values sits, and two sets can sit in the same place while behaving quite differently, so a measure of spread has to be worked out as well. Working: Ben's marks add to 62 + 64 + 65 + 66 + 68 = 325 and 325 ÷ 5 = 65; Chloe's add to 40 + 52 + 65 + 78 + 90 = 325 and 325 ÷ 5 = 65, so the two means agree, as the question says. The ranges do not: Ben's is 68 − 62 = 6 marks, while Chloe's is 90 − 40 = 50 marks. Ben's five marks all sit within 3 marks of 65; Chloe's lowest is 25 marks below it and her highest 25 marks above it. Answer: Chloe's marks are far more spread out than Ben's, which is exactly what the mean cannot show. The distractors: saying Ben's marks are more spread out comes from subtracting in the order the values are written, 62 − 68 = −6 against 40 − 90 = −50, and then reading −6 as the larger spread; saying Chloe scored far more marks in total assumes a wider set of marks must add to more, when both totals are 325; saying the two sets vary by the same amount assumes that equal means force equal spread, when the two ranges are 6 and 50.
- (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.
- (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) 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.
- (d) 648 kg — Method: to estimate a total from a sample, multiply the sample's mean by the number of items in the whole population, then check the units the question asks for. Working: 32.4 g × 20,000 = 648,000 g. Converting to kilograms, 648,000 ÷ 1,000 = 648 kg. This is only an estimate, not an exact total, because it assumes every one of the 20,000 packets has exactly the sample mean mass, when in reality individual packets vary above and below it. Giving 1.62 kg multiplies the mean by 50, the SAMPLE size, instead of by 20,000, the number of packets actually made that day — this finds the total mass of the 50 sampled packets, not the day's production. Giving 32.4 kg treats the sample mean itself, in grams, as if it already were the day's total mass in kilograms, skipping the scaling up altogether. Giving 648,000 kg correctly scales the mean up to the whole day's production but never converts the answer from grams to kilograms, leaving it 1,000 times too large. Always scale a sample's mean up by the SIZE OF THE WHOLE POPULATION, and always finish by checking the units the question asks for.
- (a) £16,000 — Method: first find the quarter with the highest sales figure, then subtract quarter 1's sales from it — remembering that every figure is given in THOUSANDS of pounds. Working: the highest sales figure is quarter 2, at £34,000 (34 thousand pounds). The increase from quarter 1 is £34,000 − £18,000 = £16,000. Giving £34,000 reads off the highest sales figure on its own, without subtracting quarter 1's sales — that is the highest quarter's total, not the increase. Giving £12,000 uses quarter 3's sales, 30, the SECOND-highest figure, instead of quarter 2's 34, the actual highest — 30 − 18 = 12, but quarter 3 is not the quarter with the highest sales. Giving £16 gets the subtraction right, 34 − 18 = 16, but forgets that every figure in the question is in thousands of pounds, so the increase is £16,000, not £16. Always identify the correct quarter FIRST, and always check the units the numbers are given in before writing your final answer.
- (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.
Build your own mix at the worksheet builder.