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
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- 1.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.
- 2.In a spelling test the 20 pupils in Group A had a mean mark of 80, and the 30 pupils in Group B had a mean mark of 70. Work out the mean mark of all 50 pupils.
- 3.Work out the median of these five numbers: 2, 4, 7, 12, 26
- 4.A shop sold seven pairs of shoes in these sizes: 4, 4, 5, 6, 6, 6, 9. Write down the modal size.
- 5.A two-way table records the favourite subject, Maths or Art, of 60 pupils in Year 10, and whether each pupil is left-handed or right-handed. 9 of the 60 pupils are left-handed, and 6 of those left-handed pupils prefer Art. In total, 24 of the 60 pupils prefer Art. A pupil is chosen at random from the 60. Work out the probability that the pupil is right-handed and prefers Art.
- 6.Priya wants to find out the favourite sport of the 900 pupils at her school. She asks 5 pupils, chosen at random. Give a reason why her sample may not give a reliable result.
- 7.A scatter graph shows the number of hours of sunshine, x, and the number of visitors, y, at an outdoor swimming pool in Torquay on each of 15 days. The line of best fit passes through the points (5, 150) and (15, 350). Work out the estimated number of visitors on a day with 8 hours of sunshine, using the line of best fit.
- 8.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.
- 9.For each of 30 fires in one county, the number of fire engines sent and the cost of the damage were recorded. The scatter graph of the data shows strong positive correlation. A newspaper prints the headline "Sending more fire engines causes more damage". Is the newspaper right? Give a reason for your answer.
- 10.A survey of 25 pupils in Derby records how many siblings each has: 0 siblings — 6 pupils, 1 sibling — 10 pupils, 2 siblings — 6 pupils, 3 siblings — 3 pupils. Calculate the mean number of siblings.
- 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.A shop records the colour of every car that uses its car park one Tuesday. Its frequency table reads: black 44, silver 37, blue 26, red 18, white 12. Write down the frequency of black cars.
- 13.A school has 1200 pupils. A teacher wants to take a random sample of 60 of them. Write down which of these methods gives a random sample.
- 14.Write down the statement that correctly describes the difference between correlation and causation.
- 15.A stacked bar for one day at a café in Norwich shows total drink sales of 50 drinks, split into three parts: 22 were tea, 15 were coffee and the rest were hot chocolate. Work out the number of hot chocolates sold.
Answer key
- (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.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (c) 7 — Method: the median is the middle value when the data are written in order of size, and with an odd number of values there is exactly one middle value. Working: the numbers are already in order, 2, 4, 7, 12, 26, and there are 5 of them, so the middle position is the third and the value sitting there is 7. Answer: 7, with two values below it and two above it. The distractors: 10.2 comes from working out the mean, 51 ÷ 5, instead of the median; 14 comes from taking the value halfway between the smallest and the largest, (2 + 26) ÷ 2; 24 comes from working out the range, 26 − 2, which measures spread rather than centre.
- (d) 6 — Method: the mode, or modal value, is the value that occurs most often in the data set, and it is a value from the data rather than a count. Working: size 4 occurs twice, size 5 occurs once, size 6 occurs three times and size 9 occurs once, so the highest frequency is three and the size it belongs to is 6. Answer: 6. The distractors: 3 comes from writing down the frequency of the most common size instead of the size itself; 9 comes from picking the largest size in the list, which confuses the mode with the maximum; 4 comes from stopping at the first size that repeats rather than checking which size repeats most often.
- (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.
- (a) 5 pupils are far too few to represent 900 pupils — Method: a sample can only support a claim about a population if it is chosen fairly and if it is large enough for the pattern in it to be more than chance. Working: Priya's method of choosing is fair, because the 5 pupils were picked at random, so every pupil had the same chance of being asked. The difficulty is the size: 900 ÷ 5 = 180, so each pupil she asks stands for 180 pupils. If two of the five happen to play in the same netball team, netball takes 40% of her sample on the strength of two answers, and a second sample of 5 could easily give a different favourite sport. Answer: 5 pupils are far too few to represent 900 pupils. The distractors: saying the pupils were not chosen at random contradicts the question, which states that they were; saying the 5 may each name a different sport describes what often happens in a small sample, but disagreement is not the fault, since 5 pupils who all named the same sport would be just as weak a basis for a claim about 900; saying a sample must hold at least half of the population is an invented rule, and a properly chosen sample of a few hundred can describe a population of many thousands.
- (d) 210 — 350 − 150 = 200. 200 ÷ 10 = 20, so the gradient is 20. Using the point (5, 150): 20 × 5 = 100, so 150 − 100 = 50 is the intercept, giving the line y = 20x + 50. At x = 8: 20 × 8 = 160, and 160 + 50 = 210, so the estimated number of visitors is 210. Choosing 160 stops after 20 × 8 = 160 and forgets to add the intercept of 50. Choosing 250 comes from averaging the two given y-values: 150 + 350 = 500, and 500 ÷ 2 = 250, instead of using the line's equation. Choosing 240 assumes the visitors are directly proportional to the hours of sunshine using the first point, 150 × 8 ÷ 5 = 240, which ignores that the line does not pass through the origin.
- (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.
- (a) No, the size of the fire affects both of the quantities — Method: correlation says that two quantities change together; a claim that one of them produces the other is a further claim, and it needs evidence that a scatter graph on its own cannot give. Working: the graph does show strong positive correlation, so more engines did go with greater damage. But neither quantity was set by the researchers: both were decided by how large the fire was. A large blaze brings many appliances and also destroys a great deal, while a small one brings few and destroys little, so a third quantity is driving both of the recorded ones. Answer: no, because the size of the fire affects both of the quantities. The distractors: saying the correlation is negative contradicts the graph, which shows the two quantities rising together, and reaching the right verdict from a false reading of the data is not the reason the mark is for; saying that strong positive correlation shows one quantity causes the other is the assumption the question exists to test, and no strength of correlation can establish cause; saying the points lie close to the line of best fit describes how strong the correlation is, and strength and cause are different matters entirely.
- (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.
- (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.
- (d) 44 — Method: the frequency of a category in a frequency table is the number of times that category was counted, and it is read from the row for that category. Working: the rows of the table pair each colour with its count, and the row for black is paired with the count 44, so the frequency of black cars is 44. Answer: 44 cars — a frequency is a count of cars, not a colour and not a percentage. The distractors: 37 comes from reading the count paired with silver, that is from reading the wrong row of the table; 137 comes from adding every count in the table, 44 + 37 + 26 + 18 + 12, which gives the total number of cars rather than the frequency of one colour; 5 comes from counting how many different colours the table lists instead of how many cars were black.
- (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.
- (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.
- (b) 13 — The total is 50, and the two known parts are 22 (tea) and 15 (coffee), so 50 − 22 − 15 = 13 hot chocolates. Choosing 28 comes from 50 − 22, subtracting only the tea and forgetting the coffee. Choosing 35 comes from 50 − 15, subtracting only the coffee and forgetting the tea. Choosing 37 comes from 22 + 15, which finds how many drinks were tea or coffee, not the number left over for hot chocolate.
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