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.On a scatter graph of the arm span and the height of some pupils, the points rise from left to right. Write down the type of correlation this shows.
- 2.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.
- 3.A vet records the masses, m kg, of 30 dogs at a clinic in Preston: 0 < m ≤ 10 — 11 dogs, 10 < m ≤ 20 — 5 dogs, 20 < m ≤ 30 — 5 dogs, 30 < m ≤ 40 — 9 dogs. Work out an estimate for the mean mass, in kg, using the midpoint of each class interval.
- 4.A school has 1,500 pupils. The head teacher takes a random sample of 150 of them from the school register and asks how long they spend on homework. Rory says the sample is too small for the result to mean anything. Is Rory right? Give a reason for your answer.
- 5.A scatter graph shows the mass, x kg, of a parcel and the cost, y pounds, of posting it. The line of best fit is y = 1.5x + 2. Work out the estimated cost of posting a parcel with a mass of 6 kg, using the line of best fit.y = 1.5x + 2
- 6.A scatter graph of the number of ice creams sold at a seaside kiosk in Bournemouth and the number of sunburn cases treated at a nearby pharmacy, recorded on the same 30 days, shows strong positive correlation. Which statement about this correlation is correct?
- 7.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
- 8.The age, in years, and the score in a reaction test are recorded for eight members of a sports club: (14, 92), (18, 88), (23, 85), (27, 80), (31, 78), (36, 74), (42, 70), (49, 65). The eight pairs are plotted on a scatter graph. Describe the correlation between age and score.
- 9.The table shows the times, t minutes, that 30 pupils took to walk to school. 0 < t ≤ 10: 8 pupils. 10 < t ≤ 20: 12 pupils. 20 < t ≤ 30: 6 pupils. 30 < t ≤ 40: 4 pupils. Write down which average can be given exactly from this table, and give a reason for your answer.
- 10.Four pairs of variables are listed below. Write down the pair that you would expect to show no correlation.
- 11.In a random sample of 40 pupils at a school, 6 are left-handed. The school has 900 pupils. Work out an estimate for the number of left-handed pupils in the school.
- 12.A sports centre in Newcastle asks 60 members which activity they prefer: swimming 15 members, football 25 members, badminton 12 members and other activities 8 members. Work out the angle at the centre of the pie chart sector that represents badminton.
- 13.A garden centre records its total sales, in pounds, at the end of each of the twelve months of one year, to see how sales change over time. Write down the most suitable type of chart to show this time series data.
- 14.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.
- 15.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
Answer key
- (b) Positive correlation — Method: the type of correlation is named from the direction the points take as the scatter graph is read from left to right. Working: the points rise from left to right, so as the arm span read on the horizontal axis increases, the height read on the vertical axis increases as well; two quantities that increase together show positive correlation. Answer: positive correlation. The distractors: negative correlation comes from naming the direction the wrong way round, since a negative correlation needs the points to fall as the graph is read from left to right; no correlation comes from treating points that are spread out rather than sitting exactly on a line as though they showed no relationship; direct proportion comes from confusing a rising trend with proportion, which would additionally need the line through the points to pass through the origin and would mean doubling one quantity doubles the other.
- (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.
- (d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (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.
- (b) £11.00 — 1.5 × 6 = 9, and 9 + 2 = 11, so the estimated cost is £11.00. Choosing £9.00 stops after 1.5 × 6 = 9 and forgets to add the £2. Choosing £12.00 adds the mass and the constant first and then multiplies: 6 + 2 = 8, and 8 × 1.5 = 12.00. Choosing £13.50 swaps the gradient and the intercept, using y = 2x + 1.5 instead: 2 × 6 = 12, and 12 + 1.5 = 13.50.
- (c) Neither causes the other; sunshine links both. — Both ice cream sales and sunburn cases tend to rise on hot, sunny days, so the amount of sunshine is a third factor linked to both — neither variable causes the other. Saying ice cream sales cause the sunburn assumes a causal link in one direction that the correlation alone cannot establish. Saying sunburn cases cause the ice cream sales assumes the reverse causal link, which is no more justified. Saying a strong correlation always means causation is the general error this question is testing: correlation, however strong, does not by itself prove that one variable causes the other.
- (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) Strong negative correlation — Method: correlation is described by two things — the direction the points take as the graph is read from left to right, and how closely the points lie to a single straight line. Working: reading the pairs in order of age, the ages rise 14, 18, 23, 27, 31, 36, 42, 49 while the scores fall 92, 88, 85, 80, 78, 74, 70, 65; the score falls at every single step, with no reversal anywhere, so the points fall from left to right and lie close to a straight line. Answer: strong negative correlation — negative for the falling direction, strong because every point follows the pattern. The distractors: strong positive correlation comes from noticing a clear pattern and calling any clear pattern positive, without checking the direction; weak negative correlation comes from reading the direction correctly but judging points that do not lie exactly on a straight line to be only loosely related, when these eight fall without a single exception; no correlation comes from reading a falling trend as though it showed no relationship at all, when a falling trend is itself a relationship.
- (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.
- (b) A person's shoe size and their favourite colour — A person's shoe size is not linked to which colour they prefer, so these two show no correlation. The other three pairs are all genuinely correlated: distance travelled and fuel used rise together, which is positive correlation; hours of revision and test score generally rise together, which is also positive correlation; and as outdoor temperature rises, fewer woolly hats are sold, which is negative correlation. Negative correlation is still a real relationship between two variables — it is not the same thing as no relationship at all, so the temperature and hats pair is not the answer to this question.
- (b) 135 — Method: use the sample to find the PROPORTION of left-handed pupils, then apply that same proportion to the whole school population. Working: in the sample, 6 out of 40 pupils are left-handed, a proportion of 6 ÷ 40 = 0.15. Applying that proportion to the school's 900 pupils gives an estimate of 0.15 × 900 = 135 pupils. Giving 6 simply repeats the number of left-handed pupils IN THE SAMPLE, without scaling up to the whole school at all. Multiplying the population by the number of left-handed pupils in the sample without first dividing by the sample size, 900 × 6 = 5400, badly overestimates — that is more pupils than the whole school has. Dividing the population by the sample size but forgetting to multiply by the number of left-handed pupils found, 900 ÷ 40 = 22.5, finds the scale factor but stops one step short of using it. Always find the proportion in the sample first, then scale that same proportion up to the population.
- (c) 72° — The angle is 12 ÷ 60 × 360 = 72°. Choosing 150° divides football's frequency of 25 instead of badminton's 12: 25 ÷ 60 × 360 = 150. Choosing 20° finds badminton as a percentage of the members, 12 ÷ 60 × 100 = 20, rather than an angle in degrees. Choosing 90° uses 48, the total of the other three activities, as the total instead of the full 60 members: 12 ÷ 48 × 360 = 90.
- (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.
- (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.
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