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.
Calculator
Answer key: Statistics worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (d) £15.50 — 2 × 6 = 12, and 12 + 3.50 = 15.50, so the estimated fare is £15.50. Choosing £12.00 stops after 2 × 6 = 12 and forgets to add the £3.50. Choosing £19.00 adds the distance and the constant first and then multiplies: 6 + 3.50 = 9.50, and 9.50 × 2 = 19.00, applying the ×2 to the whole sum instead of only to the distance. Choosing £13.00 multiplies only the constant term by 2 instead of the distance: 2 × 3.50 = 7, and 7 + 6 = 13.00.
- (d) 20 — 0.4x + 1 = 9, so 0.4x = 9 − 1 = 8, and 8 ÷ 0.4 = 20, so 20 sessions are needed. Choosing 23 divides 9 by 0.4 without first subtracting the 1: 9 ÷ 0.4 = 22.5, rounded up to 23. Choosing 25 subtracts the wrong way, adding the 1 instead of taking it away: 9 + 1 = 10, and 10 ÷ 0.4 = 25. Choosing 2 misplaces the decimal point in the gradient, dividing by 4 instead of by 0.4: 8 ÷ 4 = 2.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (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.
- (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.
- (a) 26 — Method: a line of best fit lets one quantity be predicted from the other, so the score is substituted into the equation of the line and the resulting inequality is solved for the number of hours. Working: a score of at least 80 means 2.5x + 15 ≥ 80; taking 15 from both sides gives 2.5x ≥ 65, and dividing both sides by 2.5 gives x ≥ 26, so the least whole number of hours is 26. Checking, 2.5 × 26 + 15 = 80, which does reach the target. Answer: 26 hours — and this is only an estimate, because a line of best fit predicts a trend rather than an individual result, and a prediction made outside the range of hours the pupils actually revised for would be an extrapolation and less reliable still. The distractors: 27 comes from reaching 26 and then rounding up again, although 26 hours already gives a score of exactly 80; 32 comes from 80 ÷ 2.5, which ignores the 15 in the equation of the line; 38 comes from (80 + 15) ÷ 2.5, that is from adding the 15 instead of subtracting it when rearranging.
- (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) 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) 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) 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) 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.
- (a) No correlation — Shoe size has no real relationship with spelling ability, and the points here are scattered with no rising or falling trend, so this is no correlation. A positive correlation would show the points rising together, and a negative correlation would show them falling as one increases; neither pattern is present here. Strong correlation is not correct either, since strength only applies once a positive or negative trend exists, and there isn't one.
- (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.
- (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.
- (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.
Build your own mix at the worksheet builder.