Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Statistics worksheet — GCSE Higher
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- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (c) 76 — Method: the frequency of each class is the area of its bar, frequency density × class width, so work out all three frequencies and add them. Working: the widths are 5, 10 and 25 kg, so the frequencies are 4 × 5 = 20, 2.6 × 10 = 26 and 1.2 × 25 = 30; the total is 20 + 26 + 30 = 76. Answer: 76 dogs. The distractors: 7.8 comes from adding the three frequency densities, 4 + 2.6 + 1.2, treating each height as though it were a count; 107 comes from using each upper class boundary as the width, giving 4 × 5, 2.6 × 15 and 1.2 × 40; 39 comes from using the first class width, 5, for every bar, which ignores the unequal intervals and gives 20 + 13 + 6.
- (c) 156 cm — Method: to combine two groups' means, multiply each group's mean by its own number of pupils, add the two totals together, then divide by the total number of pupils in both groups. Working: 20 × 150 = 3,000 cm for the boys and 10 × 168 = 1,680 cm for the girls, giving a combined total of 3,000 + 1,680 = 4,680 cm. Dividing by all 30 pupils gives 4,680 ÷ 30 = 156 cm. Giving 159 cm averages the two means, (150 + 168) ÷ 2, treating the two groups as if they had the same number of pupils, when there are twice as many boys as girls. Giving 4,680 cm finds the correct combined total height but stops there, forgetting the final division by the 30 pupils. Giving 234 cm divides the combined total by 20, the number of boys only, forgetting that the total also includes the 10 girls. Always weight each mean by its own group size, and always divide by the TOTAL number of pupils in both groups combined.
- (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.
- (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) Town B — higher median and smaller IQR — Method: 'higher and more consistent' needs two comparisons — the median for typical price, and the interquartile range for spread, with a smaller interquartile range meaning more consistent. Working: Town B's median, £235,000, is higher than Town A's, £220,000. Town A's interquartile range is 310 − 180 = 130 and Town B's is 260 − 175 = 85, so Town B's interquartile range is the smaller of the two. Answer: Town B has both the higher median and the smaller interquartile range, so it is the town with higher, more consistent prices. Watch which combination of median and interquartile range each statement claims, and for which town: claiming Town A has the higher median and the smaller interquartile range gets both comparisons wrong, since Town B leads on both; claiming Town A has the higher median (still wrong) but the larger interquartile range at least reads the spread correctly, without it rescuing the false median claim; and claiming Town B has the higher median (correct) but the larger interquartile range misreads the spread — Town B's interquartile range is the smaller of the two, not the larger.
- (c) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
- (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.
- (c) Route 2, because its interquartile range is smaller — Method: for a journey where turning up on time matters, what matters is not the typical (median) time but how predictable it is — a smaller interquartile range means the middle half of journeys cluster closer together. Working: Route 1's median, 34 minutes, is in fact lower than Route 2's, 41 minutes, so Route 1 is faster on average; but Route 1's interquartile range, 22 minutes, is far larger than Route 2's, 6 minutes, so Route 1's times are much less predictable. Answer: Priya should choose Route 2, because its interquartile range is smaller, even though it is slower on average. Watch which statistic answers the question actually asked: Route 1 does not have the smaller interquartile range, Route 2 does, so picking Route 1 for that reason misreads the table; Route 1's median genuinely is the lower one, but a lower median answers 'which is faster', not 'which is more reliable'; and Route 2's median is not the lower one, so that claim about Route 2 is simply false.
- (c) Branch A waits longer, and Branch A is more consistent — Method: compare the two branches using a measure of location (the median) for who waits longer, and a measure of spread (the interquartile range) for who is more consistent — a smaller interquartile range means more consistent. Working: Branch A's median, 12 minutes, is higher than Branch B's, 9 minutes, so Branch A's customers wait longer on average. Branch A's interquartile range, 5 minutes, is smaller than Branch B's, 11 minutes, so Branch A's waiting times vary less. Answer: Branch A waits longer, and Branch A is also the more consistent of the two. Watch that each half of the comparison uses the right statistic and reads it correctly: swapping both readings gives Branch B the longer wait and the greater consistency, when neither is true; keeping the median comparison right but reading a larger interquartile range as 'more consistent' has the direction of spread backwards; and swapping only the median comparison keeps the correct branch for consistency but gives the wrong branch the longer wait.
- (b) 6 — Method: frequency density = frequency ÷ class width. Working: the class 12 ≤ h < 18 has width 18 − 12 = 6, so frequency density = 36 ÷ 6 = 6. Answer: the frequency density is 6 seedlings per cm. Watch which numbers you use: taking the lower bound, 12, as the width instead of 18 − 12 = 6 gives 36 ÷ 12 = 3; dividing the total number of seedlings, 90, rather than this class's frequency, 36, by the width gives 90 ÷ 6 = 15, a density that belongs to no single class; and multiplying instead of dividing gives 36 × 6 = 216, far too large a density for so narrow a class.
- (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.
- (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.
- (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.
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