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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Statistics worksheet — GCSE Higher
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- 1.A town council wants to find out about the eating habits of the people who live in the town. It asks only the members of a local sports club. Give a reason why this sample is biased.
- 2.Write down the statement that correctly describes the difference between correlation and causation.
- 3.A bus company records the delay, d minutes, of 250 buses: 0 ≤ d < 2, 60 buses; 2 ≤ d < 5, 90 buses; 5 ≤ d < 10, 75 buses; 10 ≤ d < 20, 25 buses. The company refunds the fare whenever a bus is more than 8 minutes late. Estimate the number of refunds it must pay.
- 4.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.
- 5.A factory makes a batch of 4,000 circuit boards. It checks a random sample of 50 boards and finds that 4 are faulty. The factory will scrap the whole batch if the estimated number of faulty boards in the batch is more than 250. Should the factory scrap the batch?
- 6.A factory made 3,000 phone cases last week. Shift A checked a random sample of 100 cases and found that 34 were scratched. Shift B checked a different random sample of 50 cases and found that 21 were scratched. Using the COMBINED results from both shifts, work out an estimate for the number of scratched cases made last week.
- 7.In a histogram of the distances, d metres, thrown by some athletes, the bar covering 20 ≤ d < 60 has a constant frequency density of 1.8 per metre. Estimate the number of throws of at least 20 metres but less than 35 metres.
- 8.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 9.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?
- 10.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.
- 11.A recruitment agency in Manchester compares the weekly pay, in pounds, of seven employees at two branches. Branch A: £480, £495, £500, £505, £510, £515, £1,200. Branch B: £480, £490, £500, £510, £520, £530, £540. An advert for the agency claims 'Branch A pays more on average.' Decide whether this claim is fairly supported, using an appropriate average, and choose the correct conclusion.
- 12.The times, t minutes, taken by 120 runners to finish a fun run are summarised by these cumulative frequencies: t < 20, 8 runners; t < 30, 26 runners; t < 40, 74 runners; t < 50, 110 runners; t < 60, 120 runners. Work out the number of runners who took 40 minutes or longer to finish.
- 13.A composite (stacked) bar for a charity bake sale in Durham shows the number of cakes sold, split into three types. The bar has a total height of 80 cakes: 34 were sponge cakes, 26 were chocolate cakes and the rest were fruit cakes. Work out the percentage of the cakes sold that were fruit cakes.
- 14.A cumulative frequency graph for the diameters, d mm, of 320 ball bearings is plotted from these points (upper class boundary, cumulative frequency): (10, 30), (20, 90), (30, 190), (40, 280), (50, 320). Estimate the diameter below which 90% of the ball bearings measure.
- 15.A shop compares customer waiting times, in minutes, at two branches over one week. Branch A had a median of 12 minutes and an interquartile range of 5 minutes. Branch B had a median of 9 minutes and an interquartile range of 11 minutes. Compare the waiting times at the two branches.
Answer key
- (c) Club members probably eat differently from most people — Method: a sample is biased when the group it is drawn from differs from the population in the very thing the survey is measuring, so compare the subgroup with the population on that quantity. Working: the survey measures eating habits, and people who join a sports club take more exercise than average and are known to eat differently from the town as a whole, so their replies pull the results away from the true picture for the town however many of them are asked. Answer: club members probably eat differently from most people. The distractors: the reply about the number of members treats bias as a question of size, but a large biased sample is still biased; the reply that the members were picked at random is false, since the council picked a club rather than picking residents, and it confuses bias with non-response; the reply that everyone asked lives in the town notes something true of the members but draws the false conclusion that the sample therefore covers the town, when a sample must reflect a population and not merely be taken from inside it.
- (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) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.
- (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) Yes — with an estimate of 320, above the 250 limit. — Method: scale the sample proportion up to the whole batch to get an estimate, then compare that estimate with the 250 limit to reach a decision. Working: in the sample, 4 out of 50 boards are faulty, a proportion of 4 ÷ 50 = 0.08. Applying that proportion to the batch of 4,000 gives an estimate of 0.08 × 4000 = 320 faulty boards. Since 320 is more than 250, the factory should scrap the batch. Inverting the proportion, 50 ÷ 4 = 12.5, and treating that as a percentage of the batch, 12.5% × 4000 = 500, still gives 'yes' but from the wrong fraction, so it overstates the estimate. Comparing the raw number of faulty boards found in the sample, 4, directly with the 250 limit skips the scaling up to the batch altogether, and 4 is nowhere near 250, so that route wrongly says 'no'. Dividing the batch by the sample size, 4000 ÷ 50 = 80, finds how many samples of 50 fit into the batch but stops before multiplying by the 4 faulty boards found, so it also wrongly says 'no'. Always find the proportion in the sample first, scale it up to the whole batch, and only then compare the estimate with the limit given.
- (d) 1100 — Method: to combine two samples of different sizes, add the faulty counts together and add the sample sizes together before scaling up, rather than treating the two samples separately. Working: the combined sample found 34 + 21 = 55 scratched cases out of 100 + 50 = 150 cases checked, a proportion of 55 ÷ 150. Applying that proportion to the week's production of 3,000 gives an estimate of 55 ÷ 150 × 3000 = 1100 scratched cases. Averaging the two shifts' proportions instead of combining their totals, (34 ÷ 100 + 21 ÷ 50) ÷ 2 = 0.38, gives 0.38 × 3000 = 1140 — this treats the two samples as equally weighted even though Shift A checked twice as many cases as Shift B. Using only Shift A's sample, 34 ÷ 100 × 3000 = 1020, ignores Shift B's cases completely. Using only Shift B's sample, 21 ÷ 50 × 3000 = 1260, ignores Shift A's cases completely. When two samples are different sizes, combine their totals before finding the proportion — do not average the two proportions, and do not use only one shift's sample.
- (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (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.
- (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.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
- (c) 46 — Method: the cumulative frequency table gives the number of runners below each time; to find the number at or above a time, subtract that cumulative frequency from the total. Working: the cumulative frequency for t < 40 is 74, so 120 runners in total take away the 74 who finished in under 40 minutes: 120 − 74 = 46. Answer: 46 runners took 40 minutes or longer. Watch which boundary and which subtraction you use: reading off t < 50 instead of t < 40 and subtracting, 120 − 110 = 10, answers a different question, '50 minutes or longer'; giving 74 itself as the answer reports how many finished below 40 minutes, the opposite of what was asked; and subtracting the two nearby cumulative frequencies, 110 − 74 = 36, finds how many took between 40 and 50 minutes, not everyone from 40 minutes upward.
- (d) 25% — First find the number of fruit cakes: 80 − 34 − 26 = 20. Then write this as a percentage of the total: 20 ÷ 80 × 100 = 25%. Giving 20% comes from reporting the count of fruit cakes, 20, directly as a percentage, without dividing by the total of 80 first. Giving 32.5% computes the percentage of chocolate cakes instead of fruit cakes: 26 ÷ 80 × 100 = 32.5%. Giving 42.5% computes the percentage of sponge cakes instead of fruit cakes: 34 ÷ 80 × 100 = 42.5%.
- (c) 42 — Method: find the target cumulative frequency, 90% of the total, locate the class it falls in from the plotted points, then interpolate: lower boundary, plus the extra distance needed into the class divided by the class's frequency, times its width. Working: 90% of 320 is 0.9 × 320 = 288. The plotted points show a cumulative frequency of 280 at d = 40 and 320 at d = 50, so the class 40 ≤ d < 50 has frequency 320 − 280 = 40 and width 50 − 40 = 10, and 288 falls inside it. The extra distance needed into the class is 288 − 280 = 8, and 8 ÷ 40 × 10 = 2, so the diameter is 40 + 2 = 42. Answer: the estimated diameter is 42 mm. Watch which point and which class the interpolation actually uses: reading off d = 40, the plotted point just below the target, instead of interpolating the extra 8 ball bearings into the next 10 mm, stops one step short of the true answer; finding the diameter below which only 10% lie instead of 90% gives a target of 0.1 × 320 = 32, which falls in the class 10 ≤ d < 20 — the extra distance into that class is 32 − 30 = 2, and 2 ÷ 60 × 10 = 0.3, so this route gives 10 + 0.3 = 10.3, the bottom decile rather than the top 90%; and interpolating within the class 30 ≤ d < 40 instead of 40 ≤ d < 50, as though 288 had not yet reached a cumulative frequency of 280, treats the extra distance as 288 − 190 = 98, and 98 ÷ 90 × 10 = 10.9, giving 30 + 10.9 = 40.9, one class too early.
- (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.
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