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.
Statistics worksheet — GCSE Higher
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- 1.A histogram shows the speeds, v mph, of 100 vehicles passing a checkpoint. The bar for 0 ≤ v < 20 has a frequency density of 1 vehicle per mph, the bar for 20 ≤ v < 30 has a frequency density of 3 vehicles per mph, the bar for 30 ≤ v < 50 has a frequency density of 2 vehicles per mph, and the bar for 50 ≤ v < 70 has a frequency density of 0.5 vehicles per mph. Estimate the mean speed of the vehicles.
- 2.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
- 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 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.
- 5.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.
- 6.The marks scored by 11 pupils in a test are given in order: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42. Work out the lower quartile of these marks.
- 7.The delivery times, in minutes, of 15 parcels are given in order: 5, 8, 10, 12, 14, 15, 17, 19, 21, 23, 25, 27, 29, 31, 34. Work out the interquartile range of these times.
- 8.A scatter graph of the number of hours, x, that pupils revised against their test score, y, has the line of best fit y = 2.5x + 15. Amelia wants a score of at least 80. Work out the least whole number of hours of revision the line of best fit suggests she needs.y = 2.5x + 15
- 9.In a histogram of the times, t minutes, taken by some people to complete a task, the class 15 ≤ t < 30 contains 24 people. Work out the frequency density for this class.
- 10.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.
- 11.The midday temperature in Leeds was recorded on each day of one week: 22 °C, 24 °C, 23 °C, 25 °C, 26 °C, 21 °C, 24 °C. Work out the mean midday temperature, giving your answer to 2 decimal places.
- 12.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?
- 13.A sports shop in Cardiff sold 40 pairs of football boots last month: 4 pairs of size 6, 5 pairs of size 7, 8 pairs of size 8, 13 pairs of size 9 and 10 pairs of size 10. The manager will order 40 pairs for next month and wants as many pairs as possible to be in a size customers will buy. Work out the mean size and the modal size, and write down which of the two he should use.
- 14.The masses, m kg, of 160 fish caught by a trawler in one day are grouped into classes of unequal width: 0 ≤ m < 10, 40 fish; 10 ≤ m < 30, 60 fish; 30 ≤ m < 45, 30 fish; 45 ≤ m < 50, 30 fish. A histogram is to be drawn from this table. Which set of frequency densities, listed in the same order as the classes above, is correct?
- 15.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.
Answer key
- (a) 31.5 — Method: to estimate the mean from a histogram, first turn each bar into a frequency (frequency density × class width), then use mean = Σ(frequency × midpoint) ÷ Σfrequency, with the midpoint standing in for every value in that class. Working: the four classes have widths 20, 10, 20 and 20, so their frequencies are 1 × 20 = 20, 3 × 10 = 30, 2 × 20 = 40 and 0.5 × 20 = 10, which do add to the 100 vehicles stated. Their midpoints are 10, 25, 40 and 60, so Σfx = 20 × 10 + 30 × 25 + 40 × 40 + 10 × 60 = 200 + 750 + 1600 + 600 = 3150, and the mean is 3150 ÷ 100 = 31.5. Answer: the estimated mean speed is 31.5 mph. Watch which numbers you treat as the frequencies and which as the values: using the frequency densities themselves as the frequencies, without multiplying by the class widths first, gives 1 × 10 + 3 × 25 + 2 × 40 + 0.5 × 60 = 195 spread over 1 + 3 + 2 + 0.5 = 6.5, and 195 ÷ 6.5 = 30, a mean built from the wrong 'frequencies' altogether; averaging the four midpoints on their own, (10 + 25 + 40 + 60) ÷ 4 = 33.75, ignores how many vehicles are actually in each class; and using each class's lower boundary in place of its midpoint, 20 × 0 + 30 × 20 + 40 × 30 + 10 × 50 = 2300 and 2300 ÷ 100 = 23, systematically underestimates every class by roughly half its width.
- (c) The median, £160,000, as one very high price lifts the mean — Method: find both averages, then choose the one that sits closer to the bulk of the data. Working: in order the prices are 140,000, 150,000, 160,000, 170,000 and 580,000, so the median is the third of the five, £160,000. For the mean, 140,000 + 150,000 + 160,000 + 170,000 + 580,000 = 1,200,000 and 1,200,000 ÷ 5 = 240,000, so the mean is £240,000. Four of the five houses sold for £170,000 or less, so a reader told that a typical price is £240,000 would expect to pay at least £70,000 more than any of those four cost. Answer: the median, £160,000, as one very high price lifts the mean. The distractors: £580,000 is the middle value of the list as it is printed, which is the median only when the values have first been put in order; £240,000 is the mean, chosen on the ground that a median ignores three of the five prices, but a median uses all five to find which one is central and is then untroubled by how extreme the outer values are; £155,000 comes from deleting the £580,000 house and taking the mean of what is left, since 140,000 + 150,000 + 160,000 + 170,000 = 620,000 and 620,000 ÷ 4 = 155,000, but a real sale may not be thrown away merely for being large.
- (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) 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) 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.
- (c) 18 — Method: for n ordered values, GCSE convention places the lower quartile at position (n + 1) ÷ 4, counting from the smallest value. Working: there are 11 marks, so n + 1 = 11 + 1 = 12 and 12 ÷ 4 = 3, so the lower quartile is the 3rd value in the ordered list 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, which is 18. Answer: the lower quartile is 18 marks. Watch the position you count to: dividing 11 ÷ 4 = 2.75 without adding 1 first, then rounding down, lands on the 2nd value, 15, not the 3rd; reaching for the middle of the whole list instead gives the median, 27, a different statistic; and averaging the 3rd and 4th values, 18 + 21 = 39 and 39 ÷ 2 = 19.5, borrows a method for an even split where it is not needed here.
- (c) 15 — Method: the interquartile range is the upper quartile take away the lower quartile, IQR = Q3 − Q1. Working: n + 1 = 15 + 1 = 16, so the lower quartile sits at position 16 ÷ 4 = 4, the 4th value in the list, which is 12; the upper quartile sits at position 3 × 4 = 12, the 12th value, which is 27. So 27 − 12 = 15. Answer: the interquartile range is 15 minutes. Watch which values you use and which way round: taking the smallest time away from the largest, 34 − 5 = 29, finds the range, which uses every value between the extremes rather than just the middle half; taking the median away from the upper quartile instead of the lower quartile, 27 − 19 = 8, swaps the median in for the lower quartile; and reversing the subtraction, 12 − 27 = −15, finds the right two values but in the wrong order — an interquartile range is never negative.
- (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) 1.6 — Method: on a histogram the height of a bar is the frequency density, and frequency density = frequency ÷ class width. Working: the class 15 ≤ t < 30 runs from 15 to 30, so its width is 30 − 15 = 15 minutes; the frequency is 24, so the frequency density is 24 ÷ 15 = 1.6. Answer: 1.6 people per minute. The distractors: 360 comes from multiplying the frequency by the class width, 24 × 15, which uses the area rule backwards — area gives the frequency, so the frequency must be divided by the width to give the height; 0.625 comes from dividing the class width by the frequency, 15 ÷ 24, reversing the formula; 0.8 comes from dividing by the upper class boundary, 24 ÷ 30, instead of by the width of the class.
- (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.
- (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (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.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (d) 4, 3, 2, 6 — Method: frequency density = frequency ÷ class width for each class in turn; do not assume the classes are all the same width. Working: the four classes have widths 10 − 0 = 10, 30 − 10 = 20, 45 − 30 = 15 and 50 − 45 = 5. Dividing each frequency by its own width gives 40 ÷ 10 = 4, 60 ÷ 20 = 3, 30 ÷ 15 = 2 and 30 ÷ 5 = 6. Answer: the frequency densities, in order, are 4, 3, 2 and 6. Watch the width of each class separately: treating the last class as if it were also 10 units wide, like the first, gives 30 ÷ 10 = 3 instead of 30 ÷ 5 = 6 — the classes here are deliberately unequal, so no width can be borrowed from another class; dividing the width by the frequency instead of the frequency by the width for the third class gives 15 ÷ 30 = 0.5 in place of 2, the formula the wrong way round; and reading the frequency column straight off the table, 40, 60, 30, 30, skips the division by width altogether and reports how many fish are in each class rather than how densely packed each bar is.
- (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.
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