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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- (b) £1,000 — Wages take up 150° out of 360°, so the amount spent on wages is 150 ÷ 360 × 2400 = £1,000. Choosing £600 uses the repairs angle, 90°, instead of the wages angle: 90 ÷ 360 × 2400 = 600. Choosing £3,600 treats the angle in degrees as if it were a percentage, 150 ÷ 100 × 2400 = 3600, instead of dividing by 360°. Choosing £800 uses the angle for the 'other costs' sector, 360 − 90 − 150 = 120°, instead of the wages sector: 120 ÷ 360 × 2400 = 800.
- (d) 16 — Method: rearrange interquartile range = upper quartile − lower quartile to make the lower quartile the subject: lower quartile = upper quartile − interquartile range, then check the answer sits in the right place in the list. Working: there are 11 values, so 11 + 1 = 12; the upper quartile sits at position 3 × 12 ÷ 4 = 9, which is 35, and x sits at position 12 ÷ 4 = 3, which is the lower quartile. So x = 35 − 19 = 16, and 16 does sit between the 2nd value, 9, and the 4th value, 17, as it should. Answer: x = 16. Watch how you rearrange and where you count to: adding instead of subtracting, 35 + 19 = 54, treats the interquartile range as something added on rather than a gap taken away; subtracting in the wrong order, 19 − 35 = −16, finds the right two numbers but flips the sign; and counting to the 8th value instead of the 9th treats 31 as the upper quartile, giving 31 − 19 = 12, one position short of where the upper quartile actually sits.
- (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.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (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.
- (c) Class X has the higher median and the wider spread — Method: compare the two box plots statistic by statistic — median for location, and the interquartile range for spread — checking the true value of each rather than assuming a pattern. Working: Class X's median is 60 and Class Y's is 58, so Class X's median is the higher one. Class X's interquartile range is 70 − 45 = 25 and Class Y's is 65 − 50 = 15 (and the ranges follow the same order: 95 − 20 = 75 against 80 − 35 = 45), so Class X also has the wider spread. Answer: Class X has both the higher median and the wider spread. Watch that each half of a compound statement is checked separately: claiming Class Y has the higher median and the wider spread gets both comparisons backwards; claiming Class X has the higher median but the narrower spread keeps the median right while reading the spread the wrong way round; and claiming Class Y has the higher median but the narrower spread swaps the median comparison while getting the spread right.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (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) 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.
- (a) 49 — Method: add the frequencies of the classes that lie wholly below 35 seconds, then use linear interpolation for the class that 35 cuts through, assuming the calls in that class are spread evenly across it. Working: below 30 seconds there are 15 + 24 = 39 calls; the value 35 lies in the class 30 ≤ t < 50, which is 20 seconds wide and holds 40 calls, and 35 is 35 − 30 = 5 seconds into it, so the estimated share is (5 ÷ 20) × 40 = 10 calls; the estimate is 39 + 10 = 49. Answer: about 49 calls met the target. The distractors: 79 comes from adding the whole of the class 30 ≤ t < 50, 39 + 40, and so counting calls of up to 50 seconds as being under 35; 39 comes from stopping at the class boundary 30 and ignoring the part class altogether; 69 comes from measuring the part of the class from 35 up to 50 instead of from 30 up to 35, giving (15 ÷ 20) × 40 = 30 and then 39 + 30.
- (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.
- (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%.
- (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.
- (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.
- (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.
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