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 histogram shows the times, t minutes, taken by 120 visitors to complete an escape room. The bar for 0 ≤ t < 10 has a frequency density of 5 visitors per minute, the bar for 10 ≤ t < 20 has a frequency density of 2 visitors per minute, the bar for 20 ≤ t < 40 has a frequency density of 1.5 visitors per minute, and the bar for 40 ≤ t < 60 has a frequency density of 1 visitor per minute. Work out which class contains the median time.
- 2.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.
- 3.A Year 10 class has 20 boys with a mean height of 150 cm and 10 girls with a mean height of 168 cm. Work out the mean height of all 30 pupils in the class.
- 4.Forty pupils in class P and forty pupils in class Q each solved a puzzle. The times, in seconds, were summarised using cumulative frequency. For class P the lower quartile is 24, the median is 38 and the upper quartile is 46. For class Q the lower quartile is 30, the median is 35 and the upper quartile is 44. Write down the statement that correctly compares the two classes.
- 5.A scatter graph of taxi journeys in Bristol shows the distance, x miles, and the fare, y pounds. The line of best fit is y = 2x + 3.50. Work out the estimated fare for a journey of 6 miles, using the line of best fit.y = 2x + 3.5
- 6.A gym in Cardiff has a scatter graph of the number of training sessions, x, attended by each member and the weight lost, y kg. The line of best fit is y = 0.4x + 1. Members want to lose at least 9 kg. Using the line of best fit, work out the least whole number of sessions needed to reach this target.y = 0.4x + 1
- 7.A school has 1,500 pupils. The head teacher takes a random sample of 150 of them from the school register and asks how long they spend on homework. Rory says the sample is too small for the result to mean anything. Is Rory right? Give a reason for your answer.
- 8.In a random sample of 40 pupils at a school, 6 are left-handed. The school has 900 pupils. Work out an estimate for the number of left-handed pupils in the school.
- 9.A two-way table records the favourite subject, Maths or Art, of 60 pupils in Year 10, and whether each pupil is left-handed or right-handed. 9 of the 60 pupils are left-handed, and 6 of those left-handed pupils prefer Art. In total, 24 of the 60 pupils prefer Art. A pupil is chosen at random from the 60. Work out the probability that the pupil is right-handed and prefers Art.
- 10.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.
- 11.A histogram is drawn for the masses, m grams, of 200 letters. The bar for 0 ≤ m < 50 has a frequency density of 1.2 per gram and the bar for 50 ≤ m < 100 has a frequency density of 1.8 per gram. All the remaining letters lie in the class 100 ≤ m < 200. Work out the frequency density of the bar for 100 ≤ m < 200.
- 12.The distances, d km, cycled by 180 riders in a charity sportive are summarised by these cumulative frequencies: d < 30, 20 riders; d < 60, 60 riders; d < 80, 120 riders; d < 100, 160 riders; d < 130, 180 riders. Use interpolation to estimate the median distance cycled.
- 13.Box plots summarise the test scores of two classes. Class X: minimum 20, lower quartile 45, median 60, upper quartile 70, maximum 95. Class Y: minimum 35, lower quartile 50, median 58, upper quartile 65, maximum 80. Which statement about the two classes is correct?
- 14.The mean of five test scores is 68. Four of the scores are 55, 62, 74 and 80. Work out the fifth score.
- 15.A quality inspector weighs a random sample of 50 packets of crisps from one day's production and finds their mean mass is 32.4 g. The factory makes 20,000 packets that day. Work out an estimate for the total mass, in kg, of all the packets made that day.
Answer key
- (c) The class with times from 10 up to 20 — Method: to find the median class from a histogram, first turn each bar's frequency density into a frequency using density × class width, build up the cumulative frequency, and find the first class whose cumulative frequency reaches or passes n ÷ 2. Working: the four classes have widths 10, 10, 20 and 20, so their frequencies are 5 × 10 = 50, 2 × 10 = 20, 1.5 × 20 = 30 and 1 × 20 = 20, which add to the 120 visitors stated. The median sits at position 120 ÷ 2 = 60. The cumulative frequency is 50 after the first class and 50 + 20 = 70 after the second, so the 60th visitor is reached during the second class. Answer: the median lies in the class 10 ≤ t < 20. Watch which class each shortcut lands on: the tallest bar belongs to the first class, with the highest frequency density, 5 — but the tallest bar shows where visitors are packed most densely, not where the middle visitor falls, and picking it lands one class too early, at 0 ≤ t < 10; taking half of the total TIME span instead of half of the total NUMBER of visitors, 60 minutes ÷ 2 = 30 minutes, lands in the class 20 ≤ t < 40, confusing a value on the horizontal axis with a position in the data; and using the full 120 visitors as the target position, rather than 120 ÷ 2 = 60, reaches all the way to the last class, 40 ≤ t < 60, treating the whole data set's size as though it were the position of a single middle value.
- (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) 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) Q was faster on average and more consistent — Method: compare the medians for the average and the interquartile ranges for the spread, remembering that a shorter time is faster and a smaller interquartile range means more consistent. Working: the median for class Q is 35 seconds against 38 seconds for class P, so class Q was faster on average; the interquartile range for class P is 46 − 24 = 22 seconds and for class Q it is 44 − 30 = 14 seconds, so class Q's times are more tightly grouped. Answer: class Q was faster on average and more consistent. The distractors: calling Q slower comes from comparing the lower quartiles, 30 against 24, as though a quartile were the average; calling Q less consistent comes from using the gap between the median and the upper quartile as the spread, 44 − 35 = 9 against 46 − 38 = 8, instead of the full interquartile range; the statement that Q was both slower and less consistent comes from making both of those mistakes together.
- (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.
- (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.
- (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.
- (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.
- (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) 0.5 per gram — Method: turn the two known bars into frequencies using area, subtract from the total to find how many letters are left, then divide that frequency by the width of the last class to get its height. Working: the first bar covers 50 g at a frequency density of 1.2, giving 1.2 × 50 = 60 letters, and the second covers 50 g at 1.8, giving 1.8 × 50 = 90 letters; together that is 60 + 90 = 150 letters, so 200 − 150 = 50 letters remain; the class 100 ≤ m < 200 is 100 g wide, so its frequency density is 50 ÷ 100 = 0.5 per gram. Answer: 0.5 per gram. The distractors: 0.25 per gram comes from dividing the remaining 50 letters by the upper class boundary, 200, instead of by the class width of 100; 2 per gram comes from dividing the class width by the frequency, 100 ÷ 50, reversing the formula; 1.4 per gram comes from subtracting only the first bar's 60 letters, leaving 140, and then dividing by 100.
- (c) 70 — Method: estimate the median from the cumulative frequency table by interpolation: find its position, n ÷ 2, locate the class it falls in, then add the fraction of the way through that class (adjusted for the cumulative frequency reached before it) to the class's lower boundary. Working: there are 180 riders, so the median is at position 180 ÷ 2 = 90. Before the class 60 ≤ d < 80 the cumulative frequency is 60, and by the end of it, 120, so the 90th rider falls in this class; its frequency is 120 − 60 = 60 and its width is 80 − 60 = 20. The extra distance needed into the class is 90 − 60 = 30, and 30 ÷ 60 × 20 = 10, so the median is 60 + 10 = 70. Answer: the estimated median distance is 70 km. Watch which numbers the interpolation actually uses: reading off just the class's lower boundary, 60, ignores how far into the class the 90th rider falls; using the target position, 90, as the extra distance instead of subtracting the 60 riders already counted before the class gives 90 ÷ 60 × 20 = 30, so 60 + 30 = 90, overshooting by treating the whole position as if none of it had already been counted; and using the total number of riders, 180, instead of half of it as the target position lands in the very last class, giving an estimate of 130 km — further than any rider is known to have ridden by that point in the table.
- (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.
- (c) 69 — Method: multiply the mean by the number of values to find the total, then subtract the total of the known values. Working: the total of all five scores is 68 × 5 = 340. The total of the four known scores is 55 + 62 + 74 + 80 = 271. The fifth score is 340 − 271 = 69. Subtracting the other way round, 271 − 340 = −69, gives the right size answer with the wrong sign. Guessing that the missing score simply equals the mean, 68, ignores that the four known scores are not themselves centred on 68. Multiplying the mean by 4 instead of 5, 68 × 4 = 272, then 272 − 271 = 1, undercounts how many scores there are. Always multiply the mean by the TOTAL number of values before subtracting.
- (d) 648 kg — Method: to estimate a total from a sample, multiply the sample's mean by the number of items in the whole population, then check the units the question asks for. Working: 32.4 g × 20,000 = 648,000 g. Converting to kilograms, 648,000 ÷ 1,000 = 648 kg. This is only an estimate, not an exact total, because it assumes every one of the 20,000 packets has exactly the sample mean mass, when in reality individual packets vary above and below it. Giving 1.62 kg multiplies the mean by 50, the SAMPLE size, instead of by 20,000, the number of packets actually made that day — this finds the total mass of the 50 sampled packets, not the day's production. Giving 32.4 kg treats the sample mean itself, in grams, as if it already were the day's total mass in kilograms, skipping the scaling up altogether. Giving 648,000 kg correctly scales the mean up to the whole day's production but never converts the answer from grams to kilograms, leaving it 1,000 times too large. Always scale a sample's mean up by the SIZE OF THE WHOLE POPULATION, and always finish by checking the units the question asks for.
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