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.
Answer key: Statistics worksheet — GCSE Higher
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- (d) Route 1, as its times vary by 6 minutes rather than 20 — Method: work out an average and a measure of spread for each route, then decide which matters to a commuter who must arrive on time every day. Working: for Route 1, 22 + 23 + 24 + 24 + 25 + 25 + 26 + 26 + 27 + 28 = 250 and 250 ÷ 10 = 25, so the mean is 25 minutes, and the range is 28 − 22 = 6 minutes. For Route 2, 18 + 19 + 20 + 20 + 21 + 22 + 26 + 30 + 36 + 38 = 250 and 250 ÷ 10 = 25, so the mean is also 25 minutes, but the range is 38 − 18 = 20 minutes. The means give no reason to prefer either route; the spreads do, because a commuter who must never be late has to allow for the worst day, which is 28 minutes on Route 1 and 38 minutes on Route 2. Answer: Route 1, as its times vary by 6 minutes rather than 20. The distractors: saying Route 2 has the lower mean assumes that its quicker-looking early times must pull the average down, when both routes total 250 minutes over the ten days; choosing Route 2 for its fastest journey of 18 minutes judges a route by its best day, and the commuter has to survive its worst; saying either route will do uses the equal means and ignores the spread altogether, which is the one thing that separates the two routes.
- (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.
- (c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
- (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) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (d) 16 — Method: on a box plot the interquartile range is the width of the box itself, upper quartile take away lower quartile. Working: the upper quartile is 38 and the lower quartile is 22, so 38 − 22 = 16. Answer: the interquartile range is 16 years. Watch which part of the box plot you are reading: the whole line from whisker to whisker gives the range, 61 − 15 = 46; the left half of the box alone gives median take away lower quartile, 29 − 22 = 7; and the right half of the box alone gives upper quartile take away median, 38 − 29 = 9 — neither half is the interquartile range on its own.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (d) 45 — Method: a cumulative frequency is a running total — it counts everybody in every class up to and including the one that ends at the value given. Working: the classes that lie wholly below 30 minutes are 0 ≤ t < 10, 10 ≤ t < 20 and 20 ≤ t < 30, with frequencies 6, 14 and 25, so the running total is 6 + 14 = 20 and then 20 + 25 = 45. Answer: 45 people took less than 30 minutes. The distractors: 25 comes from quoting the frequency of the class 20 ≤ t < 30 on its own instead of the running total; 65 comes from accumulating one class too many and including 30 ≤ t < 40, which is 45 + 20; 35 comes from accumulating from the top downwards, 15 + 20, which counts the people who took 30 minutes or more rather than fewer.
- (c) 75 — Method: the number in a class is the area of its bar, frequency density × class width, so work out the frequency of each class that lies at or above 10 minutes and add them. Working: the class 10 ≤ t < 25 is 15 minutes wide with a frequency density of 3.2, giving 3.2 × 15 = 48 members; the class 25 ≤ t < 55 is 30 minutes wide with a frequency density of 0.9, giving 0.9 × 30 = 27 members; the total charged is 48 + 27 = 75. Answer: 75 members pay the extra charge. The distractors: 4.1 comes from adding the two frequency densities, 3.2 + 0.9, as though each height were a count; 93 comes from including the class 0 ≤ t < 10 as well, 1.8 × 10 = 18 added to 48 and 27, which charges every member; 27 comes from using only the class 25 ≤ t < 55 and forgetting that 10 ≤ t < 25 is also at or above 10 minutes.
- (d) 18 minutes — Method: the lower quartile is the 80 ÷ 4 = 20th value and the upper quartile is the 3 × 80 ÷ 4 = 60th value; locate each inside its class by linear interpolation, then subtract. Working: the 20th value lies between the running totals 8 and 28, so it is in the class 10 ≤ t < 20, which holds 20 journeys across 10 minutes, and it is the 20 − 8 = 12th of them, giving 10 + (12 ÷ 20) × 10 = 16 minutes; the 60th value lies between the running totals 52 and 72, so it is in the class 30 ≤ t < 40, which also holds 20 journeys across 10 minutes, and it is the 60 − 52 = 8th of them, giving 30 + (8 ÷ 20) × 10 = 34 minutes; subtracting, 34 − 16 = 18. Answer: an estimated interquartile range of 18 minutes. The distractors: 20 minutes comes from taking the lower boundaries of the two quartile classes, 30 − 10, which locates the classes but never the values inside them; 40 minutes comes from subtracting the two positions, 60 − 20, instead of the two times; 22 minutes comes from interpolating downwards from each upper boundary rather than upwards from each lower boundary, giving 20 − 6 = 14 and 40 − 4 = 36.
- (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.
- (a) (30, 60) — Method: a cumulative frequency point is plotted at the upper boundary of its class, paired with the running total of all the frequencies up to and including that class. Working: the running totals are 7, then 7 + 19 = 26, then 26 + 34 = 60, then 60 + 40 = 100; the class 20 ≤ t < 30 has upper boundary 30, and the running total there is 60. Answer: the point for that class is plotted at 30 seconds against a cumulative frequency of 60. The distractors: (25, 60) comes from plotting at the class midpoint, which is what a frequency polygon uses and not what a cumulative frequency diagram uses; (30, 34) comes from plotting the class frequency, 34, rather than the running total; (20, 60) comes from plotting at the lower boundary of the class, which would claim that 60 calls took less than 20 seconds when only 26 did.
- (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.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
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