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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- (c) 20 kg ≤ mass < 30 kg — The modal class is the class with the highest frequency. Reading the plotted points, the frequencies are 6, 10, 16, 6 and 2, so the highest frequency is 16, plotted at the midpoint 25. A class of width 10 centred on 25 runs from 25 − 5 = 20 to 25 + 5 = 30, so the modal class is 20 kg ≤ mass < 30 kg. Writing '25 kg' gives only the midpoint, not the class — the modal class is an interval, not a single value. '10 kg ≤ mass < 20 kg' is the class before the peak, centred on 15, which has frequency 10, not the highest. '30 kg ≤ mass < 40 kg' is the class after the peak, centred on 35, which has frequency 6, not the highest.
- (b) 4 — The difference, Bristol minus Leeds, on each day is: Monday 4 − 3 = 1, Tuesday 4 − 5 = −1, Wednesday 6 − 2 = 4, Thursday 2 − 4 = −2. The greatest amount by which Bristol exceeded Leeds is 4 hours, on Wednesday. Choosing 1 takes Monday's smaller positive difference instead of the greatest one. Choosing 2 takes the size of Thursday's difference, but that is the amount by which Leeds exceeded Bristol, the opposite direction to the one asked for. Choosing 6 takes Bristol's raw figure on Wednesday without subtracting Leeds's 2 hours first.
- (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.
- (a) £16,000 — Method: first find the quarter with the highest sales figure, then subtract quarter 1's sales from it — remembering that every figure is given in THOUSANDS of pounds. Working: the highest sales figure is quarter 2, at £34,000 (34 thousand pounds). The increase from quarter 1 is £34,000 − £18,000 = £16,000. Giving £34,000 reads off the highest sales figure on its own, without subtracting quarter 1's sales — that is the highest quarter's total, not the increase. Giving £12,000 uses quarter 3's sales, 30, the SECOND-highest figure, instead of quarter 2's 34, the actual highest — 30 − 18 = 12, but quarter 3 is not the quarter with the highest sales. Giving £16 gets the subtraction right, 34 − 18 = 16, but forgets that every figure in the question is in thousands of pounds, so the increase is £16,000, not £16. Always identify the correct quarter FIRST, and always check the units the numbers are given in before writing your final answer.
- (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.
- (d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
- (c) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
- (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.
- (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.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (c) No, the mode here is the lowest value of the nine — Method: an average is meant to stand for the data as a whole, so test any proposed average by asking how many values it sits near. Working: the value 4 appears three times and every other count appears once, so 4 is indeed the mode. But those three hours are the quiet ones at the start of the day, and the other six counts run from 11 up to 25; putting the nine counts in order, the middle one is the fifth, which is 13. So the mode sits at the very bottom of the data, with six of the nine hours far above it. Answer: no, because the mode here is the lowest value of the nine, so it describes the quiet opening hours rather than a typical hour. The distractors: saying the mode can only be used when no value repeats reverses the definition, since a mode exists only because a value does repeat; saying the mode is the value that occurs most often is a correct definition, but being the commonest value does not make a value typical when it lies at one end of the data; saying the mode is the best average for any list of numbers ignores the fact that mean, median and mode each describe a population well in different circumstances.
- (b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (c) 76 — Method: the frequency of each class is the area of its bar, frequency density × class width, so work out all three frequencies and add them. Working: the widths are 5, 10 and 25 kg, so the frequencies are 4 × 5 = 20, 2.6 × 10 = 26 and 1.2 × 25 = 30; the total is 20 + 26 + 30 = 76. Answer: 76 dogs. The distractors: 7.8 comes from adding the three frequency densities, 4 + 2.6 + 1.2, treating each height as though it were a count; 107 comes from using each upper class boundary as the width, giving 4 × 5, 2.6 × 15 and 1.2 × 40; 39 comes from using the first class width, 5, for every bar, which ignores the unequal intervals and gives 20 + 13 + 6.
- (c) 76 marks — Method: a mean of means only works when the groups are the same size, so rebuild each class's total mark, add the totals and divide by the number of pupils altogether. Working: Class A scored 30 × 72 = 2160 marks and Class B scored 20 × 82 = 1640 marks, giving 2160 + 1640 = 3800 marks between 50 pupils, so the overall mean is 3800 ÷ 50 = 76 marks. Answer: 76 marks. The distractors: 77 marks comes from averaging the two class means, (72 + 82) ÷ 2, which ignores the different class sizes; 78 marks comes from attaching each mean to the other class's size, (30 × 82 + 20 × 72) ÷ 50; 3800 marks comes from stopping at the combined total and never dividing by 50.
- (a) 22 — Method: the two subject totals overlap, because every pupil who passed both subjects has been counted once in the maths total and once again in the science total; adding the totals therefore counts those pupils twice, and the overlap has to be taken off once. Working: 18 + 12 = 30, and the 8 pupils who passed both have been counted twice in that 30, so the number who passed at least one subject is 30 − 8 = 22. Answer: 22 pupils, a count of pupils, and it is less than the 30 in the class, which leaves 8 pupils who passed neither. The distractors: 30 comes from adding the two subject totals and never removing the overlap, so it counts the 8 pupils twice; 14 comes from taking the 8 away twice, 18 + 12 − 8 − 8, removing an overlap that was only counted twice once too often; 18 comes from writing down the larger of the two subject totals on its own, which leaves out every pupil who passed science but not maths.
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