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.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 2.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 3.A scientist has grouped the lifetimes, in hours, of 300 batteries into classes of unequal width. She wants a diagram in which the number of batteries in a class is given by the area of its bar. Write down the type of diagram she should draw.
- 4.Two classes sat the same test. The 30 pupils in Class A had a mean mark of 72. The 20 pupils in Class B had a mean mark of 82. Work out the mean mark of all 50 pupils.
- 5.A box plot for the ages of 40 members of a gym is drawn from this five-number summary: minimum 15, lower quartile 22, median 29, upper quartile 38, maximum 61. Work out the interquartile range shown by this box plot.
- 6.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 7.The mean of 5 numbers is 8. Work out the total of the 5 numbers.
- 8.The times, t seconds, taken by 142 competitors to complete a lap are grouped like this: 0 ≤ t < 10, 20 competitors; 10 ≤ t < 25, 12 competitors; 25 ≤ t < 45, 50 competitors; 45 ≤ t < 75, 60 competitors. A histogram is drawn. Write down the class whose bar is the tallest.
- 9.The masses, m kg, of 150 boxes are summarised by these cumulative frequencies: m < 5, 18 boxes; m < 10, 52 boxes; m < 20, 96 boxes; m < 35, 130 boxes; m < 60, 150 boxes. Work out the number of boxes with a mass in the class 10 ≤ m < 20.
- 10.In a histogram of the distances, d metres, thrown by some athletes, the bar covering 20 ≤ d < 60 has a constant frequency density of 1.8 per metre. Estimate the number of throws of at least 20 metres but less than 35 metres.
- 11.Priya travels to work by one of two routes and records her journey times, in minutes, over several weeks. Route 1 has a median of 34 minutes and an interquartile range of 22 minutes. Route 2 has a median of 41 minutes and an interquartile range of 6 minutes. Priya wants the more reliable route for getting to an important meeting on time. Which route should she choose, and why?
- 12.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 13.The times, in seconds, taken by 11 runners to finish a race are given in order: 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65. Work out the upper quartile of these times.
- 14.Write down the statement that correctly describes the difference between correlation and causation.
- 15.The masses, m grams, of 100 apples are grouped like this: 100 ≤ m < 120, 10 apples; 120 ≤ m < 140, 30 apples; 140 ≤ m < 160, 40 apples; 160 ≤ m < 200, 20 apples. Estimate the median mass.
Answer key
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (c) A histogram, with frequency density up the vertical axis — Method: decide which diagram makes area stand for frequency, which is the property the question asks for. Working: on a histogram the vertical axis is frequency density, so the area of a bar is frequency density × class width, and that product is the frequency; this is exactly what is wanted, and it is what allows classes of unequal width to be shown fairly. Answer: a histogram, with frequency density up the vertical axis. The distractors: a bar chart plots frequency as the height, so with unequal widths a wide class would cover far more area than a narrow class holding the same number of batteries, and area would measure nothing; a cumulative frequency diagram plots running totals against upper class boundaries, so a point on it gives how many lie below a value rather than how many lie in a class; a pie chart shows each class as a share of the whole 300 and loses the class widths entirely, so no area on it is tied to a scale of hours.
- (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.
- (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.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (a) 40 — Method: the mean is the total divided by how many values there are, so rearranging gives total = mean × number of values. Working: the mean is 8 and there are 5 numbers, so the total is 8 × 5 = 40. Answer: 40, and checking, 40 ÷ 5 = 8, which is the mean given. The distractors: 13 comes from adding the mean and the count, 8 + 5, instead of multiplying them; 1.6 comes from dividing the mean by the count, 8 ÷ 5, which reverses the relationship; 8 comes from quoting the mean itself as the total, which is only true when there is a single number.
- (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.
- (d) 44 — Method: a cumulative frequency counts everything below a value, so the frequency of a class is the running total at the top of the class minus the running total at the bottom of it. Working: the running total below 20 kg is 96 and the running total below 10 kg is 52, so the number of boxes in the class 10 ≤ m < 20 is 96 − 52 = 44. Answer: 44 boxes. The distractors: 96 comes from quoting the running total at 20 kg itself, which counts every box below 20 kg rather than only those in this class; 34 comes from subtracting the wrong pair, 52 − 18, which gives the class 5 ≤ m < 10 instead; 54 comes from subtracting from the grand total, 150 − 96, which gives the boxes of 20 kg or more.
- (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.
- (c) Route 2, because its interquartile range is smaller — Method: for a journey where turning up on time matters, what matters is not the typical (median) time but how predictable it is — a smaller interquartile range means the middle half of journeys cluster closer together. Working: Route 1's median, 34 minutes, is in fact lower than Route 2's, 41 minutes, so Route 1 is faster on average; but Route 1's interquartile range, 22 minutes, is far larger than Route 2's, 6 minutes, so Route 1's times are much less predictable. Answer: Priya should choose Route 2, because its interquartile range is smaller, even though it is slower on average. Watch which statistic answers the question actually asked: Route 1 does not have the smaller interquartile range, Route 2 does, so picking Route 1 for that reason misreads the table; Route 1's median genuinely is the lower one, but a lower median answers 'which is faster', not 'which is more reliable'; and Route 2's median is not the lower one, so that claim about Route 2 is simply false.
- (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.
- (b) 60 — Method: for n ordered values, the upper quartile sits at position 3(n + 1) ÷ 4, counting from the smallest. Working: n + 1 = 11 + 1 = 12; 3 × 12 = 36 and 36 ÷ 4 = 9, so the upper quartile is the 9th value in the list 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65, which is 60. Answer: the upper quartile is 60 seconds. Watch which quartile you find: counting to the 6th value gives the median, 52, not the upper quartile; using 3 × 11 = 33 and 33 ÷ 4 = 8.25 without adding 1 to n first, then rounding down, reaches the 8th value, 58, not the 9th; and counting to the 3rd value uses the lower quartile's position, 45, the wrong end of the list.
- (d) Correlation is a link; causation is one causing the other — Method: the two words describe different claims — one is about a pattern in the data, the other is about what produced that pattern. Working: correlation says only that two quantities tend to change together, which is something a scatter graph can display; causation says that a change in one quantity actually brings about the change in the other, which needs evidence a scatter graph cannot supply, because a third quantity may be driving both. Answer: correlation is a link between the quantities, while causation is one quantity causing the change in another. The distractors: the statement giving causation as the link and correlation as the cause simply swaps the two words over; the statement that the words mean the same thing is the classic error of reading a correlation as proof of cause; the statement that a scatter graph shows causation but not correlation reverses what a scatter graph can do, since the pattern it displays is exactly the correlation.
- (c) 145 g — Method: find the position of the median from the total frequency, locate the class that contains it, then use linear interpolation inside that class, assuming the apples in it are spread evenly. Working: the median is the 100 ÷ 2 = 50th apple; the running totals are 10, then 10 + 30 = 40, then 40 + 40 = 80, so the 50th apple lies in the class 140 ≤ m < 160; it is the 50 − 40 = 10th of the 40 apples in that class, and the class is 20 g wide, so the median is 140 + (10 ÷ 40) × 20 = 140 + 5 = 145. Answer: an estimated median of 145 g. The distractors: 150 g comes from giving the midpoint of the class that contains the median instead of interpolating inside it; 140 g comes from stopping at the lower boundary of that class, which locates the class but not the value; 155 g comes from measuring the 5 g step down from the upper boundary, 160 − 5, instead of up from the lower boundary.
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