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.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.
- 2.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
- 3.Five friends have heights, in cm, of 150, 152, 155, 158 and 160. A sixth friend, with a height of 170 cm, joins the group. Write down what happens to the mean and the range of the heights once this sixth friend is included.
- 4.For each of 30 fires in one county, the number of fire engines sent and the cost of the damage were recorded. The scatter graph of the data shows strong positive correlation. A newspaper prints the headline "Sending more fire engines causes more damage". Is the newspaper right? Give a reason for your answer.
- 5.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.
- 6.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
- 7.A pictogram shows the number of cars sold by a garage in Plymouth each month, where each whole car symbol represents 8 cars sold and a half symbol represents 4 cars. March shows 3 whole symbols and one half symbol. Work out how many cars were sold in March.
- 8.A bus company records the delay, d minutes, of 250 buses: 0 ≤ d < 2, 60 buses; 2 ≤ d < 5, 90 buses; 5 ≤ d < 10, 75 buses; 10 ≤ d < 20, 25 buses. The company refunds the fare whenever a bus is more than 8 minutes late. Estimate the number of refunds it must pay.
- 9.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.
- 10.A garden centre's sales, in thousands of pounds, at the end of each quarter last year were: quarter 1 — 18, quarter 2 — 34, quarter 3 — 30, quarter 4 — 22. Work out the increase in sales from quarter 1 to the quarter with the highest sales.
- 11.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 12.A sports centre in Ipswich has 2,000 members. It wants to know what its members think of its opening hours, so it asks a random sample of 100 of them. Write down what the population is in this survey.
- 13.A stacked bar for one day at a café in Norwich shows total drink sales of 50 drinks, split into three parts: 22 were tea, 15 were coffee and the rest were hot chocolate. Work out the number of hot chocolates sold.
- 14.There are 10,000 pupils in a city. A researcher picks a sample of 500 of them by drawing names at random from a list of every pupil in the city. Give the reason why this method gives a representative sample.
- 15.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
Answer key
- (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.
- (c) The median, £160,000, as one very high price lifts the mean — Method: find both averages, then choose the one that sits closer to the bulk of the data. Working: in order the prices are 140,000, 150,000, 160,000, 170,000 and 580,000, so the median is the third of the five, £160,000. For the mean, 140,000 + 150,000 + 160,000 + 170,000 + 580,000 = 1,200,000 and 1,200,000 ÷ 5 = 240,000, so the mean is £240,000. Four of the five houses sold for £170,000 or less, so a reader told that a typical price is £240,000 would expect to pay at least £70,000 more than any of those four cost. Answer: the median, £160,000, as one very high price lifts the mean. The distractors: £580,000 is the middle value of the list as it is printed, which is the median only when the values have first been put in order; £240,000 is the mean, chosen on the ground that a median ignores three of the five prices, but a median uses all five to find which one is central and is then untroubled by how extreme the outer values are; £155,000 comes from deleting the £580,000 house and taking the mean of what is left, since 140,000 + 150,000 + 160,000 + 170,000 = 620,000 and 620,000 ÷ 4 = 155,000, but a real sale may not be thrown away merely for being large.
- (d) Both the mean and the range increase. — The original mean is 150 + 152 + 155 + 158 + 160 = 775, and 775 ÷ 5 = 155 cm; the original range is 160 − 150 = 10 cm. Including the new height of 170 cm gives a new total of 775 + 170 = 945, and 945 ÷ 6 = 157.5 cm, which is higher than 155 cm, and a new range of 170 − 150 = 20 cm, which is higher than 10 cm, so both the mean and the range increase. Saying the range stays the same ignores that 170 cm is a new, higher maximum than the old 160 cm. Saying the mean stays the same ignores that 170 cm is above the original mean of 155 cm, which pulls the average up. Saying both decrease is the opposite of what happens here.
- (a) No, the size of the fire affects both of the quantities — Method: correlation says that two quantities change together; a claim that one of them produces the other is a further claim, and it needs evidence that a scatter graph on its own cannot give. Working: the graph does show strong positive correlation, so more engines did go with greater damage. But neither quantity was set by the researchers: both were decided by how large the fire was. A large blaze brings many appliances and also destroys a great deal, while a small one brings few and destroys little, so a third quantity is driving both of the recorded ones. Answer: no, because the size of the fire affects both of the quantities. The distractors: saying the correlation is negative contradicts the graph, which shows the two quantities rising together, and reaching the right verdict from a false reading of the data is not the reason the mark is for; saying that strong positive correlation shows one quantity causes the other is the assumption the question exists to test, and no strength of correlation can establish cause; saying the points lie close to the line of best fit describes how strong the correlation is, and strength and cause are different matters entirely.
- (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.
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.
- (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.
- (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.
- (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.
- (b) All 2,000 members of the sports centre. — Method: in a survey, the population is the whole group the survey is trying to find out about, and the sample is the smaller group actually asked. Working: this survey wants to know what the sports centre's members think, so the population is every one of the 2,000 members — whether or not they were personally asked. Saying the population is the 100 members who were asked names the sample, not the population; the sample is drawn FROM the population, so it is smaller than it, not the same as it. Saying the population is everybody who lives in Ipswich widens the group far beyond who the survey is actually about — plenty of Ipswich residents are not members of the sports centre at all, so they are outside this survey altogether. Saying the population is the members who say they are unhappy confuses the population with a result of the survey: whether a member turns out to be happy or unhappy is something the survey finds out, not part of the definition of who is being studied. The population is always the whole group the question is about, before any sampling or any results come in.
- (b) 13 — The total is 50, and the two known parts are 22 (tea) and 15 (coffee), so 50 − 22 − 15 = 13 hot chocolates. Choosing 28 comes from 50 − 22, subtracting only the tea and forgetting the coffee. Choosing 35 comes from 50 − 15, subtracting only the coffee and forgetting the tea. Choosing 37 comes from 22 + 15, which finds how many drinks were tea or coffee, not the number left over for hot chocolate.
- (c) Because every pupil has an equal chance of being picked — Method: whether a sample represents its population is decided by the selection method, not by the size of the sample, so ask whether the method gives every member of the population the same chance of being chosen. Working: the names are drawn at random from a list of all 10,000 pupils, so each pupil has the same chance, 500 out of 10,000, of being drawn, and no group of pupils is more likely to appear than any other; that is what keeps bias out of the sample. Answer: because every pupil has an equal chance of being picked. The distractors: the reply about 5% treats the sampling fraction as the test of fairness, but a badly chosen 5% is still biased and a well chosen 1% is not; the reply about 500 being large enough makes size the test instead, which is the same mistake in another form, since a large sample drawn from one school would still misrepresent the city; the reply about the most willing pupils describes self-selection, which hands the choice of who is in the sample to the pupils who feel most strongly about the question.
- (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
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