Printable · GCSE Foundation · ages 14-16
Statistics worksheet — GCSE Foundation
Fifteen questions across the statistics statements at Foundation 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 Foundation
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- (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.
- (a) 12 mm — Method: the range is the highest value minus the lowest value. Working: the highest rainfall is 15 mm and the lowest is 3 mm, so the range is 15 − 3 = 12 mm. Giving 15 mm alone states the highest value, not the range. Giving 3 mm alone states the lowest value, not the range. Sorting the six values, 3, 5, 9, 11, 12 and 15, and averaging the middle two, (9 + 11) ÷ 2 = 10 mm, finds the median, a completely different statistic. The range always needs BOTH the highest and the lowest value — never just one of them.
- (d) The mode, because colours cannot be added or ordered — Method: an average can only be used on data that supports the operation it needs. A mean needs the values to be added and divided, a median needs them to be placed in order, and a range needs one value to be taken away from another; a mode needs only counting, so it is the average available when the data are categories rather than numbers. Working: the data collected here are colours, silver, black, blue and red. The numbers 74, 52, 40 and 34 count the cars of each colour, they do not measure them, and 74 + 52 + 40 + 34 = 200 simply returns the size of the survey. No colour can be added to another, and there is no order that puts blue before red, so of the four averages only the one found by counting survives. Answer: the mode, because colours cannot be added or ordered, and the mode is silver. The distractors: the mean is said to use all 200 colours, and a mean of the four frequencies, 200 ÷ 4 = 50, is a number of cars rather than a colour, so it describes nothing about a typical car; the median is said to put the colours in order, but ordering the frequencies 34, 40, 52, 74 orders the counts, not the colours, and gives 46, again a number of cars; the range is not an average at all, and 74 − 34 = 40 measures the gap between the commonest and rarest counts, which is a measure of spread.
- (d) 24 — Method: for a sample in proportion to the population, apply the same fraction that each group makes up of the whole population to the size of the sample. Working: women make up 300 out of the 500 members, a fraction of 300 ÷ 500 = 0.6. Applying that fraction to the sample of 40 gives 0.6 × 40 = 24 women. Splitting the sample evenly, 40 ÷ 2 = 20, ignores that the club has more women than men and treats the two groups as equal in size, which they are not. Misreading the sample size as 50 instead of 40, then applying the 3:2 ratio of women to men, 3 ÷ 5 × 50 = 30, uses the right ratio but the wrong sample total. Working out the number of MEN instead of women, 200 ÷ 500 × 40 = 16, answers a different question — how many men, not how many women, belong in the sample. Always apply each group's own share of the population to the sample size, and check which group the question is actually asking about.
- (a) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (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.
- (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) Positive correlation — Method: the type of correlation is named from the direction the points take as the scatter graph is read from left to right. Working: the points rise from left to right, so as the arm span read on the horizontal axis increases, the height read on the vertical axis increases as well; two quantities that increase together show positive correlation. Answer: positive correlation. The distractors: negative correlation comes from naming the direction the wrong way round, since a negative correlation needs the points to fall as the graph is read from left to right; no correlation comes from treating points that are spread out rather than sitting exactly on a line as though they showed no relationship; direct proportion comes from confusing a rising trend with proportion, which would additionally need the line through the points to pass through the origin and would mean doubling one quantity doubles the other.
- (a) Negative correlation — As the age of the car increases, the points fall towards a lower value, so the value decreases as the age increases. This falling pattern is a negative correlation. A positive correlation would show the points rising together instead. No correlation would apply only if the points showed no pattern at all, and correlation is not the same as causation — strong causation is not a type of correlation.
- (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) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
- (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.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (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.
- (c) Thursday — Method: on a bar chart the tallest bar belongs to the greatest value, so compare the five temperatures and then read off the day that the greatest one belongs to. Working: the temperatures are 20 °C, 22 °C, 18 °C, 25 °C and 23 °C; in order of size these are 18, 20, 22, 23 and 25, so the greatest temperature is 25 °C, and the day recorded with 25 °C is Thursday. Answer: Thursday — the answer is a day, not a temperature. The distractors: Friday comes from stopping at the final temperature listed instead of comparing all five; Wednesday comes from picking out the shortest bar, 18 °C, and so answering for the lowest temperature rather than the highest; Monday comes from writing down the first day in the chart without comparing any of the temperatures at all.
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