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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- (c) mean = 8, median = 8, mode = 8 — Method: work out each measure separately — the mean is the total divided by how many values there are, the median is the middle value once the data are in order, and the mode is the value that occurs most often. Working: the total is 8 + 8 + 8 + 8 = 32 and there are 4 marks, so the mean is 32 ÷ 4 = 8; in order the marks read 8, 8, 8, 8, and the mean of the middle pair is (8 + 8) ÷ 2 = 8; the value 8 occurs 4 times and no other value occurs at all, so the mode is 8. Answer: mean = 8, median = 8, mode = 8 — when every value in a data set is the same, all three measures of central tendency take that value. The distractors: a mean of 32 comes from stopping at the total and never dividing by 4; a mode of 4 comes from writing down how many times 8 occurs instead of the value that occurs; a mean of 2 comes from dividing a single value, 8, by the 4 marks instead of dividing the total by 4.
- (a) 5 pupils are far too few to represent 900 pupils — Method: a sample can only support a claim about a population if it is chosen fairly and if it is large enough for the pattern in it to be more than chance. Working: Priya's method of choosing is fair, because the 5 pupils were picked at random, so every pupil had the same chance of being asked. The difficulty is the size: 900 ÷ 5 = 180, so each pupil she asks stands for 180 pupils. If two of the five happen to play in the same netball team, netball takes 40% of her sample on the strength of two answers, and a second sample of 5 could easily give a different favourite sport. Answer: 5 pupils are far too few to represent 900 pupils. The distractors: saying the pupils were not chosen at random contradicts the question, which states that they were; saying the 5 may each name a different sport describes what often happens in a small sample, but disagreement is not the fault, since 5 pupils who all named the same sport would be just as weak a basis for a claim about 900; saying a sample must hold at least half of the population is an invented rule, and a properly chosen sample of a few hundred can describe a population of many thousands.
- (c) 10 — Method: work out the mean that can be found straight away, then use total = mean × number of values on the group of three to find the missing number. Working: the mean of 15 and 25 is (15 + 25) ÷ 2 = 40 ÷ 2 = 20, so the group of three must also have a mean of 20; three numbers with a mean of 20 have a total of 20 × 3 = 60, and 20 + 30 = 50 of that total is already accounted for, so x = 60 − 50 = 10. Answer: 10, and checking, (10 + 20 + 30) ÷ 3 = 20. The distractors: 20 comes from working out the mean the two groups share and writing that down as x; −10 comes from dividing the group of three by 2 instead of by 3, which gives x + 50 = 40; 70 comes from reading the total 15 + 25 = 40 as the mean of the pair, which sets the target total at 120 and leaves x = 70.
- (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) The relationship between two variables — Method: what a diagram shows is decided by what has to be known before a single mark can be plotted on it. Working: every point on a scatter graph is plotted from a pair of measurements taken from the same person or object, one read on the horizontal axis and one on the vertical axis; having two measurements for each point is what makes it possible to look for a pattern between them, and the pattern between two variables is what the graph displays. Answer: a scatter graph shows the relationship between two variables. The distractors: the frequency of each single value is what a bar chart or a vertical line chart shows, and it needs only one list of values; how a total is shared between categories is what a pie chart shows; how one quantity changes over time is what a time series line graph shows, in which one of the two axes is always time.
- (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.
- (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) 35 minutes — Chloe arrives at 09:30, so she needs the first departure time that is at or after 09:30. The bus at 09:25 has already left five minutes before she arrives, so it is missed, not caught; the next one on the timetable is 10:05. From 09:30 to 10:00 is 30 minutes, then a further 5 minutes to 10:05, so the wait is 35 minutes. Giving 40 minutes measures the gap from the missed 09:25 bus to the 10:05 bus, counting from the bus Chloe missed instead of from the time she arrived. Giving 5 minutes uses the missed 09:25 bus, working out 09:30 − 09:25 the wrong way round. Giving 45 minutes comes from treating the times as decimals — 10.05 − 9.30 = 0.75, then × 60 = 45 — instead of working in hours and minutes, where 05 minutes past the hour is not 0.05 of an hour.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (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) 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.
- (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) 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.
- (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.
- (a) A straight line sloping down from left to right — Method: a line of best fit is a straight line drawn to follow the trend of the points, so its slope is decided by the direction of the relationship between the two quantities. Working: as the price rises, the satisfaction score falls, so the points start high on the left of the graph and finish low on the right; the straight line that follows them therefore slopes downwards as the graph is read from left to right, which is the line of a negative correlation. Answer: a straight line sloping down from left to right. The distractors: a line sloping up comes from reading a falling relationship as a rising one; a horizontal line comes from expecting no correlation, since a horizontal line says the satisfaction score does not change as the price changes; a curve passing through every point comes from thinking a line of best fit has to touch all of the plotted points, when it is a single straight line drawn through the middle of them.
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