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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- (d) 75 — Method: to find a total from a bar chart, add together the height of every bar. Working: 12 + 18 + 9 + 15 + 21 = 75 books. Leaving out Wednesday's bar by mistake, 12 + 18 + 15 + 21 = 66, misses one day out of the total. Giving 21 states only Friday's bar, the tallest one, not the total of all five days. Dividing the total by the number of days, 75 ÷ 5 = 15, finds the mean number of books per day, not the total borrowed. Add up every single bar — do not stop at the biggest one, and do not divide once you have added them all.
- (a) The temperature is rising steadily — Method: the trend of a line graph is the overall direction of the readings as time goes on, found by comparing each reading with the one before it. Working: from 10 °C to 14 °C is a rise of 4 °C, and the same comparison from 14 °C to 18 °C, from 18 °C to 22 °C, from 22 °C to 26 °C and from 26 °C to 30 °C gives a rise of 4 °C every time; every reading is greater than the one before it and none of them falls. Answer: the temperature is rising steadily — steadily because the rise is the same size each hour. The distractors: falling steadily comes from reading the six values from right to left, which reverses the direction of time; stays the same comes from noticing that the step of 4 °C is the same each hour and describing the step as constant instead of the temperature; rises and then falls comes from assuming that a line graph has to turn at some point rather than reading the values that are actually given.
- (a) 26 — Method: a line of best fit lets one quantity be predicted from the other, so the score is substituted into the equation of the line and the resulting inequality is solved for the number of hours. Working: a score of at least 80 means 2.5x + 15 ≥ 80; taking 15 from both sides gives 2.5x ≥ 65, and dividing both sides by 2.5 gives x ≥ 26, so the least whole number of hours is 26. Checking, 2.5 × 26 + 15 = 80, which does reach the target. Answer: 26 hours — and this is only an estimate, because a line of best fit predicts a trend rather than an individual result, and a prediction made outside the range of hours the pupils actually revised for would be an extrapolation and less reliable still. The distractors: 27 comes from reaching 26 and then rounding up again, although 26 hours already gives a score of exactly 80; 32 comes from 80 ÷ 2.5, which ignores the 15 in the equation of the line; 38 comes from (80 + 15) ÷ 2.5, that is from adding the 15 instead of subtracting it when rearranging.
- (a) Chloe's marks are far more spread out than Ben's — Method: a mean reports where a set of values sits, and two sets can sit in the same place while behaving quite differently, so a measure of spread has to be worked out as well. Working: Ben's marks add to 62 + 64 + 65 + 66 + 68 = 325 and 325 ÷ 5 = 65; Chloe's add to 40 + 52 + 65 + 78 + 90 = 325 and 325 ÷ 5 = 65, so the two means agree, as the question says. The ranges do not: Ben's is 68 − 62 = 6 marks, while Chloe's is 90 − 40 = 50 marks. Ben's five marks all sit within 3 marks of 65; Chloe's lowest is 25 marks below it and her highest 25 marks above it. Answer: Chloe's marks are far more spread out than Ben's, which is exactly what the mean cannot show. The distractors: saying Ben's marks are more spread out comes from subtracting in the order the values are written, 62 − 68 = −6 against 40 − 90 = −50, and then reading −6 as the larger spread; saying Chloe scored far more marks in total assumes a wider set of marks must add to more, when both totals are 325; saying the two sets vary by the same amount assumes that equal means force equal spread, when the two ranges are 6 and 50.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (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) 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) 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) Only internet users reach the website; others are excluded. — Method: a sample is biased when it systematically leaves out part of the population, or systematically over-represents another part. Working: anyone without internet access, or who does not visit the council's website, has NO chance of being included — the sample is drawn only from internet-using residents, which is not the whole town. Saying too many people might respond because the survey is free confuses bias with sample size — bias is about who CAN be reached, not how many respond. Saying people might lie describes a different problem, response honesty, not who was sampled in the first place. Saying online surveys cannot be anonymous is not a reason connected to bias at all. A sample is biased when part of the population has no chance of being included, whatever the reason for that.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (b) Interpolation — The salary is being estimated for a value of x between 1 and 15, which is inside the range of x-values that were actually plotted, so this is interpolation. Extrapolation would apply if the estimate used a value of x below 1 or above 15, outside the plotted range. Correlation describes the relationship between the two variables, not the reliability of an estimate, and causation describes one variable actually causing a change in the other, which is a different idea altogether — neither is the word being asked for here.
- (c) Strong negative correlation — Method: correlation is described by two things — the direction the points take as the graph is read from left to right, and how closely the points lie to a single straight line. Working: reading the pairs in order of age, the ages rise 14, 18, 23, 27, 31, 36, 42, 49 while the scores fall 92, 88, 85, 80, 78, 74, 70, 65; the score falls at every single step, with no reversal anywhere, so the points fall from left to right and lie close to a straight line. Answer: strong negative correlation — negative for the falling direction, strong because every point follows the pattern. The distractors: strong positive correlation comes from noticing a clear pattern and calling any clear pattern positive, without checking the direction; weak negative correlation comes from reading the direction correctly but judging points that do not lie exactly on a straight line to be only loosely related, when these eight fall without a single exception; no correlation comes from reading a falling trend as though it showed no relationship at all, when a falling trend is itself a relationship.
- (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) 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) 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.
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