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.
Answer key: Statistics worksheet — GCSE Foundation
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- (b) Ethan, 10 seconds — Method: over the same distance the fastest runner is the one who takes the least time, so the smallest time in the table is found first and the name is then read from the same row. Working: the four times are 12 seconds, 15 seconds, 10 seconds and 14 seconds; in order of size these are 10, 12, 14 and 15, so the least time is 10 seconds, and the row holding 10 seconds is the row for Ethan. Answer: Ethan, 10 seconds — the time is in seconds, and a smaller time means a faster runner. The distractors: Grace with 15 seconds comes from taking the largest number in the table to mean the fastest runner, which reverses the relationship between time and speed over a fixed distance; Oliver with 12 seconds comes from writing down the first row of the table without comparing the four times; Ethan with 15 seconds comes from identifying the right runner but then reading the time from a different row of the table.
- (a) 144° — Method: a pie chart shares the 360° at its centre between the categories in proportion to their frequencies, so the angle of a sector is that category's fraction of the total multiplied by 360°. Working: the sector stands for 40 pupils out of 100, which is the fraction 40/100; one pupil is worth 360 ÷ 100 = 3.6°, so 40 pupils are worth 40 × 3.6° = 144°. Answer: 144°, and the answer is an angle in degrees, not a number of pupils. The distractors: 40° comes from sharing out 100 instead of 360, so the percentage is written straight down as a number of degrees; 216° comes from working out the angle for the other 60 pupils, 60 × 3.6°, which is the rest of the pie chart; 180° comes from assuming that the tallest line must stand for half of the pupils and so take half of the chart.
- (b) 1 — The four frequencies are 4, 7, 6 and 3 matches, and the largest of these is 7, which corresponds to 1 goal, so the modal number of goals is 1. Choosing 7 confuses the frequency, how many matches, with the number of goals itself. Choosing 2 uses the second-largest frequency, 6 matches, instead of the largest. Choosing 3 uses the smallest frequency, which belongs to the fewest matches, not the most.
- (a) A vertical line chart (discrete numerical data) — The number of pets is discrete numerical data — whole-number values such as 0, 1, 2, 3 or 4 — recorded for one variable, so a vertical line chart is the chart specified for this kind of data. A bar chart is used for categorical data, such as favourite colour, not numerical values counted like this. A pie chart shows proportions of a whole and does not show the frequency of each separate value. A scatter graph compares two different variables against each other, and only one variable, the number of pets, is recorded here.
- (d) 0 — Method: the range is the largest value minus the smallest value, whatever those two values turn out to be. Working: every value is 10, so the largest value is 10 and the smallest value is 10 as well, and the range is 10 − 10 = 0. Answer: 0 — a range of nothing says the data do not vary at all. The distractors: 10 comes from writing down the repeated value itself instead of the difference between the extremes; 20 comes from adding the largest and the smallest, 10 + 10, instead of subtracting; 40 comes from adding all four values, which gives the total sold and not a measure of spread.
- (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.
- (c) 40 minutes — Method: for grouped data, estimate the mean using the midpoint of each class — multiply each midpoint by its frequency, add the results, then divide by the total frequency. Working: the midpoints are 10, 30, 50 and 70 minutes. 10 × 5 = 50. 30 × 10 = 300. 50 × 10 = 500. 70 × 5 = 350. Σfx = 50 + 300 + 500 + 350 = 1200. Σf = 5 + 10 + 10 + 5 = 30. Estimated mean = 1200 ÷ 30 = 40 minutes. Using the upper boundary of each class instead of the midpoint — 20 × 5 = 100, 40 × 10 = 400, 60 × 10 = 600, 80 × 5 = 400 — gives a total of 1500 and an estimate of 1500 ÷ 30 = 50 minutes, too high because a boundary is not the middle of the class. Averaging the frequencies themselves, 5, 10, 10 and 5, ignores the times altogether and gives 7.5. Stopping after Σfx = 1200 without dividing by the total frequency gives a number far too large to be a time in minutes. Always find the midpoint of each class before multiplying by the frequency, and always divide by Σf at the end.
- (d) 648 kg — Method: to estimate a total from a sample, multiply the sample's mean by the number of items in the whole population, then check the units the question asks for. Working: 32.4 g × 20,000 = 648,000 g. Converting to kilograms, 648,000 ÷ 1,000 = 648 kg. This is only an estimate, not an exact total, because it assumes every one of the 20,000 packets has exactly the sample mean mass, when in reality individual packets vary above and below it. Giving 1.62 kg multiplies the mean by 50, the SAMPLE size, instead of by 20,000, the number of packets actually made that day — this finds the total mass of the 50 sampled packets, not the day's production. Giving 32.4 kg treats the sample mean itself, in grams, as if it already were the day's total mass in kilograms, skipping the scaling up altogether. Giving 648,000 kg correctly scales the mean up to the whole day's production but never converts the answer from grams to kilograms, leaving it 1,000 times too large. Always scale a sample's mean up by the SIZE OF THE WHOLE POPULATION, and always finish by checking the units the question asks for.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
- (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.
- (a) 23 minutes — Method: the equation of a line of best fit converts a value of x into a predicted value of y, so substitute the known number of pages for x and evaluate. Working: x is the number of pages, so put x = 10 into y = 2x + 3. Multiplication is carried out before addition, so 2 × 10 + 3 = 23. Answer: 23 minutes, and it is a prediction of the trend rather than a promise about any one chapter. The distractors: 20 minutes comes from working out 2 × 10 and stopping there, leaving out the 3 that the line adds; 26 minutes comes from reading the equation as y = 2(x + 3), adding first and then doubling, so 2 × 13 = 26; 13 minutes comes from adding 10 and 3 and never using the gradient at all, which treats the 2 as though it were not there.
- (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) 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.
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