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) 72 — Method: when a pie chart is divided into equal sectors, each sector stands for the same share of the people asked, so the fraction of the sectors that are shaded is also the fraction of the people. Working: 6 sectors out of 10 are purple, which is the fraction 6/10 of the whole pie chart; one tenth of the 120 people is 120 ÷ 10 = 12 people, so six tenths is 6 × 12 = 72 people. Answer: 72 people, a count of people rather than a number of sectors. The distractors: 48 comes from working with the 4 sectors that are not purple, 4 × 12, and so answering for the wrong part of the chart; 60 comes from turning the fraction 6/10 into 60% and then writing the 60 down as though it were a number of people; 6 comes from writing down the number of purple sectors instead of the number of people those sectors stand for.
- (d) Correlation is a link; causation is one causing the other — Method: the two words describe different claims — one is about a pattern in the data, the other is about what produced that pattern. Working: correlation says only that two quantities tend to change together, which is something a scatter graph can display; causation says that a change in one quantity actually brings about the change in the other, which needs evidence a scatter graph cannot supply, because a third quantity may be driving both. Answer: correlation is a link between the quantities, while causation is one quantity causing the change in another. The distractors: the statement giving causation as the link and correlation as the cause simply swaps the two words over; the statement that the words mean the same thing is the classic error of reading a correlation as proof of cause; the statement that a scatter graph shows causation but not correlation reverses what a scatter graph can do, since the pattern it displays is exactly the correlation.
- (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) 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.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (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.
- (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.
- (d) 30 — Method: on a dual bar chart each bar is a separate frequency, so a total for the whole class is found by combining the two frequencies the bars show. Working: the bars show 14 boys and 16 girls, and 14 + 16 = 30. Answer: 30 pupils, a count of pupils in the class. The distractors: 32 comes from doubling the taller bar, 16 + 16, as though the two bars were equal; 28 comes from doubling the shorter bar, 14 + 14, in the same way; 2 comes from finding the difference between the two bars, 16 − 14, which answers how many more girls there are rather than how many pupils there are.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (d) 44 — Method: the frequency of a category in a frequency table is the number of times that category was counted, and it is read from the row for that category. Working: the rows of the table pair each colour with its count, and the row for black is paired with the count 44, so the frequency of black cars is 44. Answer: 44 cars — a frequency is a count of cars, not a colour and not a percentage. The distractors: 37 comes from reading the count paired with silver, that is from reading the wrong row of the table; 137 comes from adding every count in the table, 44 + 37 + 26 + 18 + 12, which gives the total number of cars rather than the frequency of one colour; 5 comes from counting how many different colours the table lists instead of how many cars were black.
- (b) A line graph — Sales recorded at the end of each of the twelve months are time series data, and a line graph is the chart built to show how a value changes over time, with the points usually joined in order. A pie chart is for showing categorical data as shares of a whole, not a trend over time. A pictogram shows a frequency for separate categories using symbols, not a continuous trend. A scatter graph is for comparing two different variables against each other, not one variable over time.
- (a) Priya — A line of best fit should be drawn so that the plotted points are roughly balanced above and below it. Work out the difference between the two counts for each pupil: Amir 10 − 2 = 8; Kofi 10 − 2 = 8 (10 below and 2 above); Leah 12 − 0 = 12; Priya 6 − 6 = 0. Priya's line has the smallest difference, an exact balance of 6 above and 6 below, so her line is drawn correctly. Amir's line has 10 of the 12 points above it, so it is drawn too low. Kofi's line has 10 of the 12 points below it, so it is drawn too high. Leah's line has every single point below it, so it is not a line of best fit at all.
- (c) A line of best fit — Method: the straight line drawn on a scatter graph is named from the job it does — it is chosen so that it follows the whole set of points as closely as possible. Working: the line passes through the middle of the points, with roughly as many points above it as below it, and it need not pass through any of the plotted points at all; the name given to the straight line chosen in that way is a line of best fit. Answer: a line of best fit. The distractors: a line of symmetry comes from confusing a trend with symmetry, which is a property of a shape rather than of a set of data; a horizontal line through the mean comes from thinking the trend is shown by an average, when a horizontal line would say that the vertical quantity does not change and so show no correlation; a line joining the first and last points comes from thinking the line must join the two extreme points, which lets two points decide a trend that all of the points should share in.
- (c) 156 cm — Method: to combine two groups' means, multiply each group's mean by its own number of pupils, add the two totals together, then divide by the total number of pupils in both groups. Working: 20 × 150 = 3,000 cm for the boys and 10 × 168 = 1,680 cm for the girls, giving a combined total of 3,000 + 1,680 = 4,680 cm. Dividing by all 30 pupils gives 4,680 ÷ 30 = 156 cm. Giving 159 cm averages the two means, (150 + 168) ÷ 2, treating the two groups as if they had the same number of pupils, when there are twice as many boys as girls. Giving 4,680 cm finds the correct combined total height but stops there, forgetting the final division by the 30 pupils. Giving 234 cm divides the combined total by 20, the number of boys only, forgetting that the total also includes the 10 girls. Always weight each mean by its own group size, and always divide by the TOTAL number of pupils in both groups combined.
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