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) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (b) 15 pupils — The frequencies are 5, 4, 3 and 3 pupils, and 5 + 4 + 3 + 3 = 15, so there are 15 pupils in the class. Choosing 6 pupils comes from adding the numbers of pets, 0 + 1 + 2 + 3 = 6, instead of the frequencies. Choosing 10 pupils comes from leaving out the '0 pets' row: 4 + 3 + 3 = 10. Choosing 18 pupils comes from counting the '3 pets' row twice: 5 + 4 + 3 + 3 + 3 = 18.
- (c) The frequency of that category — Method: a bar chart for categorical data has one bar for each category, and the vertical scale on which the bars are measured is a count. Working: a bar drawn twice as tall as another tells you that twice as many items of data fell into its category, so the height measures how many items of data belong to that one category, which is exactly what a frequency is. Answer: the height of each bar is the frequency of that category — a count of items of data. The distractors: the number of different categories comes from reading the vertical scale as though it counted the bars, which is shown along the horizontal axis instead; the total of all the data comes from treating one bar as though it stood for the whole data set rather than for one category; the mean of all the data comes from confusing a bar chart with a measure of average, which no single bar can show.
- (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.
- (a) No correlation — Shoe size has no real relationship with spelling ability, and the points here are scattered with no rising or falling trend, so this is no correlation. A positive correlation would show the points rising together, and a negative correlation would show them falling as one increases; neither pattern is present here. Strong correlation is not correct either, since strength only applies once a positive or negative trend exists, and there isn't one.
- (a) 3/10 — There are 60 − 9 = 51 right-handed pupils. Of the 24 pupils who prefer Art, 6 are left-handed, so 24 − 6 = 18 are right-handed and prefer Art. The probability that a randomly chosen pupil is right-handed and prefers Art is 18/60, which simplifies to 3/10. Giving 2/5 is 24/60 simplified — the probability of preferring Art, ignoring the right-handed condition entirely. Giving 17/20 is 51/60 simplified — the probability of being right-handed, ignoring the Art condition entirely. Giving 1/10 is 6/60 simplified — the probability of being left-handed and preferring Art, the wrong hand condition.
- (c) 72° — The angle is 12 ÷ 60 × 360 = 72°. Choosing 150° divides football's frequency of 25 instead of badminton's 12: 25 ÷ 60 × 360 = 150. Choosing 20° finds badminton as a percentage of the members, 12 ÷ 60 × 100 = 20, rather than an angle in degrees. Choosing 90° uses 48, the total of the other three activities, as the total instead of the full 60 members: 12 ÷ 48 × 360 = 90.
- (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.
- (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.
- (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (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.
- (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.
- (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.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
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