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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- (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.
- (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.
- (d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
- (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.
- (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.
- (c) Because every pupil has an equal chance of being picked — Method: whether a sample represents its population is decided by the selection method, not by the size of the sample, so ask whether the method gives every member of the population the same chance of being chosen. Working: the names are drawn at random from a list of all 10,000 pupils, so each pupil has the same chance, 500 out of 10,000, of being drawn, and no group of pupils is more likely to appear than any other; that is what keeps bias out of the sample. Answer: because every pupil has an equal chance of being picked. The distractors: the reply about 5% treats the sampling fraction as the test of fairness, but a badly chosen 5% is still biased and a well chosen 1% is not; the reply about 500 being large enough makes size the test instead, which is the same mistake in another form, since a large sample drawn from one school would still misrepresent the city; the reply about the most willing pupils describes self-selection, which hands the choice of who is in the sample to the pupils who feel most strongly about the question.
- (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 modal class, as the class with most pupils is shown — Method: a grouped frequency table records how many values fall into each class, but not the values themselves, so any average that needs the individual times can only be estimated from it. Working: the four frequencies are 8, 12, 6 and 4, and 8 + 12 + 6 + 4 = 30, so every pupil is counted. The largest frequency is 12, which belongs to the class 10 < t ≤ 20, and that class can be written down exactly, because finding it needs nothing but the counts the table already gives. Answer: the modal class, as the class with most pupils is shown. The distractors: the mean is said to use all 30 times, but the table does not hold them; the usual method replaces each class by its midpoint, 5, 15, 25 and 35, which gives an estimate of the mean and not its true value; the median is said to be shown, but the table locates only the class holding the 15th and 16th times, which is 10 < t ≤ 20, without saying what either time was; the range is said to be shown, but 0 and 40 are the boundaries of the first and last classes, not the fastest and slowest times actually recorded.
- (a) No, the size of the fire affects both of the quantities — Method: correlation says that two quantities change together; a claim that one of them produces the other is a further claim, and it needs evidence that a scatter graph on its own cannot give. Working: the graph does show strong positive correlation, so more engines did go with greater damage. But neither quantity was set by the researchers: both were decided by how large the fire was. A large blaze brings many appliances and also destroys a great deal, while a small one brings few and destroys little, so a third quantity is driving both of the recorded ones. Answer: no, because the size of the fire affects both of the quantities. The distractors: saying the correlation is negative contradicts the graph, which shows the two quantities rising together, and reaching the right verdict from a false reading of the data is not the reason the mark is for; saying that strong positive correlation shows one quantity causes the other is the assumption the question exists to test, and no strength of correlation can establish cause; saying the points lie close to the line of best fit describes how strong the correlation is, and strength and cause are different matters entirely.
- (b) 120, unreliable — x = 25 is outside 1 to 12 — The line of best fit is y = 4x + 20. 4 × 25 = 100, and 100 + 20 = 120, so the estimate is y = 120. But x = 25 lies far outside the plotted range of 1 to 12 years, so this is an extrapolation, and the estimate is not reliable. Reaching 100 instead of 120 comes from 4 × 25 = 100 with the intercept of 20 left out — still correctly flagged as unreliable, but the wrong value. Calling the estimate reliable simply because it was calculated correctly, giving 120, wrongly assumes that a correct calculation is automatically trustworthy, ignoring that x = 25 lies far beyond the data actually collected. Reaching 68, from 4 × 12 = 48 and 48 + 20 = 68, substitutes x = 12, the top of the plotted range, instead of the assistant's actual x = 25, and wrongly calls that reliable because 12 lies inside the range.
- (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, the mode here is the lowest value of the nine — Method: an average is meant to stand for the data as a whole, so test any proposed average by asking how many values it sits near. Working: the value 4 appears three times and every other count appears once, so 4 is indeed the mode. But those three hours are the quiet ones at the start of the day, and the other six counts run from 11 up to 25; putting the nine counts in order, the middle one is the fifth, which is 13. So the mode sits at the very bottom of the data, with six of the nine hours far above it. Answer: no, because the mode here is the lowest value of the nine, so it describes the quiet opening hours rather than a typical hour. The distractors: saying the mode can only be used when no value repeats reverses the definition, since a mode exists only because a value does repeat; saying the mode is the value that occurs most often is a correct definition, but being the commonest value does not make a value typical when it lies at one end of the data; saying the mode is the best average for any list of numbers ignores the fact that mean, median and mode each describe a population well in different circumstances.
- (d) £15.50 — 2 × 6 = 12, and 12 + 3.50 = 15.50, so the estimated fare is £15.50. Choosing £12.00 stops after 2 × 6 = 12 and forgets to add the £3.50. Choosing £19.00 adds the distance and the constant first and then multiplies: 6 + 3.50 = 9.50, and 9.50 × 2 = 19.00, applying the ×2 to the whole sum instead of only to the distance. Choosing £13.00 multiplies only the constant term by 2 instead of the distance: 2 × 3.50 = 7, and 7 + 6 = 13.00.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
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