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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Statistics worksheet — GCSE Foundation
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- 1.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
- 2.A scatter graph plots the number of years, x, that 20 employees have worked at a company against their salary, y. All the plotted points lie between x = 1 and x = 15. Write down the word used to describe an estimate for y made using a value of x that lies between 1 and 15.
- 3.Write down the statement that correctly describes the difference between correlation and causation.
- 4.A council wants to find out what all residents of a town think about a new cycle lane. It posts a survey only on its website and asks people to fill it in online. Give a reason why this sample is likely to be biased.
- 5.A shop records the colour of every car that uses its car park one Tuesday. Its frequency table reads: black 44, silver 37, blue 26, red 18, white 12. Write down the frequency of black cars.
- 6.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
- 7.A study found that people who drink more coffee tend to concentrate better at work. A coffee company says that this shows that drinking coffee improves concentration. Give the reason why this conclusion cannot be drawn.
- 8.A vet records the weights, in kilograms, of 7 dogs, in the order she weighs them: 18, 12, 50, 10, 20, 14, 16. Work out the median weight.
- 9.A straight line is drawn on a scatter graph to show the trend of the points. Write down the name given to this line.
- 10.A charity shop in Bath holds 2,000 books. Volunteer A checks a random sample of 50 books and finds 35 paperbacks. Volunteer B checks a different random sample of 50 books and finds 31 paperbacks. Work out the estimate each sample gives for the whole stock, and write down what the shop should do next.
- 11.In a vertical line chart of test results, the tallest line stands for 40 pupils. The same results are to be shown in a pie chart of all 100 pupils. Work out the angle at the centre of the sector that will stand for those 40 pupils.
- 12.Two classes sat the same test. The 30 pupils in Class A had a mean mark of 72. The 20 pupils in Class B had a mean mark of 82. Work out the mean mark of all 50 pupils.
- 13.A scatter graph of the number of hours, x, that pupils revised against their test score, y, has the line of best fit y = 2.5x + 15. Amelia wants a score of at least 80. Work out the least whole number of hours of revision the line of best fit suggests she needs.y = 2.5x + 15
- 14.A gardener plants 600 daffodil bulbs. In March she digs up 20 of them, chosen at random, and finds that 17 have flowered. Write down what she can conclude about the 600 bulbs.
- 15.A scatter graph has 50 points. Most of them lie close to a rising line of best fit, but two of them lie a long way from that line. Write down how those two points should be treated.
Answer key
- (c) The median, £160,000, as one very high price lifts the mean — Method: find both averages, then choose the one that sits closer to the bulk of the data. Working: in order the prices are 140,000, 150,000, 160,000, 170,000 and 580,000, so the median is the third of the five, £160,000. For the mean, 140,000 + 150,000 + 160,000 + 170,000 + 580,000 = 1,200,000 and 1,200,000 ÷ 5 = 240,000, so the mean is £240,000. Four of the five houses sold for £170,000 or less, so a reader told that a typical price is £240,000 would expect to pay at least £70,000 more than any of those four cost. Answer: the median, £160,000, as one very high price lifts the mean. The distractors: £580,000 is the middle value of the list as it is printed, which is the median only when the values have first been put in order; £240,000 is the mean, chosen on the ground that a median ignores three of the five prices, but a median uses all five to find which one is central and is then untroubled by how extreme the outer values are; £155,000 comes from deleting the £580,000 house and taking the mean of what is left, since 140,000 + 150,000 + 160,000 + 170,000 = 620,000 and 620,000 ÷ 4 = 155,000, but a real sale may not be thrown away merely for being large.
- (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) 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.
- (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.
- (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) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (c) The data show a link only; a third factor may affect both — Method: a study of this kind measures two quantities and reports how they change together; deciding that one of them produces the other is a further claim, and it needs evidence that the measurements alone cannot give. Working: the study shows that more coffee goes with better concentration, which is a positive correlation; but a third factor that was never measured, such as how motivated someone is, could raise both the coffee drinking and the concentration, and the concentration could equally be what leads to the extra coffee. Answer: the data show a link only, because a third factor may be affecting both quantities, so no claim about cause can be made. The distractors: calling the conclusion safe because the correlation is positive treats the direction of a correlation as proof of cause, which no direction can give; calling it wrong because the correlation is negative misreads the direction of the relationship, since the study reports both quantities rising together; saying the two quantities are not linked denies the correlation the study actually found, when what fails is only the claim about cause.
- (c) 16 kg — Method: sort the seven weights before finding the middle value. Working: in order, the weights are 10, 12, 14, 16, 18, 20 and 50 kg. There are 7 values, so the median is the 4th one: 16 kg. Reading off the 4th weight in the order the vet recorded them, 10 kg, skips the sorting step and is not the median. Working out the mean, 140 ÷ 7 = 20 kg, finds a different average altogether. Working out the range, 50 − 10 = 40 kg, finds the spread, not the middle value. Always sort your data first — the median lives in the ordered list, not the collection order.
- (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.
- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (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.
- (c) 76 marks — Method: a mean of means only works when the groups are the same size, so rebuild each class's total mark, add the totals and divide by the number of pupils altogether. Working: Class A scored 30 × 72 = 2160 marks and Class B scored 20 × 82 = 1640 marks, giving 2160 + 1640 = 3800 marks between 50 pupils, so the overall mean is 3800 ÷ 50 = 76 marks. Answer: 76 marks. The distractors: 77 marks comes from averaging the two class means, (72 + 82) ÷ 2, which ignores the different class sizes; 78 marks comes from attaching each mean to the other class's size, (30 × 82 + 20 × 72) ÷ 50; 3800 marks comes from stopping at the combined total and never dividing by 50.
- (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.
- (d) About 510 of the 600 bulbs are likely to have flowered — Method: the proportion found in a random sample is used as an estimate of the proportion in the whole population, and the conclusion is stated as an estimate, never as a fact about every member. Working: 17 of the 20 bulbs dug up had flowered, so the sample proportion is 17 ÷ 20 = 0.85, and applying that proportion to the whole planting gives 0.85 × 600 = 510 bulbs. A different random sample of 20 would very probably give a slightly different figure, so 510 is an estimate. Answer: about 510 of the 600 bulbs are likely to have flowered. The distractors: saying exactly 510 have flowered takes an estimate from a sample of 20 as a count of all 600, which no sample can deliver; saying exactly 17 of the 600 have flowered reports the sample count as though it were the population count, leaving the other 580 bulbs out of the answer altogether; saying about 20 have flowered uses the size of the sample as the estimate, when 20 is the number of bulbs she dug up rather than a number that flowered.
- (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.
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