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.
Statistics worksheet — GCSE Foundation
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- 1.A vertical line chart shows the number of pets owned by each pupil in a class: 0 pets — 5 pupils, 1 pet — 4 pupils, 2 pets — 3 pupils, 3 pets — 3 pupils. Work out the total number of pupils in the class.
- 2.The times taken, in minutes, by 30 runners in a Portsmouth fun run are grouped in this table: 0 < t ≤ 20 — 5 runners, 20 < t ≤ 40 — 10 runners, 40 < t ≤ 60 — 10 runners, 60 < t ≤ 80 — 5 runners. Work out an estimate for the mean time, in minutes.
- 3.A vet records the masses, m kg, of 30 dogs at a clinic in Preston: 0 < m ≤ 10 — 11 dogs, 10 < m ≤ 20 — 5 dogs, 20 < m ≤ 30 — 5 dogs, 30 < m ≤ 40 — 9 dogs. Work out an estimate for the mean mass, in kg, using the midpoint of each class interval.
- 4.A scatter graph shows the number of hours of sunshine, x, and the number of visitors, y, at an outdoor swimming pool in Torquay on each of 15 days. The line of best fit passes through the points (5, 150) and (15, 350). Work out the estimated number of visitors on a day with 8 hours of sunshine, using the line of best fit.
- 5.A straight line is drawn on a scatter graph to show the trend of the points. Write down the name given to this line.
- 6.Write down the statement that correctly describes the difference between correlation and causation.
- 7.A scatter graph shows the height, x cm, and the mass, y kg, of 20 pupils in Year 10. The heights on the graph run from 150 cm to 180 cm, and the line of best fit is y = 0.9x − 85. Nadia puts x = 90 into this equation to estimate the mass of a two-year-old child who is 90 cm tall. Is her estimate reliable? Give a reason for your answer.y = 0.9x − 85
- 8.Five friends have heights, in cm, of 150, 152, 155, 158 and 160. A sixth friend, with a height of 170 cm, joins the group. Write down what happens to the mean and the range of the heights once this sixth friend is included.
- 9.Amelia's mean mark over four tests is 29. Her first three marks are 31, 26 and 26. Work out her fourth mark.
- 10.The rainfall, in millimetres, was recorded in Cambridge on six days: 12, 5, 9, 15, 3 and 11. Work out the range of the rainfall.
- 11.A table shows the time each of four pupils took to run 100 m: Oliver 12 seconds, Grace 15 seconds, Ethan 10 seconds, Freya 14 seconds. Write down the name of the fastest runner and the time taken.
- 12.A pie chart of the favourite hobby of a group of pupils has three sectors: reading 50%, sport 25% and music 25%. Write down, as a fraction in its simplest form, the share of the pie chart taken up by the reading sector.
- 13.Write down what a scatter graph is used to show.
- 14.A bus company runs two routes into the centre of Exeter. On ten weekdays the journey time on Route 1 was, in minutes: 22, 23, 24, 24, 25, 25, 26, 26, 27 and 28. On Route 2 it was: 18, 19, 20, 20, 21, 22, 26, 30, 36 and 38. A commuter must reach the centre on time every day. Work out the mean and the range for each route, and write down which route she should take.
- 15.A school posts a questionnaire about school dinners to the families of all 200 pupils. Only 30 families send theirs back, and 27 of those 30 say they are unhappy with school dinners. Give a reason why this result may not represent all 200 families.
Answer key
- (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) 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) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (d) 210 — 350 − 150 = 200. 200 ÷ 10 = 20, so the gradient is 20. Using the point (5, 150): 20 × 5 = 100, so 150 − 100 = 50 is the intercept, giving the line y = 20x + 50. At x = 8: 20 × 8 = 160, and 160 + 50 = 210, so the estimated number of visitors is 210. Choosing 160 stops after 20 × 8 = 160 and forgets to add the intercept of 50. Choosing 250 comes from averaging the two given y-values: 150 + 350 = 500, and 500 ÷ 2 = 250, instead of using the line's equation. Choosing 240 assumes the visitors are directly proportional to the hours of sunshine using the first point, 150 × 8 ÷ 5 = 240, which ignores that the line does not pass through the origin.
- (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.
- (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) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
- (d) Both the mean and the range increase. — The original mean is 150 + 152 + 155 + 158 + 160 = 775, and 775 ÷ 5 = 155 cm; the original range is 160 − 150 = 10 cm. Including the new height of 170 cm gives a new total of 775 + 170 = 945, and 945 ÷ 6 = 157.5 cm, which is higher than 155 cm, and a new range of 170 − 150 = 20 cm, which is higher than 10 cm, so both the mean and the range increase. Saying the range stays the same ignores that 170 cm is a new, higher maximum than the old 160 cm. Saying the mean stays the same ignores that 170 cm is above the original mean of 155 cm, which pulls the average up. Saying both decrease is the opposite of what happens here.
- (c) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
- (a) 12 mm — Method: the range is the highest value minus the lowest value. Working: the highest rainfall is 15 mm and the lowest is 3 mm, so the range is 15 − 3 = 12 mm. Giving 15 mm alone states the highest value, not the range. Giving 3 mm alone states the lowest value, not the range. Sorting the six values, 3, 5, 9, 11, 12 and 15, and averaging the middle two, (9 + 11) ÷ 2 = 10 mm, finds the median, a completely different statistic. The range always needs BOTH the highest and the lowest value — never just one of them.
- (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.
- (d) 1/2 — Method: a percentage is turned into a fraction by writing it over 100 and then cancelling the fraction down to its simplest form. Working: the reading sector is 50% of the pie chart, so as a fraction it is 50/100; dividing the numerator and the denominator by 50 gives 1/2. Answer: 1/2 of the pie chart — a fraction of the chart, not a number of pupils. The distractors: 1/4 comes from reading the share of the sport sector, 25%, instead of the share of the reading sector; 1/3 comes from counting the three sectors and assuming that three sectors must each take a third of the chart, which is only true when the sectors are equal; 3/4 comes from adding the reading and sport sectors together, 50% + 25% = 75%, instead of taking the reading sector on its own.
- (a) The relationship between two variables — Method: what a diagram shows is decided by what has to be known before a single mark can be plotted on it. Working: every point on a scatter graph is plotted from a pair of measurements taken from the same person or object, one read on the horizontal axis and one on the vertical axis; having two measurements for each point is what makes it possible to look for a pattern between them, and the pattern between two variables is what the graph displays. Answer: a scatter graph shows the relationship between two variables. The distractors: the frequency of each single value is what a bar chart or a vertical line chart shows, and it needs only one list of values; how a total is shared between categories is what a pie chart shows; how one quantity changes over time is what a time series line graph shows, in which one of the two axes is always time.
- (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.
- (d) Only families with strong feelings bothered to reply. — Method: a survey has non-response bias when only some of the people asked actually reply, and those who do are not a typical cross-section of everyone who was asked. Working: only 30 of the 200 families sent back their questionnaire, and 27 of those 30 — the great majority — said they were unhappy. Families who feel strongly about an issue, particularly those with a complaint, are far more likely to make the effort to reply than families who are simply satisfied and see no need to say anything, so the 30 replies over-represent unhappy families. Saying the families who replied were picked at random by the school gets the sampling the wrong way round: nobody picked them — they picked themselves by deciding to reply, and that is precisely why they are not a typical cross-section of all 200. Saying postal surveys always have low response rates restates that the response was low without explaining why a low response rate, on its own, makes a result unrepresentative — it is the reason FOR the low response, not the low response itself, that causes the bias here. Saying that the 27 unhappy replies show most families are unhappy is exactly the mistake the question is warning against: it treats the loudest 30 replies as if they stood for the other 170 who never sent theirs back. A low response rate is a warning sign only because the people who bother to reply are rarely typical of everyone who was asked.
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