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.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.
- 2.A pictogram shows the number of cars sold by a garage in Plymouth each month, where each whole car symbol represents 8 cars sold and a half symbol represents 4 cars. March shows 3 whole symbols and one half symbol. Work out how many cars were sold in March.
- 3.Two classes sat the same maths test, both marked out of 100. Class A had a mean mark of 70 and a range of 30 marks. Class B had a mean mark of 70 and a range of 10 marks. Compare the marks of the two classes.
- 4.In a spelling test the 20 pupils in Group A had a mean mark of 80, and the 30 pupils in Group B had a mean mark of 70. Work out the mean mark of all 50 pupils.
- 5.A composite (stacked) bar for a charity bake sale in Durham shows the number of cakes sold, split into three types. The bar has a total height of 80 cakes: 34 were sponge cakes, 26 were chocolate cakes and the rest were fruit cakes. Work out the percentage of the cakes sold that were fruit cakes.
- 6.The mean of the four numbers 10, 15, 20 and x is 18. Work out the value of x.
- 7.A company found that the higher the price it charged for a product, the lower the satisfaction score its customers gave. The price and the satisfaction score are plotted on a scatter graph. Describe the line of best fit that would be drawn on that graph.
- 8.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.
- 9.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.
- 10.A recruitment agency in Manchester compares the weekly pay, in pounds, of seven employees at two branches. Branch A: £480, £495, £500, £505, £510, £515, £1,200. Branch B: £480, £490, £500, £510, £520, £530, £540. An advert for the agency claims 'Branch A pays more on average.' Decide whether this claim is fairly supported, using an appropriate average, and choose the correct conclusion.
- 11.Priya wants to find out the favourite sport of the 900 pupils at her school. She asks 5 pupils, chosen at random. Give a reason why her sample may not give a reliable result.
- 12.The daily high temperatures, in °C, in Leeds over five days were 14, 16, 18, 19 and 23. In York over the same five days they were 15, 17, 17, 18 and 18. Write a sentence comparing Leeds and York using both the median and the range.
- 13.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 14.The number of pets owned by each of 19 pupils in a class is recorded: 0 pets — 7 pupils, 1 pet — 3 pupils, 2 pets — 4 pupils, 3 pets — 5 pupils. Work out the median number of pets.
- 15.A school has 1200 pupils. A teacher wants to take a random sample of 60 of them. Write down which of these methods gives a random sample.
Answer key
- (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.
- (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.
- (a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (d) 25% — First find the number of fruit cakes: 80 − 34 − 26 = 20. Then write this as a percentage of the total: 20 ÷ 80 × 100 = 25%. Giving 20% comes from reporting the count of fruit cakes, 20, directly as a percentage, without dividing by the total of 80 first. Giving 32.5% computes the percentage of chocolate cakes instead of fruit cakes: 26 ÷ 80 × 100 = 32.5%. Giving 42.5% computes the percentage of sponge cakes instead of fruit cakes: 34 ÷ 80 × 100 = 42.5%.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (a) A straight line sloping down from left to right — Method: a line of best fit is a straight line drawn to follow the trend of the points, so its slope is decided by the direction of the relationship between the two quantities. Working: as the price rises, the satisfaction score falls, so the points start high on the left of the graph and finish low on the right; the straight line that follows them therefore slopes downwards as the graph is read from left to right, which is the line of a negative correlation. Answer: a straight line sloping down from left to right. The distractors: a line sloping up comes from reading a falling relationship as a rising one; a horizontal line comes from expecting no correlation, since a horizontal line says the satisfaction score does not change as the price changes; a curve passing through every point comes from thinking a line of best fit has to touch all of the plotted points, when it is a single straight line drawn through the middle of them.
- (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.
- (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.
- (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.
- (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.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (b) 1 — Method: for data in a frequency table, find the position of the median using (n + 1) ÷ 2, then read off the value at that position from the cumulative frequencies. Working: there are 19 pupils, so the median is the 10th value. The cumulative frequencies are 7 (up to 0 pets), 10 (up to 1 pet), 14 (up to 2 pets) and 19 (up to 3 pets). The 10th value falls at the end of the '1 pet' group, so the median is 1 pet. Giving 0 pets is the mode — the category with the highest frequency, 7 — not the median. Giving 3, the highest number of pets minus the lowest, finds the range, a different statistic entirely. Giving 19 states the total number of pupils, not a number of pets at all. Find the middle POSITION first, then read off the value it belongs to — do not confuse it with the mode, the range or the total.
- (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.
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