Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher 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 Higher
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- 1.The masses, m kg, of 160 fish caught by a trawler in one day are grouped into classes of unequal width: 0 ≤ m < 10, 40 fish; 10 ≤ m < 30, 60 fish; 30 ≤ m < 45, 30 fish; 45 ≤ m < 50, 30 fish. A histogram is to be drawn from this table. Which set of frequency densities, listed in the same order as the classes above, is correct?
- 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 scatter graph shows the number of years of experience, x, of 18 sales assistants and their monthly sales, y hundred pounds. The plotted points run from x = 1 to x = 12 years, and the line of best fit is y = 4x + 20. A new assistant has 25 years of experience. Use the line of best fit to estimate a value of y for this assistant, and decide whether the estimate would be reliable.y = 4x + 20
- 4.A histogram shows the ages, in years, of 250 members of a running club. The bar for the class 10 ≤ age < 20 has a frequency density of 4.5 members per year, the bar for 20 ≤ age < 35 has a frequency density of 6 members per year, and the bar for 50 ≤ age < 70 has a frequency density of 2.75 members per year. Work out the frequency of the remaining class, 35 ≤ age < 50.
- 5.In a histogram of the heights, h cm, of 90 seedlings, the class 12 ≤ h < 18 contains 36 seedlings. Work out the frequency density for this class.
- 6.A town council wants to find out about the eating habits of the people who live in the town. It asks only the members of a local sports club. Give a reason why this sample is biased.
- 7.Four pairs of variables are listed below. Write down the pair that you would expect to show no correlation.
- 8.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 9.A box plot for the amount of pocket money, in pounds, saved by 30 students in a month is based on a lower quartile of £18 and an upper quartile of £42. A value above upper quartile + 1.5 × interquartile range is considered an outlier. Amara saved £80 in the month. Determine whether Amara's saving is an outlier.
- 10.A café owner in Brighton records the midday temperature, x °C, and the number of hot chocolates sold, y, on 12 days. The temperatures recorded run from 4 °C to 18 °C, and the line of best fit is y = −3x + 74. The forecast for tomorrow gives a midday temperature of 12 °C. Work out the number the line of best fit predicts, and write down how much confidence the owner can have in it.y = -3x + 74
- 11.The times, t minutes, of 80 journeys are summarised by these cumulative frequencies: t < 10, 8 journeys; t < 20, 28 journeys; t < 30, 52 journeys; t < 40, 72 journeys; t < 50, 80 journeys. Estimate the interquartile range.
- 12.The seven members of Team A took 20, 21, 22, 23, 24, 25 and 40 seconds to finish a task. The seven members of Team B took 20, 30, 32, 34, 36, 38 and 40 seconds. Tomás says that because the two teams have the same range, their times are spread out in the same way. Is Tomás right? Give a reason for your answer.
- 13.A scatter graph shows the number of days, x, that each of 16 tomato plants was watered and its height, y cm. The line of best fit has equation y = 1.5x + 4. Write down what the 1.5 in this equation tells you about the plants.y = 1.5x + 4
- 14.A school has 1,000 pupils. A student wants to estimate how many of them walk to school, so she asks 8 pupils in her own class. She says her sample is large enough to give a reliable estimate for the whole school. Is she right? Give a reason for your answer.
- 15.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.
Answer key
- (d) 4, 3, 2, 6 — Method: frequency density = frequency ÷ class width for each class in turn; do not assume the classes are all the same width. Working: the four classes have widths 10 − 0 = 10, 30 − 10 = 20, 45 − 30 = 15 and 50 − 45 = 5. Dividing each frequency by its own width gives 40 ÷ 10 = 4, 60 ÷ 20 = 3, 30 ÷ 15 = 2 and 30 ÷ 5 = 6. Answer: the frequency densities, in order, are 4, 3, 2 and 6. Watch the width of each class separately: treating the last class as if it were also 10 units wide, like the first, gives 30 ÷ 10 = 3 instead of 30 ÷ 5 = 6 — the classes here are deliberately unequal, so no width can be borrowed from another class; dividing the width by the frequency instead of the frequency by the width for the third class gives 15 ÷ 30 = 0.5 in place of 2, the formula the wrong way round; and reading the frequency column straight off the table, 40, 60, 30, 30, skips the division by width altogether and reports how many fish are in each class rather than how densely packed each bar is.
- (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.
- (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) 60 — Method: on a histogram the frequency of a class is its frequency density × its class width, and the frequencies of all the classes add up to the total, so turn each labelled bar into a frequency and subtract their total from 250. Working: 10 ≤ age < 20 has width 20 − 10 = 10, so its frequency is 4.5 × 10 = 45; 20 ≤ age < 35 has width 35 − 20 = 15, so its frequency is 6 × 15 = 90; 50 ≤ age < 70 has width 70 − 50 = 20, so its frequency is 2.75 × 20 = 55. Those three come to 45 + 90 + 55 = 190, and the total is 250, so the missing frequency is 250 − 190 = 60. Answer: the class 35 ≤ age < 50 has 60 members. Watch what you do with the total and the three frequencies you have found: giving the total, 250, as the answer forgets that three bars have already accounted for some of the members; giving 190, the total of the other three classes, reports how many members are not in this class rather than how many are; and leaving one of the three out of the subtraction, for example 45 + 90 = 135 and 250 − 135 = 115, still owes the class at 50 ≤ age < 70 its 55 members.
- (b) 6 — Method: frequency density = frequency ÷ class width. Working: the class 12 ≤ h < 18 has width 18 − 12 = 6, so frequency density = 36 ÷ 6 = 6. Answer: the frequency density is 6 seedlings per cm. Watch which numbers you use: taking the lower bound, 12, as the width instead of 18 − 12 = 6 gives 36 ÷ 12 = 3; dividing the total number of seedlings, 90, rather than this class's frequency, 36, by the width gives 90 ÷ 6 = 15, a density that belongs to no single class; and multiplying instead of dividing gives 36 × 6 = 216, far too large a density for so narrow a class.
- (c) Club members probably eat differently from most people — Method: a sample is biased when the group it is drawn from differs from the population in the very thing the survey is measuring, so compare the subgroup with the population on that quantity. Working: the survey measures eating habits, and people who join a sports club take more exercise than average and are known to eat differently from the town as a whole, so their replies pull the results away from the true picture for the town however many of them are asked. Answer: club members probably eat differently from most people. The distractors: the reply about the number of members treats bias as a question of size, but a large biased sample is still biased; the reply that the members were picked at random is false, since the council picked a club rather than picking residents, and it confuses bias with non-response; the reply that everyone asked lives in the town notes something true of the members but draws the false conclusion that the sample therefore covers the town, when a sample must reflect a population and not merely be taken from inside it.
- (b) A person's shoe size and their favourite colour — A person's shoe size is not linked to which colour they prefer, so these two show no correlation. The other three pairs are all genuinely correlated: distance travelled and fuel used rise together, which is positive correlation; hours of revision and test score generally rise together, which is also positive correlation; and as outdoor temperature rises, fewer woolly hats are sold, which is negative correlation. Negative correlation is still a real relationship between two variables — it is not the same thing as no relationship at all, so the temperature and hats pair is not the answer to this question.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (c) Yes — £80 is above the boundary, £78 — Method: a value counts as an outlier when it lies more than 1.5 times the interquartile range beyond the nearer quartile; here that means checking it against upper quartile + 1.5 × interquartile range. Working: the interquartile range is 42 − 18 = 24. 1.5 × 24 = 36, and 42 + 36 = 78, so any saving above £78 is an outlier. Amara saved £80, and 80 is greater than 78. Answer: yes, Amara's saving is an outlier, because £80 is above the outlier boundary, £78. Watch how you build the boundary and what you compare it with: adding the two quartiles instead of subtracting them, 42 + 18 = 60, gives an interquartile range three times too big, and 42 + 1.5 × 60 = 42 + 90 = 132 puts the boundary so far out that £80 wrongly looks ordinary; comparing £80 with the upper quartile alone, £42, checks only that it lies in the top quarter of the data, which every value above £42 does, not that it lies unusually far beyond it; and adding the interquartile range on once instead of one and a half times, 42 + 24 = 66, uses the wrong multiplier, even though £80 still happens to clear that lower boundary too.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (d) 18 minutes — Method: the lower quartile is the 80 ÷ 4 = 20th value and the upper quartile is the 3 × 80 ÷ 4 = 60th value; locate each inside its class by linear interpolation, then subtract. Working: the 20th value lies between the running totals 8 and 28, so it is in the class 10 ≤ t < 20, which holds 20 journeys across 10 minutes, and it is the 20 − 8 = 12th of them, giving 10 + (12 ÷ 20) × 10 = 16 minutes; the 60th value lies between the running totals 52 and 72, so it is in the class 30 ≤ t < 40, which also holds 20 journeys across 10 minutes, and it is the 60 − 52 = 8th of them, giving 30 + (8 ÷ 20) × 10 = 34 minutes; subtracting, 34 − 16 = 18. Answer: an estimated interquartile range of 18 minutes. The distractors: 20 minutes comes from taking the lower boundaries of the two quartile classes, 30 − 10, which locates the classes but never the values inside them; 40 minutes comes from subtracting the two positions, 60 − 20, instead of the two times; 22 minutes comes from interpolating downwards from each upper boundary rather than upwards from each lower boundary, giving 20 − 6 = 14 and 40 − 4 = 36.
- (d) No, the range uses only the fastest and slowest time — Method: check what the range is built from, then look at what it leaves out. Working: both teams have a fastest time of 20 seconds and a slowest of 40 seconds, so both ranges are 40 − 20 = 20 seconds and Tomás has that part right. But the range is calculated from those two values alone. Six of Team A's seven times lie between 20 and 25 seconds, with a single time far out at 40; Team B is the other way round, with six of its seven times at 30 seconds or more and a single time far out at 20. So Team A bunches at the fast end and Team B at the slow end. The two patterns are quite different, and the range cannot see the difference because the five middle times never enter the calculation. Answer: no, because the range uses only the fastest and slowest time. The distractors: comparing the means answers a different question, since a mean measures position rather than spread, and two sets with the same spread can have different means; saying that equal ranges mean equal spread is the very assumption that fails here; saying that seven times each forces the spreads to match confuses the size of a data set with how its values are arranged inside it.
- (b) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (c) No — 8 from one class is too small to represent the school. — Method: judge reliability by asking whether the sample is both large enough, and spread across the population, relative to what it is meant to represent. Working: 8 pupils is a tiny fraction of the school's 1,000 pupils, and all 8 come from a single class rather than a range of year groups, so the sample is both too small and too narrow to represent the whole school reliably. She is not right. Saying any sample size gives an equally reliable estimate ignores that reliability generally improves with a larger, more representative sample. Saying the method is unreliable because it was not done online is not a reason connected to sample size or representativeness at all. Saying 8 is reliable because it is more than half her class compares the sample to the wrong population — the school has 1,000 pupils, not one class. Always judge a sample's size against the population it is meant to represent, not against a smaller group within it.
- (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.
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