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.
Answer key: Statistics worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (b) 4 — The difference, Bristol minus Leeds, on each day is: Monday 4 − 3 = 1, Tuesday 4 − 5 = −1, Wednesday 6 − 2 = 4, Thursday 2 − 4 = −2. The greatest amount by which Bristol exceeded Leeds is 4 hours, on Wednesday. Choosing 1 takes Monday's smaller positive difference instead of the greatest one. Choosing 2 takes the size of Thursday's difference, but that is the amount by which Leeds exceeded Bristol, the opposite direction to the one asked for. Choosing 6 takes Bristol's raw figure on Wednesday without subtracting Leeds's 2 hours first.
- (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.
- (c) 0.5 per gram — Method: turn the two known bars into frequencies using area, subtract from the total to find how many letters are left, then divide that frequency by the width of the last class to get its height. Working: the first bar covers 50 g at a frequency density of 1.2, giving 1.2 × 50 = 60 letters, and the second covers 50 g at 1.8, giving 1.8 × 50 = 90 letters; together that is 60 + 90 = 150 letters, so 200 − 150 = 50 letters remain; the class 100 ≤ m < 200 is 100 g wide, so its frequency density is 50 ÷ 100 = 0.5 per gram. Answer: 0.5 per gram. The distractors: 0.25 per gram comes from dividing the remaining 50 letters by the upper class boundary, 200, instead of by the class width of 100; 2 per gram comes from dividing the class width by the frequency, 100 ÷ 50, reversing the formula; 1.4 per gram comes from subtracting only the first bar's 60 letters, leaving 140, and then dividing by 100.
- (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
- (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.
- (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.
- (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.
- (a) 40 — Method: the mean is the total divided by how many values there are, so rearranging gives total = mean × number of values. Working: the mean is 8 and there are 5 numbers, so the total is 8 × 5 = 40. Answer: 40, and checking, 40 ÷ 5 = 8, which is the mean given. The distractors: 13 comes from adding the mean and the count, 8 + 5, instead of multiplying them; 1.6 comes from dividing the mean by the count, 8 ÷ 5, which reverses the relationship; 8 comes from quoting the mean itself as the total, which is only true when there is a single number.
- (c) The class with times from 10 up to 20 — Method: to find the median class from a histogram, first turn each bar's frequency density into a frequency using density × class width, build up the cumulative frequency, and find the first class whose cumulative frequency reaches or passes n ÷ 2. Working: the four classes have widths 10, 10, 20 and 20, so their frequencies are 5 × 10 = 50, 2 × 10 = 20, 1.5 × 20 = 30 and 1 × 20 = 20, which add to the 120 visitors stated. The median sits at position 120 ÷ 2 = 60. The cumulative frequency is 50 after the first class and 50 + 20 = 70 after the second, so the 60th visitor is reached during the second class. Answer: the median lies in the class 10 ≤ t < 20. Watch which class each shortcut lands on: the tallest bar belongs to the first class, with the highest frequency density, 5 — but the tallest bar shows where visitors are packed most densely, not where the middle visitor falls, and picking it lands one class too early, at 0 ≤ t < 10; taking half of the total TIME span instead of half of the total NUMBER of visitors, 60 minutes ÷ 2 = 30 minutes, lands in the class 20 ≤ t < 40, confusing a value on the horizontal axis with a position in the data; and using the full 120 visitors as the target position, rather than 120 ÷ 2 = 60, reaches all the way to the last class, 40 ≤ t < 60, treating the whole data set's size as though it were the position of a single middle value.
- (c) 10 — Method: work out the mean that can be found straight away, then use total = mean × number of values on the group of three to find the missing number. Working: the mean of 15 and 25 is (15 + 25) ÷ 2 = 40 ÷ 2 = 20, so the group of three must also have a mean of 20; three numbers with a mean of 20 have a total of 20 × 3 = 60, and 20 + 30 = 50 of that total is already accounted for, so x = 60 − 50 = 10. Answer: 10, and checking, (10 + 20 + 30) ÷ 3 = 20. The distractors: 20 comes from working out the mean the two groups share and writing that down as x; −10 comes from dividing the group of three by 2 instead of by 3, which gives x + 50 = 40; 70 comes from reading the total 15 + 25 = 40 as the mean of the pair, which sets the target total at 120 and leaves x = 70.
- (c) Route 2, because its interquartile range is smaller — Method: for a journey where turning up on time matters, what matters is not the typical (median) time but how predictable it is — a smaller interquartile range means the middle half of journeys cluster closer together. Working: Route 1's median, 34 minutes, is in fact lower than Route 2's, 41 minutes, so Route 1 is faster on average; but Route 1's interquartile range, 22 minutes, is far larger than Route 2's, 6 minutes, so Route 1's times are much less predictable. Answer: Priya should choose Route 2, because its interquartile range is smaller, even though it is slower on average. Watch which statistic answers the question actually asked: Route 1 does not have the smaller interquartile range, Route 2 does, so picking Route 1 for that reason misreads the table; Route 1's median genuinely is the lower one, but a lower median answers 'which is faster', not 'which is more reliable'; and Route 2's median is not the lower one, so that claim about Route 2 is simply false.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (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.
- (d) 1100 — Method: to combine two samples of different sizes, add the faulty counts together and add the sample sizes together before scaling up, rather than treating the two samples separately. Working: the combined sample found 34 + 21 = 55 scratched cases out of 100 + 50 = 150 cases checked, a proportion of 55 ÷ 150. Applying that proportion to the week's production of 3,000 gives an estimate of 55 ÷ 150 × 3000 = 1100 scratched cases. Averaging the two shifts' proportions instead of combining their totals, (34 ÷ 100 + 21 ÷ 50) ÷ 2 = 0.38, gives 0.38 × 3000 = 1140 — this treats the two samples as equally weighted even though Shift A checked twice as many cases as Shift B. Using only Shift A's sample, 34 ÷ 100 × 3000 = 1020, ignores Shift B's cases completely. Using only Shift B's sample, 21 ÷ 50 × 3000 = 1260, ignores Shift A's cases completely. When two samples are different sizes, combine their totals before finding the proportion — do not average the two proportions, and do not use only one shift's sample.
- (a) 15 — Year 11 has 50 − 28 = 22 pupils in total. Of the 22 pupils who walk in total, 15 are in Year 10, so 22 − 15 = 7 Year 11 pupils walk. Subtracting that from the Year 11 total gives 22 − 7 = 15 Year 11 pupils who are driven. Choosing 28 takes the whole school's driven total, 50 − 22 = 28, and treats it as if it were Year 11's alone, without separating the year groups. Choosing 7 correctly finds how many Year 11 pupils walk but stops there, giving that figure instead of the number who are driven. Choosing 35 comes from 50 − 15, subtracting the Year 10 walkers from the whole school total rather than working within Year 11.
Build your own mix at the worksheet builder.