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.
Statistics worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 2.A council in Leeds wants to know what local people think about letting shops stay open later in the evening. It rings landline telephone numbers between 10 am and 2 pm on a Tuesday. Write down which group is most likely to be under-represented in the sample, and give a reason for your answer.
- 3.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.
- 4.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
- 5.Amelia's mean mark over four tests is 29. Her first three marks are 31, 26 and 26. Work out her fourth mark.
- 6.Box plots summarise house prices, in thousands of pounds, in two towns. Town A: minimum 120, lower quartile 180, median 220, upper quartile 310, maximum 520. Town B: minimum 95, lower quartile 175, median 235, upper quartile 260, maximum 340. A buyer wants the town where house prices are typically higher but more consistent. Which statement is correct?
- 7.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.
- 8.Box plots summarise the test scores of two classes. Class X: minimum 20, lower quartile 45, median 60, upper quartile 70, maximum 95. Class Y: minimum 35, lower quartile 50, median 58, upper quartile 65, maximum 80. Which statement about the two classes is correct?
- 9.The distances, d km, cycled by 180 riders in a charity sportive are summarised by these cumulative frequencies: d < 30, 20 riders; d < 60, 60 riders; d < 80, 120 riders; d < 100, 160 riders; d < 130, 180 riders. Use interpolation to estimate the median distance cycled.
- 10.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 11.For each of 30 fires in one county, the number of fire engines sent and the cost of the damage were recorded. The scatter graph of the data shows strong positive correlation. A newspaper prints the headline "Sending more fire engines causes more damage". Is the newspaper right? Give a reason for your answer.
- 12.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.
- 13.A gym in Cardiff has a scatter graph of the number of training sessions, x, attended by each member and the weight lost, y kg. The line of best fit is y = 0.4x + 1. Members want to lose at least 9 kg. Using the line of best fit, work out the least whole number of sessions needed to reach this target.y = 0.4x + 1
- 14.A call centre records the length, t seconds, of 100 calls: 0 ≤ t < 20, 15 calls; 20 ≤ t < 30, 24 calls; 30 ≤ t < 50, 40 calls; 50 ≤ t < 80, 21 calls. The manager's target is for a call to be finished in under 35 seconds. Estimate the number of calls that met the target.
- 15.A stacked bar for one day at a café in Norwich shows total drink sales of 50 drinks, split into three parts: 22 were tea, 15 were coffee and the rest were hot chocolate. Work out the number of hot chocolates sold.
Answer key
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (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.
- (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.
- (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.
- (b) Town B — higher median and smaller IQR — Method: 'higher and more consistent' needs two comparisons — the median for typical price, and the interquartile range for spread, with a smaller interquartile range meaning more consistent. Working: Town B's median, £235,000, is higher than Town A's, £220,000. Town A's interquartile range is 310 − 180 = 130 and Town B's is 260 − 175 = 85, so Town B's interquartile range is the smaller of the two. Answer: Town B has both the higher median and the smaller interquartile range, so it is the town with higher, more consistent prices. Watch which combination of median and interquartile range each statement claims, and for which town: claiming Town A has the higher median and the smaller interquartile range gets both comparisons wrong, since Town B leads on both; claiming Town A has the higher median (still wrong) but the larger interquartile range at least reads the spread correctly, without it rescuing the false median claim; and claiming Town B has the higher median (correct) but the larger interquartile range misreads the spread — Town B's interquartile range is the smaller of the two, not the larger.
- (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.
- (c) Class X has the higher median and the wider spread — Method: compare the two box plots statistic by statistic — median for location, and the interquartile range for spread — checking the true value of each rather than assuming a pattern. Working: Class X's median is 60 and Class Y's is 58, so Class X's median is the higher one. Class X's interquartile range is 70 − 45 = 25 and Class Y's is 65 − 50 = 15 (and the ranges follow the same order: 95 − 20 = 75 against 80 − 35 = 45), so Class X also has the wider spread. Answer: Class X has both the higher median and the wider spread. Watch that each half of a compound statement is checked separately: claiming Class Y has the higher median and the wider spread gets both comparisons backwards; claiming Class X has the higher median but the narrower spread keeps the median right while reading the spread the wrong way round; and claiming Class Y has the higher median but the narrower spread swaps the median comparison while getting the spread right.
- (c) 70 — Method: estimate the median from the cumulative frequency table by interpolation: find its position, n ÷ 2, locate the class it falls in, then add the fraction of the way through that class (adjusted for the cumulative frequency reached before it) to the class's lower boundary. Working: there are 180 riders, so the median is at position 180 ÷ 2 = 90. Before the class 60 ≤ d < 80 the cumulative frequency is 60, and by the end of it, 120, so the 90th rider falls in this class; its frequency is 120 − 60 = 60 and its width is 80 − 60 = 20. The extra distance needed into the class is 90 − 60 = 30, and 30 ÷ 60 × 20 = 10, so the median is 60 + 10 = 70. Answer: the estimated median distance is 70 km. Watch which numbers the interpolation actually uses: reading off just the class's lower boundary, 60, ignores how far into the class the 90th rider falls; using the target position, 90, as the extra distance instead of subtracting the 60 riders already counted before the class gives 90 ÷ 60 × 20 = 30, so 60 + 30 = 90, overshooting by treating the whole position as if none of it had already been counted; and using the total number of riders, 180, instead of half of it as the target position lands in the very last class, giving an estimate of 130 km — further than any rider is known to have ridden by that point in the table.
- (d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
- (a) No, the size of the fire affects both of the quantities — Method: correlation says that two quantities change together; a claim that one of them produces the other is a further claim, and it needs evidence that a scatter graph on its own cannot give. Working: the graph does show strong positive correlation, so more engines did go with greater damage. But neither quantity was set by the researchers: both were decided by how large the fire was. A large blaze brings many appliances and also destroys a great deal, while a small one brings few and destroys little, so a third quantity is driving both of the recorded ones. Answer: no, because the size of the fire affects both of the quantities. The distractors: saying the correlation is negative contradicts the graph, which shows the two quantities rising together, and reaching the right verdict from a false reading of the data is not the reason the mark is for; saying that strong positive correlation shows one quantity causes the other is the assumption the question exists to test, and no strength of correlation can establish cause; saying the points lie close to the line of best fit describes how strong the correlation is, and strength and cause are different matters entirely.
- (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.
- (d) 20 — 0.4x + 1 = 9, so 0.4x = 9 − 1 = 8, and 8 ÷ 0.4 = 20, so 20 sessions are needed. Choosing 23 divides 9 by 0.4 without first subtracting the 1: 9 ÷ 0.4 = 22.5, rounded up to 23. Choosing 25 subtracts the wrong way, adding the 1 instead of taking it away: 9 + 1 = 10, and 10 ÷ 0.4 = 25. Choosing 2 misplaces the decimal point in the gradient, dividing by 4 instead of by 0.4: 8 ÷ 4 = 2.
- (a) 49 — Method: add the frequencies of the classes that lie wholly below 35 seconds, then use linear interpolation for the class that 35 cuts through, assuming the calls in that class are spread evenly across it. Working: below 30 seconds there are 15 + 24 = 39 calls; the value 35 lies in the class 30 ≤ t < 50, which is 20 seconds wide and holds 40 calls, and 35 is 35 − 30 = 5 seconds into it, so the estimated share is (5 ÷ 20) × 40 = 10 calls; the estimate is 39 + 10 = 49. Answer: about 49 calls met the target. The distractors: 79 comes from adding the whole of the class 30 ≤ t < 50, 39 + 40, and so counting calls of up to 50 seconds as being under 35; 39 comes from stopping at the class boundary 30 and ignoring the part class altogether; 69 comes from measuring the part of the class from 35 up to 50 instead of from 30 up to 35, giving (15 ÷ 20) × 40 = 30 and then 39 + 30.
- (b) 13 — The total is 50, and the two known parts are 22 (tea) and 15 (coffee), so 50 − 22 − 15 = 13 hot chocolates. Choosing 28 comes from 50 − 22, subtracting only the tea and forgetting the coffee. Choosing 35 comes from 50 − 15, subtracting only the coffee and forgetting the tea. Choosing 37 comes from 22 + 15, which finds how many drinks were tea or coffee, not the number left over for hot chocolate.
Build your own mix at the worksheet builder.