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.A factory made 3,000 phone cases last week. Shift A checked a random sample of 100 cases and found that 34 were scratched. Shift B checked a different random sample of 50 cases and found that 21 were scratched. Using the COMBINED results from both shifts, work out an estimate for the number of scratched cases made last week.
- 2.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
- 3.A teacher records the number of pets owned by each of 25 pupils in a class; each pupil owns 0, 1, 2, 3 or 4 pets. The teacher wants to show how many pupils own each number of pets. Write down the most suitable type of chart for this data, and give a reason for your answer.
- 4.The heights, h cm, of 80 plants are grouped like this: 0 ≤ h < 20, 14 plants; 20 ≤ h < 40, 22 plants; 40 ≤ h < 50, 16 plants; 50 ≤ h < 80, 28 plants. Write down the class interval that contains the lower quartile.
- 5.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.
- 6.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?
- 7.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.
- 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.In a histogram of the times, t minutes, taken by some people to complete a task, the class 15 ≤ t < 30 contains 24 people. Work out the frequency density for this class.
- 10.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.
- 11.The midday temperature in Leeds was recorded on each day of one week: 22 °C, 24 °C, 23 °C, 25 °C, 26 °C, 21 °C, 24 °C. Work out the mean midday temperature, giving your answer to 2 decimal places.
- 12.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.
- 13.The masses of eight school bags, in kilograms, are 3, 4, 4, 5, 6, 7, 8 and 11. Work out the median mass.
- 14.A two-way table records the favourite subject, Maths or Art, of 60 pupils in Year 10, and whether each pupil is left-handed or right-handed. 9 of the 60 pupils are left-handed, and 6 of those left-handed pupils prefer Art. In total, 24 of the 60 pupils prefer Art. A pupil is chosen at random from the 60. Work out the probability that the pupil is right-handed and prefers Art.
- 15.A box plot for the ages of 40 members of a gym is drawn from this five-number summary: minimum 15, lower quartile 22, median 29, upper quartile 38, maximum 61. Work out the interquartile range shown by this box plot.
Answer key
- (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) 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.
- (a) A vertical line chart (discrete numerical data) — The number of pets is discrete numerical data — whole-number values such as 0, 1, 2, 3 or 4 — recorded for one variable, so a vertical line chart is the chart specified for this kind of data. A bar chart is used for categorical data, such as favourite colour, not numerical values counted like this. A pie chart shows proportions of a whole and does not show the frequency of each separate value. A scatter graph compares two different variables against each other, and only one variable, the number of pets, is recorded here.
- (c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
- (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) 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.
- (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) 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.
- (a) 1.6 — Method: on a histogram the height of a bar is the frequency density, and frequency density = frequency ÷ class width. Working: the class 15 ≤ t < 30 runs from 15 to 30, so its width is 30 − 15 = 15 minutes; the frequency is 24, so the frequency density is 24 ÷ 15 = 1.6. Answer: 1.6 people per minute. The distractors: 360 comes from multiplying the frequency by the class width, 24 × 15, which uses the area rule backwards — area gives the frequency, so the frequency must be divided by the width to give the height; 0.625 comes from dividing the class width by the frequency, 15 ÷ 24, reversing the formula; 0.8 comes from dividing by the upper class boundary, 24 ÷ 30, instead of by the width of the class.
- (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) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (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) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (a) 3/10 — There are 60 − 9 = 51 right-handed pupils. Of the 24 pupils who prefer Art, 6 are left-handed, so 24 − 6 = 18 are right-handed and prefer Art. The probability that a randomly chosen pupil is right-handed and prefers Art is 18/60, which simplifies to 3/10. Giving 2/5 is 24/60 simplified — the probability of preferring Art, ignoring the right-handed condition entirely. Giving 17/20 is 51/60 simplified — the probability of being right-handed, ignoring the Art condition entirely. Giving 1/10 is 6/60 simplified — the probability of being left-handed and preferring Art, the wrong hand condition.
- (d) 16 — Method: on a box plot the interquartile range is the width of the box itself, upper quartile take away lower quartile. Working: the upper quartile is 38 and the lower quartile is 22, so 38 − 22 = 16. Answer: the interquartile range is 16 years. Watch which part of the box plot you are reading: the whole line from whisker to whisker gives the range, 61 − 15 = 46; the left half of the box alone gives median take away lower quartile, 29 − 22 = 7; and the right half of the box alone gives upper quartile take away median, 38 − 29 = 9 — neither half is the interquartile range on its own.
Build your own mix at the worksheet builder.