Questions on representing and interpreting data

Question 1

In a histogram, the class 20 ≤ x < 25 has frequency 30. Find its frequency density.

Accepted Answer

Density = frequency/width = 30/5 = 6. M1 for frequency divided by width, A1 for 6. In a histogram the area of a bar carries the frequency, so its height has to be frequency per unit of width.

Question 2

Another class, 10 ≤ x < 30, is drawn with frequency density 1.5. Find its frequency.

Accepted Answer

Frequency = density × width = 1.5 × 20 = 30. M1 for density times width, A1 for 30. A low wide bar can hold exactly as much frequency as a tall narrow one. The height on its own tells you only how concentrated the data are.

Question 3

Accepted Answer

There are 3 + 5 + 5 + 2 = 15 leaves, so the median is the 8th value counted from the smallest. B1 for the position. The first row supplies 3 values and the second supplies 5, reaching 8 exactly, so the median is the last leaf of the second row: 28 years. B1 for 28. The range is the largest age minus the smallest, 45 − 14 = 31 years. B1 for 31. Read the key before anything else. Without it the leaves are unlabelled digits, and the same diagram carrying the key 1 | 4 means 1.4 would describe an entirely different set.

Question 4

A data set has mean 34 and median 40. Describe the skew of the distribution, with a reason.

Accepted Answer

Negative skew. B1. The mean lies below the median, and that happens when a tail of low values drags the mean down while the middle value stays where it is: most of the data sit at the high end with a straggle of low values below them. B1 for the reason.

Question 5

In a histogram, the class 0 ≤ x < 20 has frequency 40. The bar for 20 ≤ x < 25 is drawn 1.6 times as tall. Find the frequency of the second class.

Accepted Answer

The first bar's density is 40/20 = 2, so the second bar's density is 2 × 1.6 = 3.2 and its frequency is 3.2 × 5 = 16. M1 for the first bar's density, M1 for scaling it by 1.6, A1 for 16. The second bar is the taller of the two yet holds well under half the frequency, because it is only a quarter as wide. Answering 64 by scaling the frequency directly is the trap.

Question 6

A data set has Q₁ = 20 and Q₃ = 36. Using the 1.5 × IQR rule, determine whether the values 62 and 58 are outliers.

Accepted Answer

IQR = 36 − 20 = 16, so 1.5 × IQR = 24 and the fences sit at 20 − 24 = −4 and 36 + 24 = 60. The value 62 lies beyond the upper fence, so it is an outlier. The value 58 does not, so it is large but within bounds. M1 for 1.5 × IQR, A1 for both fences, B1 for the two verdicts.

Question 7

On a box plot, the median line sits much closer to Q₁ than to Q₃. Describe the skew, and what it says about the data.

Accepted Answer

Positive skew. The lower half of the middle 50% is squashed and the upper half stretched, so values above the median straggle further than values below it. B1 for positive skew, B1 for what it says about the data. Most observations sit low with a long tail of high values, which is the shape typical of income data.

Question 8

Two box plots show daily journey times for two routes. Route A: median 30 minutes, IQR 6 minutes. Route B: median 28 minutes, IQR 18 minutes. Compare the routes in context.

Accepted Answer

Route B is typically two minutes quicker, since its median is 28 minutes against route A's 30. Route B is far less predictable, though. Its middle half spans 18 minutes against route A's 6. A commuter who has to arrive on time would sensibly take route A and trade the two minutes for the consistency. B1 for comparing the medians, B1 for comparing the spread and B1 for the recommendation, all in minutes and all naming the routes.

Question 9

A cumulative frequency diagram for the times, in minutes, of 80 runners passes through the points (20, 0), (25, 10), (30, 34), (35, 62), (40, 74) and (45, 80). Estimate the median and the interquartile range.

Accepted Answer

On a cumulative frequency diagram the median is read at n/2 = 40, Q₁ at n/4 = 20 and Q₃ at 3n/4 = 60. M1 for the three positions. For the median, 40 falls between (30, 34) and (35, 62), which is 6 runners into a step of 28 spread across 5 minutes: 30 + (6/28) × 5 = 31.1 minutes. A1 for 31.1. For Q₁, 20 falls between (25, 10) and (30, 34), giving 25 + (10/24) × 5 = 27.1; for Q₃, 60 falls between (30, 34) and (35, 62), giving 30 + (26/28) × 5 = 34.6. M1 for both quartiles. The interquartile range is 34.64 − 27.08 = 7.56 minutes. A1 for 7.56. A cumulative frequency curve is read at n/2 rather than (n + 1)/2, because it plots a running total against the upper end of each class and not a list of ranked values.

Question 10

A club has recorded the ages of its 200 members in classes of unequal width. State which diagram should be used to show how the ages are distributed, with a reason, and explain why a bar chart with equal-width bars would misrepresent the data.

Accepted Answer

A histogram, with frequency density on the vertical axis. B1 for the histogram named together with frequency density. In a histogram the area of each bar carries the frequency, so classes of different widths are drawn at different widths and can still be compared fairly. B1 for area representing frequency. A bar chart with equal-width bars invites a comparison of heights alone, so a class spanning 20 years and a class spanning 5 years are drawn the same width, and the wide class then looks four times as common as the narrow one even when its ages are no more common, year for year. B1 for the comparison being distorted. Bar charts belong to categorical data, where there are no widths to get wrong.

Question 11

The reaction times, in seconds, of 40 people have mean 0.32 and standard deviation 0.05. A value more than 3 standard deviations from the mean is to be treated as an outlier. Determine whether the recorded values 0.02 s and 0.46 s are outliers, and state how a recorded value of −0.05 s should be dealt with.

Accepted Answer

The fences sit at 0.32 ± 3 × 0.05, that is at 0.32 − 0.15 = 0.17 and 0.32 + 0.15 = 0.47. M1 for the rule applied, A1 for both fences. The value 0.02 lies below 0.17, so it is an outlier; 0.46 lies below 0.47, so it is not, even though it is the larger of the two. B1 for both verdicts. The value −0.05 is not an outlier but an error: a reaction time cannot be negative, so it is a recording or entry fault, and it should be corrected if the original record can be found and otherwise removed, with the removal stated. B1.

Question 12

A histogram of 70 observations has four classes: 0 ≤ x < 10 with frequency density 1.2, 10 ≤ x < 25 with 2.4, 25 ≤ x < 40 with 1.0, and 40 ≤ x < 60 with 0.35. Estimate the number of observations below 20, and estimate the median.

Accepted Answer

Turn every density back into a frequency first, using frequency = density × width: 1.2 × 10 = 12, 2.4 × 15 = 36, 1.0 × 15 = 15 and 0.35 × 20 = 7. These total 70, which confirms the readings. For the count below 20, take the whole first class and the part of the second class up to 20, which is 10 of its 15 units of width: 12 + (10/15) × 36 = 12 + 24 = 36 observations. For the median, the position is 70/2 = 35. The first class carries 12, so the median lies 23 into the second class: 10 + (23/36) × 15 = 19.6. M1 for converting the densities, A1 for the four frequencies, M1 for the proportion of the second class below 20, A1 for 36 observations, M1 for interpolating for the median, A1 for 19.6. Both parts assume the data are spread evenly within each class, which is the standard assumption behind interpolation and worth stating. As a check, 36 of the 70 observations fall below 20 and that is just over half, so a median a little under 20 is exactly right.

Question 13

A newspaper prints a diagram of the cups of coffee sold at a café, in which one drawn cup stands for 20 cups sold. The rows show 3 drawn cups for Monday, 3½ for Tuesday, 5 for Wednesday, 4½ for Thursday and 7 for Friday. The caption states that 450 cups were sold over the five days. Use the key to find the number sold on each day, and state whether the caption agrees with the diagram.

Accepted Answer

Read the key first: one drawn cup stands for 20 cups sold, so a row of n drawn cups means 20n sold and a half cup means 10. B1 for using the key. Monday is 3 × 20 = 60. M1 for one row converted. Tuesday is 3.5 × 20 = 70, Wednesday 5 × 20 = 100, Thursday 4.5 × 20 = 90 and Friday 7 × 20 = 140. A1 for all five daily figures. The total is 60 + 70 + 100 + 90 + 140 = 460 cups. M1 for adding, A1 for 460. The caption says 450, so it does not agree with the diagram; one of the two is wrong, and nothing in the diagram alone settles which. B1 for the comparison and the conclusion.

Question 14

A box plot summarises the amounts, in thousands of pounds, for which 120 houses sold. It has lowest value 120, Q₁ = 180, median 210, Q₃ = 290 and highest value 640. A value more than 1.5 × IQR beyond a quartile is to be treated as an outlier. Determine whether 640 is an outlier, describe the skew from the quartiles, and state which of the mean and the median a report should quote as a typical amount, with a reason.

Accepted Answer

The interquartile range is 290 − 180 = 110, so 1.5 × IQR = 165 and the upper fence sits at 290 + 165 = 455. M1 for the fence. Since 640 > 455, the highest value is an outlier. A1 for 455 with the verdict. For the skew, compare the two halves of the box: the median lies 210 − 180 = 30 above Q₁ and 290 − 210 = 80 below Q₃, so the upper half of the middle 50% is far more stretched than the lower. B1 for the two gaps. That is positive skew. B1. A report should quote the median, £210 000: the long right tail and the value at 640 pull the mean above the amount most of these houses fetched, while the median takes no account of how far out an extreme value sits. B1 for the median with that reason.

Question 15

A company report shows a bar chart of annual profit whose vertical axis runs from 4.2 to 4.6 million pounds. The four bars stand at 4.25, 4.30, 4.34 and 4.45 million pounds, and the caption reads 'profits have soared'. Explain what the truncated axis does to a reader's impression, calculate the percentage change over the four years, and state one change that would make the diagram fair.

Accepted Answer

The axis starts at 4.2 rather than at zero, so the drawn height of each bar shows only the amount by which that year's profit exceeds 4.2 million. B1 for identifying the truncated axis. The first bar then rises 4.25 − 4.2 = 0.05 above the axis and the last rises 4.45 − 4.2 = 0.25, so the last bar is drawn five times the height of the first and the picture suggests a fivefold rise. B1 for the distortion with the two heights compared. The real change is (4.45 − 4.25)/4.25 = 0.0471, an increase of 4.71% across the four years. M1 for the percentage change, A1 for 4.71%. Drawing the vertical axis from zero would make the four bars almost the same height, which is what the figures say. B1 for a fair alternative. A truncated axis is not wrong in itself, since it can show a small change that matters, but it has to be labelled plainly and never described in words the numbers do not support.

Question 16

The histogram shows the times taken by the members of a club to complete a puzzle. Use the histogram to find the total number of members in the club, to estimate the number of members who took between 25 and 40 minutes, and to estimate the median time.

Accepted Answer

In a histogram the frequency of a class is the area of its bar, so frequency = frequency density × class width. Reading the five heights and the five boundaries off the diagram gives 0.6 × 10 = 6, 2.4 × 5 = 12, 3.0 × 5 = 15, 1.5 × 10 = 15 and 0.6 × 20 = 12. M1 for frequency = density × width, A1 for the five frequencies. Adding them, the club has 6 + 12 + 15 + 15 + 12 = 60 members. For the members taking between 25 and 40 minutes, 25 to 30 is half of the class 20 to 30 and contributes 5 × 1.5 = 7.5, while 30 to 40 is half of the class 30 to 50 and contributes 10 × 0.6 = 6. M1 for taking each part-class as an area of its own. That is 7.5 + 6 = 13.5, so about 14 members. A1 for 13.5 or 14. For the median, the position is 60/2 = 30. The running totals are 6 up to 10 minutes, 18 up to 15 and 33 up to 20, so the 30th member lies in the class 15 to 20, which holds 15 members spread across 5 minutes: 15 + (30 − 18)/15 × 5 = 15 + 4 = 19. M1 for the median position with the interpolation, A1 for 19 minutes. Two of the bars are drawn at exactly the same height, 0.6, and yet the wider of them holds twice the frequency of the narrower, 12 members against 6, because frequency is area and not height. Both estimates assume the members are spread evenly within each class, which is the standard assumption behind reading part of a bar.