Understand · explore · practise
Independent mixed questions on outliers, box plots, cumulative frequency, histograms, scale calibration and contextual comparisons, with worked solutions and assumption checks.
Before you startThe preceding Representations of data lessons.
01 / Choose the representation
Identify the variable, units, grouping, sample size and requested quantity.
Write down any interpolation or scale assumption before calculating.
Try each question before opening its hint. These original examples mix methods so that recognising what to do is part of the practice. An unsupported exact answer is not improved by extra decimal places.
Q₁ = 12, Q₃ = 20. Is 32 an outlier under a strict 1.5 IQR rule?
IQR 8 gives upper fence 32. Equality is not beyond the fence, so 32 is not flagged. A flag would require investigation, not automatic deletion.
02 / Flag, then investigate
Q₁ = 12 and Q₃ = 20. Using 1.5 IQR fences, classify observations 0, 32 and 33.
Compute the IQR before applying the multiplier.
IQR = 8, fences 0 and 32. Under a strict beyond-fence rule, 0 and 32 are not flagged; 33 is flagged.
Mean 50, variance 9. Flag values strictly more than 2 SD from the mean among 44, 43, 56 and 57.
SD is √9 = 3, not 9.
Limits 44 and 56. Flag 43 and 57; equality at 44 or 56 is not flagged.
An unusually high measurement appears in the source log with matching units and a documented unusual event. Should it be replaced by the mean?
Distinguish genuine unusual values from verified recording errors.
No. Retain a verified in-scope observation and document the context. Replacing it by the mean would alter valid data without justification.
03 / Separate fences and whiskers
Ordered data: 2,4,6,8,10,12,14,30. Use medians of the lower and upper halves for quartiles and 1.5 IQR fences. Give the box, whiskers and any outlier.
Q₁ is the mean of 4 and 6; Q₃ the mean of 12 and 14.
Median 9; Q₁ = 5; Q₃ = 13; IQR 8; fences −7 and 25. Box from 5 to 13, median at 9. Whiskers at observed 2 and 14; 30 is separately plotted.
Sample A has median 18 minutes and IQR 6 minutes; B has median 21 and IQR 4. Give two comparisons.
There is a trade-off in centre and spread.
A has the shorter median journey time by 3 minutes. B’s middle half is less spread out, with IQR 2 minutes smaller. Neither group is uniformly better on both summaries.
A grouped cumulative graph begins at (0,0) and ends at (40,80). Must 0 and 40 be observed values?
They may be class boundaries.
No. They bound the grouped data under its class convention. Exact raw minimum and maximum are not recovered from these points.
04 / Read counts and values
Classes [0,10), [10,20), [20,40) have counts 5, 9 and 6. Give cumulative points including the starting point.
Accumulate at upper boundaries.
(0,0), (10,5), (20,14), (40,20).
For question 7, estimate the median using straight within-class interpolation.
Target cumulative count 10 lies in [10,20).
10 + (10 − 5)/9×10 = 140/9 ≈ 15.56. It is a grouped estimate under uniformity within the class.
For question 7, estimate the count at least 30.
Half of the last class remains above 30 under uniformity.
6×10/20 = 3. The actual count is not fixed by the grouped table.
05 / Add areas with the right widths
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Classes [0,10), [10,30), [30,40) have counts 6, 12 and 8. Find their densities.
Divide by widths 10, 20 and 10.
0.6, 0.6 and 0.8. The first two bars have equal height but different areas and frequencies.
For question 10, estimate the count below 20, then its percentage of the sample.
Include all of the first class and half of the second.
Estimated count 6 + 12×10/20 = 12. Total 26; estimated percentage 100×12/26 ≈ 46.15%. State uniformity within the second class.
For question 10, give points joining histogram-top centres with a density axis.
Use midpoints 5, 20 and 35.
(5,0.6), (20,0.6), (35,0.8). A frequency-height polygon would instead use 6, 12, 8 as heights.
06 / Recover missing scale information
A 2 cm by 4 cm bar represents 32 observations. How many does a 3 cm by 5 cm bar represent on the same histogram?
Find observations per cm².
32/8 = 4 observations/cm². The second area is 15 cm², so its frequency is 60.
A width-5 class with frequency 20 has displayed height 2 cm. Find the displayed height of a width-10 class with frequency 30.
The known density 4 corresponds to 2 cm.
One cm represents density 2. Unknown density 30/10 = 3, so height = 1.5 cm.
07 / Keep conclusions proportionate
A has 15 of 60 values above a threshold; B has 24 of 120. Compare the sample proportions.
Divide by each sample’s own total.
A: 25%; B: 20%. B has more observations above the threshold but the smaller sample proportion.
A report supplies only mean 20 and SD 4. A student claims exactly 16% of observations exceed 24. Is that established?
No distribution or raw observations have been supplied.
No. Mean and SD alone do not determine the percentage. The student is importing an unstated distributional assumption, and even a model probability would not guarantee an exact sample percentage.
08 / Review the source of each error
If you missed a question, identify whether the cause was a boundary, quartile convention, scale, denominator, interpolation assumption or overclaim. Use the chapter links to revisit that method, then retry without the solution.
Section 1 of 8 · Choose the representation