Five-Number Summaries and Boxplots — Worked Solutions
Worked Solutions
Compare your reasoning with the worked solutions after attempting the questions. A different valid representation or explanation can earn credit when it satisfies the stated criteria.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
Name every position: worked answer
Ordered values in grams: 2,4,6,8,10,12,14,16. Use medians of halves.
Part a
Calculate the five-number summary and state the order of its components.
Minimum 2, Q1=5, median=9, Q3=13, maximum 16 grams, in that order.
| Evidence for credit | Points |
|---|---|
| All five correct positions and units. | 1 |
Box and whiskers: worked answer
A sample has minimum 2, Q1=5, median=9, Q3=13, maximum 16 grams and no outliers under its stated rule.
Part a
Construct its boxplot on a labeled scale. Identify the box, median line, and whiskers and the portions they summarize. Explain why ties or a quartile convention may prevent exact quarter counts, and why a shortened whisker in a modified boxplot need not by itself represent a whole quarter.
Box extends 5–13 grams with a median line at 9; whiskers run 2–5 and 13–16 grams. Label the gram axis and scale. The box represents the nominal middle half. With no separately marked outliers, the lower and upper regions beyond Q1 and Q3 represent roughly the lower and upper quarter, respectively. For finite data, ties and the quartile convention can prevent exact 25% counts in each region. In a modified plot, the outer observations include both the shortened whisker region and any separately marked outliers; the shortened whisker alone is not guaranteed to contain exactly 25%.
| Evidence for credit | Points |
|---|---|
| Accurate positions/scale and units. | 1 |
| Identifies the median line, nominal middle half and roughly lower/upper quarter regions, qualifying finite-data ties/quartile conventions and the shortened-whisker plus separate-outlier interpretation. | 1 |
The fence is not a whisker: worked answer
For a sample, Q1=4 and Q3=9; data below Q1 include 2 and 3, and data above Q3 include 10 and 30. All other values lie between 4 and 9.
Part a
Use the 1.5×IQR rule to determine whisker endpoints and separately plotted values. Explain why the upper whisker does not end at the fence.
IQR=5, fences −3.5 and 16.5. Lower whisker reaches 2 and upper reaches 10, the most extreme non-outliers. Plot 30 separately. The fence 16.5 is a threshold, not an observed endpoint.
| Evidence for credit | Points |
|---|---|
| Correct fences, endpoints, and outlier. | 1 |
| Threshold versus actual data-value explanation. | 1 |
A mean is additional information: worked answer
Three distributions have the following supplied centers: S mean 20, median 20; R mean 25, median 20; L mean 15, median 20. Units are minutes.
Part a
Describe the usual symmetry/skew relationship each pair suggests. Explain why these pairs or a boxplot do not uniquely determine every feature.
S is consistent with relative symmetry; R with right skew pulling mean above median; L with left skew pulling mean below median. These are usual relationships, not unique diagnoses; exact tails, modes, and clusters require more distribution information. A standard boxplot does not itself supply the mean.
| Evidence for credit | Points |
|---|---|
| All three relationships, with qualified interpretation. | 1 |
| Mean availability and nonunique-distribution limit. | 1 |
Construct two delivery-time boxplots: worked answer
Independent delivery-time samples in minutes: A=2,3,4,5,6,7,8,9,10,30; B=4,5,6,7,8,9,10,11,12,13. Use medians of halves and the 1.5×IQR rule for modified boxplots.
Part a
Calculate both five-number summaries, retaining each actual maximum.
A: 2,4,6.5,9,30. B: 4,6,8.5,11,13 minutes. Each maximum belongs to the summary even if separately plotted as an outlier.
| Evidence for credit | Points |
|---|---|
| Both full correctly ordered summaries. | 1 |
| Retains A maximum 30. | 1 |
Part b
Construct both modified boxplots on the same labeled scale. Calculate fences and identify whisker endpoints and all outlier symbols.
A IQR=5, fences −3.5/16.5: whiskers 2 and 10, box 4–9, median 6.5, separate dot 30. B IQR=5, fences −1.5/18.5: whiskers 4 and 13, box 6–11, median 8.5, no outlier dots. Use a common minutes scale.
| Evidence for credit | Points |
|---|---|
| Both fence/outlier classifications and exact marks. | 1 |
| Common labeled scale with actual non-outlier endpoints. | 1 |
Part c
Calculate the two means and describe the distributions using their mean/median relationships. State how the means were obtained.
A mean=84/10=8.4 minutes, greater than median 6.5, consistent with the high right tail/outlier. B mean=85/10=8.5, equal to median 8.5, consistent with relative symmetry. Means came from complete data sums, not from reading standard boxplot marks.
| Evidence for credit | Points |
|---|---|
| Correct means and two contextual relationships. | 1 |
| Identifies full-data basis for means and qualifies shape inference. | 1 |
Part d
For a left-skewed comparison sample, state the usual mean/median relation and explain the low-tail effect.
The mean is usually below the median because low-tail values pull the sum-based mean downward more than they shift the ordered middle. This is a usual pattern, not an unconditional ordering theorem.
| Evidence for credit | Points |
|---|---|
| Correct left-skew relation and rationale. | 1 |