HS5–HS6 - Box plots and comparing distributions

The five-number summary

The five-number summary is minimum, lower quartile $Q_1$, median, upper quartile $Q_3$, maximum, in that order. The range is maximum minus minimum; the interquartile range is $Q_3-Q_1$. The median measures centre, while the IQR measures spread across the middle half of the distribution.

Worked example 1: an even number of observations

Question. Eight waits, in minutes, are $17,9,15,6,28,11,20,13$. Find the five-number summary using the stated convention.

Plan. Order the eight observations, average the middle pair for the median, and find the middle of each four-value half.

The ordered list is

The overall median is $(13+15)/2=14$. The lower-half median is $(9+11)/2=10$; the upper-half median is $(17+20)/2=18.5$.

Answer. Minimum $6$, $Q_1=10$, median $14$, $Q_3=18.5$, maximum $28$ minutes.

Check. All eight original values occur exactly once in the ordered list. The range is $28-6=22$, and $IQR=18.5-10=8.5$, both smaller than or equal to the full spread as expected. All five summary values are correctly ordered.

Worked example 2: an odd number of observations

Question. Nine waits are $24,7,35,18,10,27,15,20,13$ minutes. Find the five-number summary using the stated median-of-halves convention.

Plan. Find the fifth ordered value, then form two four-value halves without that value.

Ordered data:

The median is eighteen. For the lower half, $Q_1=(10+13)/2=11.5$. For the upper half, $Q_3=(24+27)/2=25.5$.

Answer. $(7,11.5,18,25.5,35)$ minutes, in five-number order.

Check. Four values lie on each side of the excluded middle value; reinserting it gives nine observations, not ten. The range is $35-7=28$ and the IQR is $25.5-11.5=14$. The quartiles lie on opposite sides of the median within the observed extremes.

Watch out

Order the data before finding any median or quartile. A middle entry in an unsorted list has no statistical significance.

Box and whiskers

Draw a box from $Q_1$ to $Q_3$, with a line at the median. For the standard five-number plots here, draw whiskers from the box to the observed minimum and maximum, with end marks. Use a uniform labelled scale. The box spans the middle half by quartile position; a longer section shows greater spread across that portion.

Worked example 3: construct a box plot

Question. Construct a horizontal box plot for Sample A with summary $(6,10,14,18.5,28)$ minutes.

Plan. Put the five values on a uniform time axis, then join only the appropriate parts.

Draw the box from ten to $18.5$, the median line at fourteen, and whiskers reaching six and twenty-eight. The first row of the diagram below shows the completed plot.

Answer. A box with width $18.5-10=8.5$ minutes, median at fourteen, and whisker endpoints six and twenty-eight.

Check. The left whisker length is $10-6=4$, the box length $8.5$, and the right whisker length $28-18.5=9.5$. Their sum is $22$, matching maximum minus minimum. The median lies inside the box, as required.

Worked example 4: complete and interpret a summary

Question. Sample B has eighty waits, minimum four, lower quartile twelve, median nineteen, IQR fifteen and maximum forty-five minutes. Find its upper quartile, construct its box plot, and estimate how many waits are above that quartile.

Plan. Recover $Q_3$ from $IQR=Q_3-Q_1$, then use the upper-quarter interpretation.

Draw the box from twelve to twenty-seven, its median at nineteen, and whiskers to four and forty-five. Approximately one quarter of the observations lie in the upper quarter, so the estimated number above $Q_3$ is $80/4=20$.

Answer. Upper quartile twenty-seven minutes; box plot as in the second row; about twenty waits above it.

Check. $27-12=15$, recovering the supplied IQR, and $4<12<19<27<45$. The estimate of twenty uses quartile position; the five numbers alone do not give an exact count when values are tied at the boundary.