A survey recorded the ages of 25 customers: 20, 32, 35, 28, 32, 31, 30, 15, 28, 12, 39, 23, 43, 47, 27, 26, 23, 32, 36, 33, 35, 42, 42, 43, 47. Find the first and third quartiles using the Tukey method, and compare them with the quartiles obtained by excluding the median.

Step-by-step solution

1. 1

   Sort the data — every quartile rule is a rule about positions, so order comes first.

   $12,\ 15,\ 20,\ 23,\ 23,\ 26,\ 27,\ 28,\ 28,\ 30,\ 31,\ 32,\ \underline{32},\ 32,\ 33,\ 35,\ 35,\ 36,\ 39,\ 42,\ 42,\ 43,\ 43,\ 47,\ 47$

2. 2

   Find the median. With $n=25$ (odd) the median is the single middle value, in position $\frac{25+1}{2}=13$:

   $M=32$

   This is where the two conventions part company: the median is an actual data value, so there is a genuine choice about whether it belongs to the lower half, the upper half, both, or neither.

3. 3

   Apply Tukey's rule: the median goes into both halves. Tukey's hinges are defined as the medians of the lower and upper halves including the overall median. The lower half is then the first $13$ values and the upper half the last $13$:

   $\text{lower: } 12,\,15,\,20,\,23,\,23,\,26,\,\underline{27},\,28,\,28,\,30,\,31,\,32,\,32$

   $\text{upper: } 32,\,32,\,33,\,35,\,35,\,36,\,\underline{39},\,42,\,42,\,43,\,43,\,47,\,47$

   Each half has an odd count, so each hinge is again a single middle value — the $7$th of $13$:

   $Q_1=27,\qquad Q_3=39$

4. 4

   Compare with the exclusive convention used by most calculators and spreadsheets. Dropping the median from both halves leaves $12$ values on each side, so each quartile is an average of two neighbours:

   $Q_1=\frac{26+27}{2}=26.5,\qquad Q_3=\frac{39+42}{2}=40.5$

   Both answers are "correct" — they are different definitions, not different arithmetic. They can only disagree when $n$ is odd, because that is the only case in which a data value sits exactly at the median.

5. 5

   Compute the interquartile range under each rule and note the effect.

   $\text{IQR}_{\text{Tukey}}=39-27=12,\qquad \text{IQR}_{\text{exclusive}}=40.5-26.5=14$

   Including the median pulls both hinges toward the centre, so Tukey's rule always gives an IQR that is the same or smaller. Anything built on the IQR — a boxplot's whiskers, the $1.5\,\text{IQR}$ outlier fences — shifts accordingly, which is why a report should say which convention it used.

6. 6

   Check the fences to see the practical consequence. With Tukey's values the outlier fences are $27-1.5(12)=9$ and $39+1.5(12)=57$; with the exclusive values they are $26.5-1.5(14)=5.5$ and $40.5+1.5(14)=61.5$. Neither rule flags any of the $25$ ages as an outlier here (the extremes are $12$ and $47$), so the choice of convention does not change the conclusion in this data set — but it easily could in a set with a longer tail.