The ages of 25 customers are 19, 33, 35, 28, 32, 31, 30, 15, 29, 11, 36, 22, 43, 48, 27, 25, 22, 32, 35, 33, 36, 43, 42, 43, 48. Find Q1 and Q3 using the Tukey method.

Step-by-step solution

1. 1

   Sort the data. Every quartile method starts from the ordered list:

   $11,15,19,22,22,25,27,28,29,30,31,32,32,33,33,35,35,36,36,42,43,43,43,48,48$

2. 2

   Find the median. With $n=25$ (odd) the median is the $13^{\text{th}}$ value:

   $\text{median}=32$

3. 3

   Split into halves the Tukey way. This is the step that distinguishes the methods: Tukey's hinges keep the median in both halves, so each half has $13$ values:

   Lower half: $11,15,19,22,22,25,27,28,29,30,31,32,32$

   Upper half: $32,33,33,35,35,36,36,42,43,43,43,48,48$

4. 4

   Take the median of each half. Each half has $13$ values, so each quartile is a single $7^{\text{th}}$ value — no averaging needed:

   $Q_1=27,\qquad Q_3=36$

   $\boxed{Q_1=27,\quad Q_3=36}$

5. 5

   Compare with the exclusive method that software often uses. Dropping the median from both halves leaves $12$ values per half, so each quartile becomes an average of two values:

   $Q_1=\frac{25+27}{2}=26,\qquad Q_3=\frac{36+42}{2}=39$

   That is why a calculator can report $26$ and $39$ while the Tukey answer is $27$ and $36$. Neither is wrong — they are different conventions, and an exam question must say which one it wants.

6. 6

   Check the ordering. In both methods $Q_1<\text{median}<Q_3$ ($27<32<36$) ✓, and the Tukey interquartile range $36-27=9$ is narrower than the exclusive $39-26=13$, as expected since including the median pulls both hinges toward the centre.