Group A has data \(0,0,0,0,100\), and Group B has data \(100,100,100,100,100\). Use the median-of-halves convention.
a) Find each group's five-number summary.
b) Pool all ten observations and find the pooled five-number summary.
c) Show that averaging corresponding entries of the two group summaries does not produce the pooled summary.
Hints
- Compute each summary from its own ordered observations first.
- Pooling requires reordering all observations together.
- Compare the actual pooled positions with a componentwise numerical average.
Solution
1. Group A has summary \(0,0,0,50,100\). Group B has summary \(100,100,100,100,100\).
2. The pooled ordered data contain four \(0\)s followed by six \(100\)s.
3. The pooled median is \(100\), the lower-half median is \(0\), and the upper-half median is \(100\). Thus the pooled summary is \(0,0,100,100,100\).
4. Averaging corresponding entries gives \(50,50,50,75,100\), which differs from the pooled summary.
Answer
a) A: \(0,0,0,50,100\); B: \(100,100,100,100,100\)
b) Pooled: \(0,0,100,100,100\)
c) Componentwise averaging gives \(50,50,50,75,100\), so five-number summaries cannot be pooled by averaging their entries.