A gardener at a community allotment records the weights, in kg, of \(15\) pumpkins in the stem-and-leaf diagram below, where the stem is whole kilograms and...
Question 1 Report
A gardener at a community allotment records the weights, in kg, of \(15\) pumpkins in the stem-and-leaf diagram below, where the stem is whole kilograms and each leaf is one decimal place.
| Stem | Leaves |
|---|---|
| 1 | 2 4 7 9 |
| 2 | 0 1 3 5 6 8 9 |
| 3 | 0 2 4 6 |
Answer Details
A stem-and-leaf diagram lists every value in order once the stems and leaves are combined, which makes it straightforward to find the median (the middle value) and the range (largest minus smallest), and to check any claim made about them.
(a) Reading off all 15 weights in order: \( 1.2, 1.4, 1.7, 1.9, 2.0, 2.1, 2.3, 2.5, 2.6, 2.8, 2.9, 3.0, 3.2, 3.4, 3.6 \). With 15 values, the median is the \( \dfrac{15+1}{2} = 8 \)th value, which is \( 2.5 \) kg [1 mark].
(b) The range is the largest value minus the smallest: \( 3.6 - 1.2 = 2.4 \) kg [1 mark].
(c) Counting the values strictly greater than the median of \( 2.5 \) kg: \( 2.6, 2.8, 2.9, 3.0, 3.2, 3.4, 3.6 \), which is exactly \( 7 \) pumpkins. Since half of 15 pumpkins would be \( 7.5 \), and \( 7 \) is fewer than \( 7.5 \), the claim that half the pumpkins weigh more than the median is not correct [1 mark].
With an odd number of values, the median is itself one of the data points (here, one actual pumpkin weighs exactly 2.5 kg), so the remaining 14 values split into 7 below and 7 above, never exactly half of the full 15 above it.
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