Date | November 2008 | Marks available | 2 | Reference code | 08N.1.sl.TZ0.3 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Calculate | Question number | 3 | Adapted from | N/A |
Question
The diagram below shows the cumulative frequency distribution of the heights in metres of \(600\) trees in a wood.
Write down the median height of the trees.
Calculate the interquartile range of the heights of the trees.
Given that the smallest tree in the wood is \(3{\text{ m}}\) high and the tallest tree is \(28{\text{ m}}\) high, draw the box and whisker plot on the grid below that shows the distribution of trees in the wood.
Markscheme
Median \( = 11{\text{ m}}\) (A1) (C1)
Note: Award A0 for “\(11\)” without units; correct units must be included for the A1 to be awarded.
[1 mark]
\({\text{Interquartile range}} = 14 - 10\) (A1)
\( = 4\) (A1)(ft) (C2)
Note: (M1) for taking a sensible difference or for both correct quartile values seen.
[2 marks]
correct median (A1)(ft)
correct quartiles and box (A1)(ft)
endpoints at \(3\) and \(28\), joined to box by straight lines (A1) (C3)
Notes: Award (A0) if the lines go right through the box. Award final (A1) if the whisker goes to \(20\) with an outlier at \(28\).
[3 marks]
Examiners report
Candidates showed less facility in this question compared to question 2.
(a) was generally answered well. There were a number of inaccurate readings from the graph.
Candidates showed less facility in this question compared to question 2.
(b) Errors came from candidates who either used the \(x\) coordinates for the quartiles or who wrote the quartiles as an interval, rather than subtracting these.
Candidates showed less facility in this question compared to question 2.
(c) was well attempted from the candidates’ (a) and (b).