Date | May 2017 | Marks available | 1 | Reference code | 17M.1.sl.TZ2.6 |
Level | SL only | Paper | 1 | Time zone | TZ2 |
Command term | Write down | Question number | 6 | Adapted from | N/A |
Question
For a study, a researcher collected 200 leaves from oak trees. After measuring the lengths of the leaves, in cm, she produced the following cumulative frequency graph.
The researcher finds that 10% of the leaves have a length greater than \(k\) cm.
Write down the median length of these leaves.
Write down the number of leaves with a length less than or equal to 8 cm.
Use the graph to find the value of \(k\).
Before measuring, the researcher estimated \(k\) to be approximately 9.5 cm. Find the percentage error in her estimate.
Markscheme
9 (cm) (A1) (C1)
[1 mark]
40 (leaves) (A1) (C1)
[1 mark]
\((200 \times 0.90 = ){\text{ }}180\) or equivalent (M1)
Note: Award (M1) for a horizontal line drawn through the cumulative frequency value of 180 and meeting the curve (or the corresponding vertical line from 10.5 cm).
\((k = ){\text{ }}10.5{\text{ (cm)}}\) (A1) (C2)
Note: Accept an error of ±0.1.
[2 marks]
\(\left| {\frac{{9.5 - 10.5}}{{10.5}}} \right| \times 100\% \) (M1)
Notes: Award (M1) for their correct substitution into the percentage error formula.
\({\text{9.52 (% ) }}\left( {{\text{9.52380}} \ldots {\text{ (% )}}} \right)\) (A1)(ft) (C2)
Notes: Follow through from their answer to part (c)(i).
Award (A1)(A0) for an answer of \( - 9.52\) with or without working.
[2 marks]