Date | November 2014 | Marks available | 2 | Reference code | 14N.2.hl.TZ0.2 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The wingspans of a certain species of bird can be modelled by a normal distribution with mean \(60.2\) cm and standard deviation \(2.4\) cm.
According to this model, \(99\% \) of wingspans are greater than \(x\) cm.
Find the value of \(x\).
In a field experiment, a research team studies a large sample of these birds. The wingspans of each bird are measured correct to the nearest \(0.1\) cm.
Find the probability that a randomly selected bird has a wingspan measured as \(60.2\) cm.
Markscheme
\({\text{P}}(X > x) = 0.99\;\;\;\left( { = {\text{P}}(X < x) = 0.01} \right)\) (M1)
\( \Rightarrow x = 54.6{\text{ (cm)}}\) A1
[2 marks]
\({\text{P}}(60.15 \le X \le 60.25)\) (M1)(A1)
\( = 0.0166\) A1
[3 marks]
Total [5 marks]
Examiners report
Many candidates did not use the symmetry of the normal curve correctly. Many, for example, calculated the value of \(x\) for which \({\text{P}}(X < x) = 0.99\) rather than \({\text{P}}(X < x) = 0.01\).
Most candidates did not recognize that the required probability interval was \({\text{P}}(60.15 \le X \le 60.25)\). A large number of candidates simply stated that \({\text{P}}(X = 60.2) = 0.166\). Some candidates used \({\text{P}}(60.1 \le X \le 60.3)\) while a number of candidates bizarrely used probability intervals not centred on \(60.2\), for example, \({\text{P}}(60.15 \le X \le 60.24)\).