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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 \(x\).

[2]
a.

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.

[3]
b.

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]

a.

\({\text{P}}(60.15 \le X \le 60.25)\)     (M1)(A1)

\( = 0.0166\)     A1

[3 marks]

Total [5 marks]

b.

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 \({\text{P}}(X < x) = 0.01\).

a.

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, \({\text{P}}(60.15 \le X \le 60.24)\).

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.7 » Normal distribution.
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