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.

${\text{P}}(X > x) = 0.99\;\;\;\left( { = {\text{P}}(X < x) = 0.01} \right)$     (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]

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