Find \({\rm{P}}(X \le 475)\) .

Given that \({\rm{P}}(X > a) = 0.27\) , find \(a\).

evidence of attempt to find \({\rm{P}}(X \le 475)\)     (M1)

e.g. \({\rm{P}}(Z \le 1.25)\)

\({\rm{P}}(X \le 475) = 0.894\)     A1     N2

[2 marks]

evidence of using the complement     (M1)

e.g. 0.73, \(1 - p\)

\(z = 0.6128\)     (A1)

setting up equation     (M1)

e.g. \(\frac{{a - 450}}{{20}} = 0.6128\)

\(a = 462\)     A1     N3

[4 marks] 

It remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause for concern. Most candidates had been taught the topic but many had difficulty understanding the difference between \(z\), \(F(z)\), \(a\) and \(x\) . Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates often worked with the wrong tail and did not write their answers correct to 3 significant figures.

It remains very clear that some centres still do not give appropriate attention to the normal distribution. This is a major cause for concern. Most candidates had been taught the topic but many had difficulty understanding the difference between \(z\), \(F(z)\), \(a\) and \(x\) . Very little working was shown which demonstrated understanding. Although the GDC was used extensively, candidates often worked with the wrong tail and did not write their answers correct to 3 significant figures.

Many candidates had trouble with part (b), a majority never found the complement, instead using their GDCs to calculate the result, which many times was finding a for \(P(X \leqslant a) = 0.27\) instead of for \(P(X \geqslant a) = 0.27\) . Many others substituted the values of \(0.27\) or \(0.73\) into the equation, instead of the \(z\)-scores.