Date | May 2009 | Marks available | 4 | Reference code | 09M.2.sl.TZ1.4 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
A random variable X is distributed normally with mean 450 and standard deviation 20.
Find P(X≤475) .
Given that P(X>a)=0.27 , find a.
Markscheme
evidence of attempt to find P(X≤475) (M1)
e.g. P(Z≤1.25)
P(X≤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. a−45020=0.6128
a=462 A1 N3
[4 marks]
Examiners report
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⩽a)=0.27 instead of for P(X⩾a)=0.27 . Many others substituted the values of 0.27 or 0.73 into the equation, instead of the z-scores.