Date | May 2013 | Marks available | 3 | Reference code | 13M.2.sl.TZ2.2 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The random variable \(X\) is normally distributed with mean \(20\) and standard deviation \(5\).
Find \({\rm{P}}(X \le 22.9)\) .
Given that \({\rm{P}}(X < k) = 0.55\) , find the value of \(k\) .
Markscheme
evidence of appropriate approach (M1)
eg \(z = \frac{{22.9 - 20}}{5}\)
\(z = 0.58\) (A1)
\({\rm{P}}(X \le 22.9) = 0.719\) A1 N3
[3 marks]
\(z\)-score for \(0.55\) is \(0.12566…\) (A1)
valid approach (must be with \(z\)-values) (M1)
eg using inverse normal, \(0.1257 = \frac{{k - 20}}{5}\)
\(k = 20.6\) A1 N3
[3 marks]
Examiners report
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and \(z\)-scores while others used the GDC directly; the GDC approach earned full marks more often than the \(z\)-score approach.
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and \(z\)-scores while others used the GDC directly; the GDC approach earned full marks more often than the \(z\)-score approach. A common error in part (b) was to set the expression for \(z\)-score equal to the probability. Many candidates had difficulty giving answers correct to three significant figures; this was particularly an issue if no working was shown.