Date | May 2016 | Marks available | 2 | Reference code | 16M.2.hl.TZ2.2 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
A random variable \(X\) is normally distributed with mean 3 and variance \({2^2}\).
Find \({\text{P}}(0 \leqslant X \leqslant 2)\).
Find \({\text{P}}(\left| X \right| > 1)\).
If \({\text{P}}(X > c) = 0.44\), find the value of \(c\).
Markscheme
\({\text{P}}(0 \leqslant X \leqslant 2) = 0.242\) (M1)A1
[2 marks]
METHOD 1
\({\text{P}}(\left| X \right| > 1) = {\text{P}}(X < - 1) + {\text{P}}(X > 1)\) (M1)
\( = 0.02275 \ldots + 0.84134 \ldots \) (A1)
\( = 0.864\) A1
METHOD 2
\({\text{P}}(\left| X \right| > 1) = 1 - {\text{P}}( - 1 < X < 1)\) (M1)
\( = 1 - 0.13590 \ldots \) (A1)
\( = 0.864\) A1
[3 marks]
\(c = 3.30\) (M1)A1
[2 marks]
Examiners report
Part (a) was generally well done. In each question part, a number of candidates could have benefited from producing a labelled sketch of the situation.
Part (b) was not well done with many candidates not knowing what \({\text{P}}(\left| X \right| > 1)\) represents. In each question part, a number of candidates could have benefited from producing a labelled sketch of the situation.
In part (c), a number of candidates did not recognise that \({\text{P}}(X > c) = 1 - {\text{P}}(X \leqslant c)\). In each question part, a number of candidates could have benefited from producing a labelled sketch of the situation.