User interface language: English | Español

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 \({\text{P}}(0 \leqslant X \leqslant 2)\).

[2]
a.

Find \({\text{P}}(\left| X \right| > 1)\).

[3]
b.

If \({\text{P}}(X > c) = 0.44\), find the value of \(c\).

[2]
c.

Markscheme

\({\text{P}}(0 \leqslant X \leqslant 2) = 0.242\)    (M1)A1

[2 marks]

a.

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]

b.

\(c = 3.30\)    (M1)A1

[2 marks]

c.

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.

a.

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.

b.

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.

c.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.7 » Normal distribution.
Show 52 related questions

View options