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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 22.

Find P(0.

[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.
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