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Date November 2016 Marks available 2 Reference code 16N.2.hl.TZ0.8
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 8 Adapted from N/A

Question

A random variable \(X\) is normally distributed with mean \(\mu \) and standard deviation \(\sigma \), such that \({\text{P}}(X < 30.31) = 0.1180\) and \({\text{P}}(X > 42.52) = 0.3060\).

Find \(\mu \) and \(\sigma \).

[6]
a.

Find \({\text{P}}\left( {\left| {X - \mu } \right| < 1.2\sigma } \right)\).

[2]
b.

Markscheme

\({\text{P}}(X < 42.52) = 0.6940\)    (M1)

either \({\text{P}}\left( {Z < \frac{{30.31 - \mu }}{\sigma }} \right) = 0.1180{\text{ or P}}\left( {Z < \frac{{42.52 - \mu }}{\sigma }} \right) = 0.6940\)     (M1)

\(\frac{{30.31 - \mu }}{\sigma } = \underbrace {{\Phi ^{ - 1}}(0.1180)}_{ - 1.1850 \ldots }\)    (A1)

\(\frac{{42.52 - \mu }}{\sigma } = \underbrace {{\Phi ^{ - 1}}(0.6940)}_{0.5072 \ldots }\)    (A1)

attempting to solve simultaneously     (M1)

\(\mu  = 38.9\) and \(\sigma  = 7.22\)     A1

[6 marks]

a.

\({\text{P}}(\mu  - 1.2\sigma  < X < \mu  + 1.2\sigma )\) (or equivalent eg. \(2{\text{P}}(\mu  < X < \mu  + 1.2\sigma )\))     (M1)

\( = 0.770\)    A1

 

Note: Award (M1)A1 for \({\text{P}}( - 1.2 < Z < 1.2) = 0.770\).

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.7 » Normal distribution.
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