16N.2.AHL.TZ0.H_8

pestleMathematicsAAHLPaper 216N· sl-4-9-normal-distribution-and-calculationssource ↗

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]

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

[2]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

${\text{P}}(\mu - 1.2\sigma < X < \mu + 1.2\sigma )$ (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]

Examiners’ report

[N/A]