19M.1.AHL.TZ1.H_6

pestleMathematicsAIHLPaper 119M· sl-4-9-normal-distribution-and-calculationssource ↗

Let $X$ be a random variable which follows a normal distribution with mean $\mu$ . Given that ${\text{P}}\left( {X < \mu - 5} \right) = 0.2$ , find

${\text{P}}\left( {X > \mu + 5} \right)$ .

[2]

${\text{P}}\left( {X < \mu + 5\,\left| {\,X > \mu - 5} \right.} \right)$ .

[5]

Markscheme / solution

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

use of symmetry eg diagram (M1)

${\text{P}}\left( {X > \mu + 5} \right) = 0.2$ A1

[2 marks]

EITHER

${\text{P}}\left( {X < \mu + 5\,\left| {\,X > \mu - 5} \right.} \right) = \frac{{{\text{P}}\left( {X > \mu - 5 \cap X < \mu + 5} \right)}}{{{\text{P}}\left( {X > \mu - 5} \right)}}$ (M1)

$= \frac{{{\text{P}}\left( {\mu - 5 < X < \mu + 5} \right)}}{{{\text{P}}\left( {X > \mu - 5} \right)}}$ (A1)

$= \frac{{0.6}}{{0.8}}$ A1A1

Note: A1 for denominator is independent of the previous A marks.

use of diagram (M1)

Note: Only award (M1) if the region $\mu - 5 < X < \mu + 5$ is indicated and used.

${\text{P}}\left( {X > \mu - 5} \right) = 0.8$ ${\text{P}}\left( {\mu - 5 < X < \mu + 5} \right) = 0.6$ (A1)

Note: Probabilities can be shown on the diagram.

$= \frac{{0.6}}{{0.8}}$ M1A1

THEN

$= \frac{3}{4}\,\,\, = \left( {0.75} \right)$ A1

[5 marks]

Examiners’ report

[N/A]