17N.2.SL.TZ0.S_7

pestleMathematicsAASLPaper 217N· sl-4-9-normal-distribution-and-calculationssource ↗

The heights of adult males in a country are normally distributed with a mean of 180 cm and a standard deviation of $\sigma {\text{ cm}}$ . 17% of these men are shorter than 168 cm. 80% of them have heights between $(192 - h){\text{ cm}}$ and 192 cm.

Find the value of $h$ .

Markscheme / solution

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

finding the $z$ -value for 0.17 (A1)

eg $\,\,\,\,\,$ $z = - 0.95416$

setting up equation to find $\sigma$ , (M1)

eg $\,\,\,\,\,$ $z = \frac{{168 - 180}}{\sigma },{\text{ }} - 0.954 = \frac{{ - 12}}{\sigma }$

$\sigma = 12.5765$ (A1)

EITHER (Properties of the Normal curve)

correct value (seen anywhere) (A1)

eg $\,\,\,\,\,$ ${\text{P}}(X < 192) = 0.83,{\text{ P}}(X > 192) = 0.17$

correct working (A1)

eg $\,\,\,\,\,$ ${\text{P}}(X < 192 - h) = 0.83 - 0.8,{\text{ P}}(X < 192 - h) = 1 - 0.8 - 0.17,$

${\text{P}}(X > 192 - h) = 0.8 + 0.17$

correct equation in $h$

eg $\,\,\,\,\,$ $\frac{{(192 - h) - 180}}{{12.576}} = - 1.88079,{\text{ }}192 - h = 156.346$ (A1)

35.6536

$h = 35.7$ A1 N3

OR (Trial and error using different values of h)

two correct probabilities whose 2 sf will round up and down, respectively, to 0.8 A2

eg $\,\,\,\,\,$ ${\text{P}}(192 - 35.6 < X < 192) = 0.799706,{\text{ P}}(157 < X < 192) = 0.796284,$

${\text{P}}(192 - 36 < X < 192) = 0.801824$

$h = 35.7$ A2

[7 marks]

Examiners’ report

[N/A]