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 $h$.

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]