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]