The duration of direct flights from London to Singapore in a particular year followed a normal distribution with mean \(\mu \) and standard deviation \(\sigma \).

92% of flights took under 13 hours, while only 12% of flights took under 12 hours 35 minutes.

Find \(\mu \) and \(\sigma \) to the nearest minute.

\({\text{P}}\left( {Z < \frac{{780 - \mu }}{\sigma }} \right) = 0.92\) and \({\text{P}}\left( {Z < \frac{{755 - \mu }}{\sigma }} \right) = 0.12\)     (M1)

use of inverse normal     (M1)

\( \Rightarrow \frac{{780 - \mu }}{\sigma } = 1.405 \ldots \) and \(\frac{{755 - \mu }}{\sigma } =  - 1.174 \ldots \)     (A1)

solving simultaneously     (M1)

Note:     Award M1 for attempting to solve an incorrect pair of equations eg, inverse normal not used.

\(\mu  = 766.385\)

\(\sigma  = 9.6897\)

\(\mu  = 12{\text{ hrs 46 mins (}} = 766{\text{ mins)}}\)     A1

\(\sigma  = 10{\text{ mins}}\)     A1

[6 marks]

Generally well done. Most candidates made correct use of the symmetry of the normal curve and the inverse normal to set up a correct pair of equations involving \(\mu \) and \(\sigma \). A few candidates expressed equations containing the GDC command term invNorm.

A few candidates did not express their answers correct to the nearest minute and a few candidates performed erroneous conversions from hours to minutes.