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.