Originally Posted by

**looi76** **Question:**

In a statistics examinations, 15% of the candidates scored more than 63 marks and 10% of the candidates scored less than 32 marks. Assuming that the marks were distributed normally find the mean marks and the standard deviation.

**Attempt:**

$\displaystyle P(X \geq 63) = 15\%$

$\displaystyle P(X \geq 63) = -85\%$████$\displaystyle 15\% - 100\% = -85\%$

$\displaystyle P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 1 - \phi^{-1}(0.85)$

$\displaystyle P\left(Z \geq \frac{63 - \mu}{\sigma} \right) = 0.036$

$\displaystyle 1 - \left( \frac{63 - \mu}{\sigma} \right) = 0.036$

$\displaystyle \left( \frac{63 - \mu}{\sigma} \right) = 0.036 + 1$

$\displaystyle 63 - \mu = 1.036\sigma$

$\displaystyle 1.036\sigma + \mu = 63$

$\displaystyle P(X \leq 32) = 10\%$

$\displaystyle P(X \leq 32) = -90\%$████$\displaystyle 10\% - 100\% = 90\%$

$\displaystyle P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = 1 - \phi^{-1}(90)$

$\displaystyle P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = {\color{red}1 - 1.282}$ Mr F says: Your mistake is in red. It should be -1.282. See main reply below.

Then $\displaystyle {\color{red}\left(\frac{32 - \mu}{\sigma}\right) = -1.282}$ etc.

[snip]

Answer in textbook is **49.1** & **13.4**... Where did I go wrong?