# Math Help - [SOLVED] Normal Distribution

1. ## [SOLVED] Normal Distribution

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:

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

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

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

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

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

$63 - \mu = 1.036\sigma$

$1.036\sigma + \mu = 63$

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

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

$P\left(Z \leq \frac{32 - \mu}{\sigma}\right) = 1 - 1.282$

$\left(\frac{32 - \mu}{\sigma}\right) = -0.282$

$32 - \mu = -0.282\sigma$

$-0.282\sigma + \mu = 32$

$1.036\sigma + \mu = 63$

$-0.282\sigma + \mu = 32$

$1.036\sigma = 63$

$-0.282\sigma = 32$

$1.318\sigma = 31$

$\sigma = \frac{31}{1.318}$

$\sigma = 23.52$

$1.036\sigma + \mu = 63$

$1.036 \times 23.52 + \mu = 63$

$\mu = 63 - (1.036\times23.52)$

$\mu = 38.44$

$\mu = 38.44$

$\sigma = 23.52$

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

2. 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:

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

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

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

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

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

$63 - \mu = 1.036\sigma$

$1.036\sigma + \mu = 63$

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

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

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

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

[snip]

Answer in textbook is 49.1 & 13.4... Where did I go wrong?
Again, well done on taking the time and trouble to show all your working.

You want the value of $z_{0.10}$ corresponding to $\Pr(Z < z_{0.10}) = 0.10$

$\Rightarrow \Pr(Z > - z_{0.10}) = 0.10$

(note that $z_{0.10} < 0 \Rightarrow -z_{0.10} > 0$ so this is not as wrong as it first looks)

$\Rightarrow \Pr(Z < - z_{0.10}) = 0.90 \Rightarrow - z_{0.10} = 1.28$

from the tables

$\Rightarrow z_{0.10} = -1.28$.

This would be a good time to re-inforce the importance of drawing simple normal distribution pictures with the required area shaded ......