Forgive me for bumping in here, without a hello-post... So consider this my hello

But I got math exam on monday and I'm getting quite desperate, since any of my friends can solve it either.

English isn't my native tongue and the exercise isn't in English but I did my best to translate it...

*The results of the entree exam for doctors is divided normally. The max score is 600. After correcting the exams it seems 15% of the participants has less than 230 and 70% has less then 425.*

What is the average of the entree exam...
The solution should be 359 according to the book, but I need to find out how you calculate it...

+Rep for everyone who helps!

Thx in advance

Edit: Just noticed I put this in the wrong place...

should be in high school section... Blame me being frustrated because I can't find the solution

Let \(\displaystyle X\) design the random variable of the marks.

Let \(\displaystyle \mu\) be the average of X.

Let \(\displaystyle \sigma^2\) be the variance of X.

From the text, we have :

\(\displaystyle P(0 \le X \le 230)=.15\)

\(\displaystyle P(0 \le X \le 425)=.7\)

corrected
Make the distribution be a standard normal distribution (transforming \(\displaystyle X\) into \(\displaystyle \frac{X-\mu}{\sigma})\)

---->

\(\displaystyle \left\{\begin{array}{lll} P \left(-\frac{\mu}{\sigma} \le \frac{X-\mu}{\sigma} \le \frac{230-\mu}{\sigma} \right) &=&.15 \\ \\

P \left(\frac{-\mu}{\sigma} \le \frac{X-\mu}{\sigma} \le \frac{425-\mu}{\sigma} \right)&=&.7 \end{array} \right.\)

But there is one information left !

\(\displaystyle P(425 \le X \le 600)=1-.7=.3\)

corrected
Transforming it like we did before :

\(\displaystyle \boxed{\left\{\begin{array}{lllll} P \left(-\frac{\mu}{\sigma} \le \frac{X-\mu}{\sigma} \le \frac{230-\mu}{\sigma} \right) &=&.15 \\ \\

P \left(\frac{-\mu}{\sigma} \le \frac{X-\mu}{\sigma} \le \frac{425-\mu}{\sigma} \right)&=&.7 \\ \\

P \left(\frac{425-\mu}{\sigma} \le \frac{X-\mu}{\sigma} \le \frac{600-\mu}{\sigma} \right) &=&.3

\end{array} \right.}\)

Now, transform it to a nicer way :

Let \(\displaystyle P(Z<z)=\pi(z)\), where \(\displaystyle Z=\frac{X-\mu}{\sigma}\), following a standard normal distribution (I hope you're a little used to it..otherwise you won't be able to follow..), and using this table :

http://sweb.cz/business.statistics/normal01.jpg
The system is now :

\(\displaystyle \left\{\begin{array}{lllll} \pi \left(\frac{230-\mu}{\sigma}\right)-\pi \left(\frac{-\mu}{\sigma}\right)=.15 \\ \\

\pi \left(\frac{425-\mu}{\sigma}\right)-\pi \left(\frac{-\mu}{\sigma}\right)=.7 \\ \\

\pi \left(\frac{600-\mu}{\sigma}\right)-\pi \left(\frac{425-\mu}{\sigma}\right)=.3 \end{array} \right.\)

You've got 4 unknowns (the pi functions) with three equations.

So try to find the values of these.. I'll give it a try after my posting...

Then, find the corresponding values of \(\displaystyle \mu\) and \(\displaystyle \sigma\), thanks to the table. Your problem is asking for \(\displaystyle \mu\)

I hope this is clear enough... English is not my native language either.