How many students would have to take the exam to ensure with probability at least $.9$ that the class average would be within $5$ of $75$? I'm having trouble solving this problem:

From past experience, a professor knows that the test score of a
   student taking her final examination is a random variable with mean
   $75$.
  How many students would have to take the examination to ensure with
   probability at least $.9$ that the class average would be within $5$ of
   $75$? Use the central limit theorem.  
The professor knows that the variance of a student's test score is $25$.

I'm not entirely sure on how to solve this problem.
Right now this is what I have:
We know: $\mu = 75$ and $\sigma^2 = 25$.
This is what I set up (by defn C.L.T): $\mathbb{P}(\frac{X_1+\cdots+X_n - n75}{5\sqrt{n}}\le\frac{.9-n75}{5\sqrt{n}}) = 1-\mathbb{P}(\frac{X_1+\cdots+X_n - n75}{5\sqrt{n}}>\frac{.9-n75}{5\sqrt{n}})$
I'm not sure how to solve for $n$. Thanks.
 A: If the students have mean $\mu$ and variance $\sigma^2$, then the distribution for the average of $n$ students is approximately normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$. 
In this case, you want to know the probability that the sample mean $\bar{X}$ is within $\pm r$ of the true mean $\mu$. So, $$P(\mu-r \leq \bar{X} \leq \mu-r) = P(-r \leq \bar{X} -\mu \leq r) = P(\frac{-r}{ \sqrt{\frac{\sigma^2}{n}}} \leq \frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}} \leq  \frac{r}{ \sqrt{\frac{\sigma^2}{n}}} ) \approx P(\frac{-r \sqrt{n}}{\sigma} \leq N(0,1) \leq \frac{r \sqrt{n}}{\sigma})$$ since $\frac{\bar{X} - \mu}{\sqrt{\frac{\sigma^2}{n}}}$ is approximated to be standard normal. 
By symmetry of the normal distribution, 
$$P(\frac{-r \sqrt{n}}{\sigma} \leq N(0,1) \leq \frac{r \sqrt{n}}{\sigma}) = 1- 2 Q(\frac{r \sqrt{n}}{\sigma}) = 1-2 \Phi(\frac{-r \sqrt{n}}{\sigma})$$
where $Q$ is the complementary standard normal CDF and $\Phi$ is the standard normal CDF. 
In this case, you want $0.9=1-2 \Phi(\frac{-r \sqrt{n}}{\sigma})$. You can solve this for $\Phi(\frac{-r \sqrt{n}}{\sigma})=0.05$, look up the value $z$ such that $\Phi(z)=0.05$ and then solve $z=\frac{-r \sqrt{n}}{\sigma}$ for $n$. 
Since you have to have an integer number of students, round the resulting $n$ up (which will give you a higher probability of being within the interval than $0.9$ if it is not an integer). 
