The average score of one lesson during the exams is $75 \%$ and variance is $25 \%$. How many students need during the exams to be the average score $75\pm5$ with probability $90 \%$.
I have tried to solve this problem with this way: We know the variance and the mean: $E(X) = 75$ and $Var(X) = 25$. We have a normal distribution and let's suppose that the students that we want are $n$. So we have $Y~N(75n,25\sqrt n)$ and also we know that $P(70<x<75)= 0.9$, $P\left(z < \dfrac{80-75n}{5\sqrt n}\right) - P\left(\dfrac{80-75n}{5\sqrt n}\right) = P(z<1.28)$
I don't know if this is true and how I can continue to find $n$.
Hint:
Let the average score be $A$ with mean $75$ and standard deviation $50$%. What we want to find is the n for which the below holds
$P(70\le A\le 80) = 0.90$
$P(\frac{70-75}{\frac{50}{\sqrt{n}}}\le \frac{A-75}{\frac{50}{\sqrt{n}}}\le\frac{80-75}{\frac{50}{\sqrt{n}}}) = .9$
$P(\frac{-5}{\frac{50}{\sqrt{n}}}\le Z\le \frac{5}{\frac{50}{\sqrt{n}}}) = 0.9$
Let $c= \frac{5}{\frac{50}{\sqrt{n}}}$
Then $P(-c\le Z\le c) = 0.9$
$P(Z\le c) - P(Z\le -c) = 0.9$
$\Phi(c) - (1-\Phi(c)) = 0.9$
$ 2\Phi(c)-1 = 0.9$
$2\Phi(c) = 1.9$
$\Phi(c) = .95$
$c = \Phi^{-1}(0.95) = 1.645$
$\frac{5}{\frac{50}{\sqrt{n}}} =1.645$
$\sqrt{n} = 1.645*10 = 16.45$
$n = 271$