I have the following question:
Use the Central Limit Theorem to show at least how big $n$ must be so that the following is true:
$P(|\bar{X}-\mu| \le 1) \ge 90\%$ with mean $= 7$ and standard deviation being $4$.
I have started out this problem by using the weak law of large numbers to find that the mean of the sample means $E[\bar{X}] = 7$ and the standard deviation of sample means is $4/\sqrt{n}$.
Using this information I have:
$P(|\bar{X} - 7| \le 1) \ge 0.9 = P(6 \le \bar{X} \le 8) \ge 0.9$
Normalizing this I get:
$P(-\sqrt{n}/4 \le Z \le \sqrt{n}/4) \ge 0.9$
Up to this point I am not sure if my steps are correct and if they are I have found $n = 24$ to suffice for $+\sqrt{n}/4$ (as it results in a probability $>0.9$) but for $-\sqrt{n}/4$ I get a value that is a lot less then $90\%$( ie: $P(Z \le -\sqrt{24}/4) = P(Z \le +\sqrt{24}/4) = 1-P(Z \le \sqrt{24}/4)$ which is alot less then $90$ percent).
Any help regarding this question would be greatly appreciated and sorry for the improper format as I am still getting used to this site.
Your analysis up to and including the normalization is right. We want $$\Pr(-\sqrt{n}/4\le Z\le \sqrt{n}/4)=0.9.$$ Thus we want $0.05$ in the right tail of the normal, and the same in the left tail. We get $0.05$ in the right tail at about $z=1.645$ (I got this from a table of the standard normal, but one can also get it from software).
So we want $n\approx (4^2)(1.645)^2$. Round up to the next integer.