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I need help in solving this question:

Let $Y_1$, $Y_2$, $...$ be $i.i.d.$ with distribution $N(2, 5)$. Prove that for some $n$, we have $P(Y_1 + Y_2 +···+ Y_n > n) > 0.999$

I know from this, I need to find the new value of $\bar{Y}$ first, which is $$N\left(2, \frac{5}{n}\right)$$

How do I proceed from here?

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  • $\begingroup$ I don´t think that this is true. Have you calculated the case if $n=1$? $\endgroup$ – callculus Feb 20 at 21:38
  • $\begingroup$ The answer should be $n \gt 5000$ $\endgroup$ – dembrownies Feb 21 at 1:17
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By Strong Law of Large Numbers $\frac {Y_1+Y_2+\cdots+Y_n} n \to 2$ almost surely . This implies $P(Y_1+Y_2+\cdots+Y_n >n) \to 1$. Note that the exact distribution of $Y_i$ is not required for this and even the variance is not required. CLT is an overkill for this question.

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  • $\begingroup$ How do you know the probability converges to 1? $\endgroup$ – dembrownies Feb 21 at 1:18
  • $\begingroup$ $P( Y_1+...+Y_n \leq n) \leq P( |\frac {Y_1+...+Y_n} n -2| \geq 1) \to 0$. $\endgroup$ – Kavi Rama Murthy Feb 21 at 5:26
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You know $\bar{Y} \sim N(2, 5/n)$. Thus $Z:=\sqrt{\frac{n}{5}}(\bar{Y} - 2) \sim N(0,1)$. Further, $$P(Y_1 + \cdots + Y_n > n) = P(\bar{Y} > 1) = P(Z > -\sqrt{n/5}).$$ As $n \to \infty$, this quantity tends to $1$.

As noted in another answer, there are more direct methods (under even more relaxed assumptions) to show that there exists an $n$ for which the inequality holds. The virtue of this method (which uses knowledge of the distribution of each $Y_i$) is that one can compute for which specific values of $n$ the inequality holds.

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