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I am setting this problem up as P(-2 < (Mn) < 2) and then plugged in the formula for mean and standard deviation. I don't feel like I am headed in the correct direction. Could someone set me straight?


You want to solve for

$$P(|M_n| \ge 2) \le \frac1{100}$$

Note that mean of $M_n=0$,

Hence this reduces to

$$P(M_n \le -2) \le \frac1{200}$$

You might like to find the standard deviation of $M_n$ to proceed with the question.

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  • $\begingroup$ Okay, I understand that. But how do you go from there to solve for n without know what the sum of the X's are? $\endgroup$ – Lucky12456 Apr 30 '18 at 20:37
  • $\begingroup$ Can you express the problem in the form of $P(Z \le \alpha_n) \le \frac1{200}$ where $\alpha_n$ depends on $n$ and $Z$ is the standard normal distribution? from the inverse CDF value, try to figure out a bound for $\alpha_n$? $\endgroup$ – Siong Thye Goh Apr 30 '18 at 20:39

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