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The question I am asked is $P(\bar{X} > 3 + 0.4984S)$, where I am additionally provided $n = 25, \mu = 3.0, \sigma^2_\text{pop} = 3.0$. $\bar{X}$ is the sample mean and $S$ is the sample variance. The population is normal.

I understand the concept of finding such probabilities; what I mean is if you ask me to graph a question like this, I can draw it out for you with every detail. I just don't understand how to get an answer to this. I am not given sample variance so I can't solve x_bar > some number.

I have also tried manipulating the equation, from which I receive $P((\bar{X} - 3)/(0.4984S) > 1)$. Then I said that $0.4984S = \sigma / \sqrt{n} = 2 / 5$. But that is also incorrect, according to my professor.

Please give me some tips numerically or conceptually as to how I should approach solving these kinds of problem.

Thanks!

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$$P(\bar{X} > 3 + 0.4984S) = 1 - P(\bar{X} \le 3 + 0.4984S)$$

where

$$P(\bar{X} \le 3 + 0.4984S) = P(\frac{\bar{X} - 3}{S} \le 0.4984)$$

$$= P(\sqrt{25}\frac{\bar{X} - 3}{S} \le 0.4984\sqrt{25})$$

3 is the true mean so we have

$$= P(Z \le 0.4984\sqrt{25}) = P(Z \le 2.492)$$

Then look up 2.492 on the Z-table or computer

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  • $\begingroup$ Wow, I was so dumb. Thanks for your help. $\endgroup$ – Harshil Dec 22 '15 at 0:39
  • $\begingroup$ @Harshil ?! I wouldn't have got this when I first took probability :P $\endgroup$ – BCLC Dec 22 '15 at 0:45
  • $\begingroup$ I have a follow-up question. We can only model X - mu / (sigma / sqrt(n)) to Z distribution if and only if we have mu and sigma at their respective places. Here, you are assuming that S is equal to sigma. Is that allowed? $\endgroup$ – Harshil Dec 22 '15 at 0:58
  • $\begingroup$ @Harshil ? $\endgroup$ – BCLC Dec 22 '15 at 0:59

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