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A SAT score is designed to have a normal distribution with mean 400 and standard deviation 200. If we take 5 independent SAT scores, what is the probability that the mean of them is greater than 500?


I'm just making sure I am thinking about this right. So I am trying to find $P(X > 500)$ so I would be using something like:

$$ P(X > 500) = 1 - \phi \frac{500-400}{200}$$

$\phi$ would be the normal integration of this number in some cases this fraction is denoted as Z = $\frac {X- \mu}{\sigma}$

I know I should alter this since I'm given that I'm using 5 independent trials and that I'm finding the probability of their mean rather than a certain score.

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Hint:

Yes, you are on the right track. You should "alter" it. Notice we take a sample, $X_1,\dotsc, X_5$. Then they are asking about the mean of the five, $$\bar X = \frac{X_1+\dotsb+X_5}{5}.$$ Specifically, they are asking for $$P(\bar X >500).$$

Now can you proceed?

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  • $\begingroup$ Thank You. But how would I input X bar into $\frac {X- \mu}{\sigma}$ if i don't know the values of $X_1, X_2, ..., X_5$ ? $\endgroup$ – Deegeeek Aug 29 '16 at 2:44
  • $\begingroup$ You are sampling from a normal $N(400, 200^2)$ distribution, so each $X_i$ follows that normal distribution. $\endgroup$ – Em. Aug 29 '16 at 2:48
  • $\begingroup$ Sorry, I'm not following. So each $X_i$ has mean 400 and standard deviation of $200^2$? $\endgroup$ – Deegeeek Aug 29 '16 at 2:54
  • $\begingroup$ The mean is correct, the variance is $200^2$, so the sd is $200$. So, you should be able to find $E[\bar X]$ and $\text{Var}(\bar X)$ so that you can do the $Z$ trick (standardization) and find $P(Z>500)$. $\endgroup$ – Em. Aug 29 '16 at 3:00

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