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.



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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.