0
$\begingroup$

This is another statistical question that I cannot fully understand:

Suppose that $100$ fair dice are tossed. Estimate the probability that the sum of the faces showing exceeds 370.

Now that the sample variables are $X_1,...,X_{100}$.

So $n=100$ and I believe that $p=\frac{7}{200}$. Am I correct here? If yes, what is the value of the standard deviation $\sigma$?

I know the formula which is to be used is $\frac{X-np}{\sqrt{np(1-p)}}$, I just need to confirm my $p$ and know $\sigma$ to proceed.

Thanks a lot!

$\endgroup$
  • $\begingroup$ Yes, was just a typo :) $\endgroup$ – Khaled Apr 6 '18 at 3:03
2
$\begingroup$

That formula is not the right one to use. That would apply if each dice had two faces, 0 and 1, and then $p$ is the probability it shows 1.

Examples of things like this:

  • tossing a coin $n$ times ($p=\frac{1}{2}$)
  • rolling a die $n$ times and couunting how many times you see the number 5 turn up ($p=\frac{1}{6}$)
  • asking $n$ voters if they would vote for Mr Snoo McSnooface ($p$ is unknown here)

You have a different situation - your dice doesn't just show "yes" or "no", it shopws a nubmer from 1 to 6. And you are not counting the "yesses", you're adding those numbers.

The central limit theorem still applies.

For one die, you have the mean is $\mu=\frac{7}{2}$. Can you find the variance $\sigma^2$?

Then for $n$ dice, the central limint theorem says the average is approximately $N(\mu,\frac{\sigma^2}{n})$, so the total is approximately $N(n\mu,n\sigma^2)$.

$\endgroup$
  • $\begingroup$ I am aware I need to use this formula $\frac{1}{\sqrt{100}} \sum_{i=0}^n \frac{X_i - 3.5}{\sqrt{\sigma^2}}$ , so is the standard deviation $1.8378$? P.s. It should have been $p=\frac{7}{200}$ in my original question. $\endgroup$ – Khaled Apr 6 '18 at 2:59
  • $\begingroup$ I got the problem now, thanks for replying! $\endgroup$ – Khaled Apr 6 '18 at 3:15

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.