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!

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

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 $N(n\mu,n\sigma^2)$.

  • $\begingroup$ 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

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