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#### CaptainBlack

##### Well-known member

- Jan 26, 2012

- 890

The sum after \(N\) throws is approximately normal with mean \(\mu_N=N \mu_1\) and with variance \(\sigma^2_N=N\sigma^2_1\), where \(\mu_1=3.5\) is the mean for a single throw, and \( \sigma^2_1=17.5\) is the variance of a single throw."A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."

I have the suspicion that if

X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance

How could I prove that?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ...

Now the probability that the process finished on or before the \(N\) throw is:

\[p_{N}\approx P\left(\frac{N\mu_1-300.5}{\sqrt{N}\sigma_1}\right) \]

where \(P(.)\) is the cumulative Standard Normal distribution function.

So the probability that at least \(N+1\) throws are necessary is \(1-p_N\)

This normal approximation method gives a probability of \(\approx 92.4\%\) while simulation gives \(\approx 93.6\%\)

Please check the arithmetic etc.

CB

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