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CaptainBlack
Well-known member
- Jan 26, 2012
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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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