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Suppose a “fair” die is rolled 100 times. Let $X_i$ be the value obtained in the i-th roll. Calculate an approximation for:

$P( \displaystyle\prod_{i=1}^{100} X_i \leq a^{100})$, where $1 < a < 6$

I know that $X_i$~Unif{1,2,3,4,5,6}, and I think I should use some kind of normal distribution to approximate the probability, but I don't know what to do with the product. Can someone help me?

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Hint: try taking the log of both sides of the inequality.

$$P(\prod_{i=1}^{100} X_i \le a^{100}) = P(\frac{1}{100} \sum_{i=1}^n \log X_i \le \log a)$$ and $\frac{1}{100} \sum_{i=1}^n \log X_i$ is approximately normal (you need to find the mean and variance).

Mean: $E \frac{1}{100} \sum_{i=1}^{100} \log X_i = E \log X_1 = \frac{1}{6} \log 720.$

Variance: $$\text{Var}(\frac{1}{100} \sum_{i=1}^{100} \log X_i) = \frac{1}{100} \text{Var}(\log X_1) = \frac{1}{600} \sum_{x=1}^6 (\log x - \frac{1}{6} \log 720)^2.$$

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  • $\begingroup$ But $Z = \frac{X-E(X)}{Var(X)}$, how does $\frac{1}{100} \sum_{i=1}^n \log X_i$ is equal to $Z$? $\endgroup$
    – Mystery
    Commented Nov 23, 2019 at 21:16
  • $\begingroup$ @Mystery I only said that its distribution is approximately normal (not standard normal). If you find the mean and variance of $\frac{1}{100} \sum_{i=1}^n \log X_i$, then you can standardize it by subtracting the mean and dividing by the standard deviation (not the variance) to get a random variable that is approximately standard normal. $\endgroup$
    – angryavian
    Commented Nov 24, 2019 at 0:11

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