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I'm looking at the diameters of trees, so let $X$ be the diameter random variable. Im told that the diameters are normally distributed with $\mu = 8.8$ and $\sigma =2.8$. Now if four trees are selected independently, what is the probability that at least one has a diameter exceeding 10 inches? I don't think this is a standard normal distribution problem but rather closer to a binomial distribution. I'm not entirely sure what it's asking.

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  • $\begingroup$ Calculate with normal distribution the P that a tree doesnt exceed 10 in. Once you have that P, you can then apply binomial distribution with n=4, the P from above to calculate that no four trees are more than 10 in. Then subtract from 1 (complement rule) $\endgroup$ – imranfat Sep 10 '13 at 23:58
  • $\begingroup$ That's what I tried and I got zero... It wouldn't make much sense that with a probability of zero that no trees out of four wouldn't even be near the mean. $\endgroup$ – TheHopefulActuary Sep 11 '13 at 0:03
  • $\begingroup$ What is the P that a single tree does not exceed 10 in? That's normal distribution. Raise that to the power 4. What did you get? $\endgroup$ – imranfat Sep 11 '13 at 0:11
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For a randomly selected tree, let $p$ be the probability it is $\le 10$. Then $$p=\Pr(X\le 10)=\Pr\left(Z\le \frac{10-8.8}{2.8}\right),$$ where $Z$ is standard normal. You can look up this probability in a table of the standard normal.

The probability all $4$ are $\le 10$ is $p^4$. So the probability at least one is $\gt 10$ is $1-p^4$.

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  • $\begingroup$ Ah I see it was much simpler than I thought. Thanks! $\endgroup$ – TheHopefulActuary Sep 11 '13 at 1:17
  • $\begingroup$ You are welcome. I we had a sample of $12$ and wanted the probability of $6$ of them being $\le 10$ we would use the binomial with the $p$ of the above answer. $\endgroup$ – André Nicolas Sep 11 '13 at 3:05

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