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Suppose that the diameter of trees of a certain type are normally distributed with $\mu = 8.8in $ and $\sigma = 1.75in$ If four trees are independently selected, what is the probability that at least one tree has a diameter exceeding than $10in$?

$X\sim{N(8.8,1.75^2)}$

So the probability for a single tree's diameter being greater than 10 is:

$P(X\ge 10) = 1 - P(X<10)$

Standardize: $ z=\frac{10-8.8}{2.8} = 0.43$

$1 - P(z<0.43) =$ 1 - pnorm(0.43) = 0.334

If $n = 4$ independent trees, then if I was to create a new distribution $Y$ with a diameter greater than 10, then $Y\sim binom(4,.334)$ I can conclude that:

$P(Y \ge 1) = 1 - P(Y \le 0) =$ 1-dbinom(0,4,.334) = 0.803

Is this correct? Would it make sense that this can be considered to be a binomial distribution? Under what conditions should I use dbinom over pbinom commands?

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    $\begingroup$ This appears to be a homework question, so please add the self-study tag. $\endgroup$ – StatsStudent Mar 5 '16 at 23:17
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    $\begingroup$ This is a minor point, but since the diameter of a tree cannot be <0, the diameters cannot be normally distributed. $\endgroup$ – gung Mar 5 '16 at 23:50
  • $\begingroup$ If the standard deviation is 1.75 as indicated, then the standardized value should be (standardized <- (10 - 8.8)/1.75) 0.6857143. Then, pnorm(10, 8.8, 1.75, lower.tail = F) = 0.2464466 getting the same result with 1- pnorm(standardized) or pnorm(standardized, lower.tail = F). $\endgroup$ – Antoni Parellada Mar 6 '16 at 13:14

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