Question

Suppose the diameter at breast height (in.) of trees of a certain type is normally dis- tributed with 88 and σ 28, as suggested in the article Simulating a Harvester- Forwarder Softwood Thinning (Forst Products J., May 1997: 3641) (a) What is the probability that the diameter of a randomly selected tree will be at least 10 in. Will exceed 10 in.? (b) What is the probability that the diameter of a randomly selected ee wl be between 5 and 10 in.? (c) If four trees are independently selected, what is the probability that at least one has a diameter exceeding 10 in.?

Answer 1

$Let~X~be~a~random~variable~denoting~the~diameter~at~breast~height~(in.)~of\\ trees~of~a~certain~type\\ X\sim Normal(8.8,~2.8^2).\\ (a)~Probability~that~the~diameter~of~a~randomly~selected~tree~will~be~at~least~10~in.=\\ P(X\geq 10)=P\left(\frac{X-8.8}{2.8}\geq \frac{10-8.8}{2.8} \right )=1-\Phi\left( \frac{10-8.8}{2.8}\right )=0.3341\\ Since~it~is~a~continuous~random~variable~so\\ P(X\geq 10)=P(X>10)\\ so\\ Probability~that~the~diameter~of~a~randomly~selected~tree~will~exceed~least~10~in.=\\ Probability~that~the~diameter~of~a~randomly~selected~tree~will~be~at~least~10~in.\\=0.3341\\ (b)~P(5\leq X\leq 10)=\Phi\left( \frac{10-8.8}{2.8}\right )-\Phi\left( \frac{5-8.8}{2.8}\right )=0.5785\\ (c)~Let~Y~be~a~random~variable~denoting~no.~of~trees~who~has~diameter~exceeding~10\\ Then~Y\sim Binomial(4,0.3341).\\ Required~probability=P(Y\geq 1)=1-P(Y=0)=1-(1-0.3341)^4\\=0.8034$