Around here, forestry might be the leading industry. \u00a0Of course, I hardly ever see evidence of it; up in the hills is where the work is done.<\/p>\n
As I understand, one angle of the forest industry is trading woodlots. The buyer may not harvest the wood, but rather hold it until a good time to sell. The key in such a market is understanding the value of the trees on a given lot – which, largely, can boil down to statistics.<\/p>\n
Case 1:<\/p>\n
Let’s imagine Bill owns Managed Woodlot A. Nine years ago he knew the mean tree height was 25m, with standard deviation 8m. Let’s imagine a mature tree commonly grows about 3% per year (likely fairly reasonable, considering my post here<\/a>).<\/p>\n Bill wonders, has the standard deviation of the trees’ heights necessarily changed, given the growth? (The reader might want a briefing on mean and standard deviation; see my post here<\/a>.)<\/p>\n Solution:<\/p>\n First, we realize the effect on a given tree’s height, xi<\/sub>, of nine years’ growth at 3% per year:<\/p>\n xi<\/sub>*(1.03)^9 ≈ 1.3xi<\/sub><\/p>\n We can assume the same effect on the mean x<\/span><\/p>\n Σ1.3xi<\/sub>\/n = 1.3x<\/span><\/p>\n Using the standard deviation formula<\/p>\n s=(Σ(xi<\/sub>–x<\/span>)^2\/(n-1))^0.5<\/p>\n we substitute the grown values:<\/p>\n s=(Σ(1.3xi<\/sub>-1.3x<\/span>)^2\/(n-1))^0.5<\/p>\n We can factor out 1.3:<\/p>\n s=(Σ1.3^2(xi<\/sub>–x<\/span>)^2\/(n-1))^0.5<\/p>\n to finally arrive at<\/p>\n s=1.3(Σ(xi<\/sub>–x<\/span>)^2\/(n-1))^0.5<\/p>\n We see that, indeed, the standard deviation of the trees’ heights has changed: like the mean height, it’s also 1.3 times what it was nine years ago.<\/p>\n Bill can proceed to other reasonable conclusions from the results here. I’ll be telling about them in future posts:)<\/p>\n Source:<\/p>\n Levin, Richard I. Statistics for Management<\/u>. Englewood Cliffs, New Jersey: Prentice-Hall, 1978.<\/p>\n