Question

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 8% of the trees in a minimal growth condition at 350 parts per million (ppm), 12% at 440 ppm (slow growth), 47% at 540 ppm (moderate growth), and 33% at 670 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees? The mean is ppm. [Round your answer to one decimal place. The standard deviation is ppm. [Round your answer to two decimal places

Answer 1

x	f(x)	xP(x)	x2P(x)
350	0.080	28.0000	9800.0000
440	0.120	52.8000	23232.0000
540	0.470	253.8000	137052.0000
670	0.330	221.1000	148137.0000
total		555.7000	318221.0000
	E(x) =μ=	ΣxP(x) =	555.70
	E(x2) =	Σx2P(x) =	318221.0000
	Var(x)=σ2 =	E(x2)-(E(x))2=	9418.5100
	std deviation=	σ= √σ2 =	97.05

from above:

the mean =555.7

The standard deviation =97.05