Question

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 4% of the trees in a minimal growth condition at 340 parts per million (ppm), 11% at 460 ppm (slow growth), 47% at 540 ppm (moderate growth), and 38% at 650 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees?

Answer 1

24.53

Explanation:

We need to define our data, that is

w -> x

0.04->340

0.11->460

0.47->540

0.38->650

We can obtain our mean,

`\bar{x} = (\sum\limit_(i=1)^n x_iw_i)/(\sum\limit_(i=1)^n w_i)`

`\bar{x} = ((0.04*340)+(0.11*460)+(0.47*540)+(650*0.38))/(1)`

`\bar{x} = 570.7070`

To find the SE we need the Variance,

`V(x) = E(x^2) - E(x)^2`

Make
`E(x)=\bar{x}`, then

`E(x^2)=((0.04*340^2)+(0.11*460^2)+(0.47*540^2)+(650^2*0.38))/(1)`

`E(x^2) = 325502`

`V(x) = 325502 - 570^2`

`V(x) = 602`

`SE = √(V(x)) = √(602) = 24.53`