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

Answer:

24.53

Step-by-step 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,

$\bar{x} = \frac{\sum\limit_{i=1}^n x_iw_i}{\sum\limit_{i=1}^n w_i}$

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

$\bar{x} = 570.7070$

To find the SE we need the Variance,

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

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

$E(x^2)=\frac{(0.04*340^2)+(0.11*460^2)+(0.47*540^2)+(650^2*0.38)}{1}$

$E(x^2) = 325502$

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

$V(x) = 602$

$SE = \sqrt{V(x)} = \sqrt{602} = 24.53$