Question

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 6% of the trees in a minimal growth condition at 370 parts per million (ppm), 13% at 470 ppm (slow growth), 46% at 560 ppm (moderate growth), and 35% at 630 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees? The mean is Enter your answer; The mean is _ ppm ppm. [Round your answer to one decimal place (e.g. 98.7).] The standard deviation is Enter your answer; The standard deviation is _ ppm ppm. [Round your answer to two decimal places (e.g. 98.76).]

Answer 1

The mean is 561.40 ppm. The standard deviation is 29.15 ppm.

Explanation:

We can calculate the mean and the standard deviation using the proportions of the population that are subjected to the different levels.

The mean can be calculated as

`\mu=\sum p_i*L_i\\\\\mu=0.06*370+0.13*470+0.46*560+0.35*630\\\\\mu=561.40 \,ppm`

The standard deviation of the mean can be calculated in a similar way

`\sigma=√(\sum p_i(L_i-\mu)^2)\\\\\sigma=√(0.06(370-561.4)^2+0.13(470-561.4)^2+0.46(560-561.4)^2+0.35(630-561.4)^2)\\\\\sigma=√(0.06(36634)+0.13(8354)+0.46(2)+0.35(4706))\\\\\sigma=√(2198.04+1086.01+0.90+1647.09)=√(4932.04)=29.15`

The mean is 561.40 ppm. The standard deviation is 29.15 ppm.