20.0k views
3 votes
Assume that trees are subjected to different levels of carbon dioxide atmosphere with 7% of the trees in a minimal growth condition at 370 parts per million (ppm), 10% at 440 ppm (slow growth), 49% at 550 ppm (moderate growth), and 34% at 670 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees

1 Answer

3 votes

Answer: The mean and standard deviation are 567.2 and 89.88 resp.

Explanation:

Since we have given that

For 370 parts per million = 7% = 0.07

For 440 parts per million = 10% = 0.10

For 550 parts per million = 49% = 0.49

For 670 parts per million = 34% = 0.34

So, Mean of the carbon dioxide atmosphere for these trees would be


E[x]=370* 0.07+440* 0.1+550* 0.49+670* 0.34=567.2

And


E[x^2]=370^2* 0.07+440^2* 0.1+550^2* 0.49+670^2* 0.34=329794

So, Variance would be


Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16

So, the standard deviation would be


\sigma=√(8078.16)=89.88

Hence, the mean and standard deviation are 567.2 and 89.88 resp.

User Tushar Sharma
by
5.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.