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

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windyyork windyyork · Answer 1 · 2020-01-31T11:23:01+01:00

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

Step-by-step 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

[tex]E[x]=370\times 0.07+440\times 0.1+550\times 0.49+670\times 0.34=567.2[/tex]

And

[tex]E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 0.34=329794[/tex]

So, Variance would be

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

So, the standard deviation would be

[tex]\sigma=\sqrt{8078.16}=89.88[/tex]

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