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navik [9.2K]
3 years ago
13

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 7% of the trees in a minimal growth condit

ion 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
Mathematics
1 answer:
taurus [48]3 years ago
3 0

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

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

And

E[x^2]=370^2\times 0.07+440^2\times 0.1+550^2\times 0.49+670^2\times 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=\sqrt{8078.16}=89.88

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

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