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.