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](https://img.qammunity.org/2021/formulas/mathematics/high-school/8inwe3k81yvw5exa4rctjjcxgpl5kkefmj.png)
And
![E[x^2]=370^2* 0.07+440^2* 0.1+550^2* 0.49+670^2* 0.34=329794](https://img.qammunity.org/2021/formulas/mathematics/high-school/rwmhrm5i2i2xnzmgyufk2iuel6sus6focg.png)
So, Variance would be
![Var\ x=E[x^2]-E[x]^2=329794-567.2^2=8078.16](https://img.qammunity.org/2021/formulas/mathematics/high-school/gr9ihuwhth4j2rh6ycph8qkpzoiibmpb0u.png)
So, the standard deviation would be

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