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4 years ago

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Assume that trees are subjected to different levels of carbon dioxide atmosphere with 8% of the trees in a minimal growth condit

ion at 350 parts per million (ppm), 11% at 440 ppm (slow growth), 48% at 560 ppm (moderate growth), and 33% at 640 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees? The mean is ppm. [Round your answer to one decimal place (e.g. 98.7).] The standard deviation is ppm. [Round your answer to two decimal places (e.g. 98.76).]

1 answer:

Daniel [21]4 years ago

7 0

Answer: The mean is <u>556.4</u> ppm.

The standard deviation is<u> 84.92</u> ppm.

Step-by-step explanation:

Let X denotes the concentration of carbon (ppm) .

Given :

X p(x)

350 0.08

440 0.11

560 0.48

640 0.33

The mean(expected value) is given by :-

$E[x]=\sum x p(x)$

$\Rightarrow\ E[x]=(350)(0.08)+(440)(0.11)+(560)(0.48)+(640)(0.33)=556.4$

Hence, the mean is <u>556.4</u> ppm.

Now, $E[x^2]=\sum x^2 p(x)$

$\Rightarrow\ E[x]=(350)^2(0.08)+(440)^2(0.11)+(560)^2(0.48)+(640)^2(0.33)=316792$

$\text{Var(x)}=E[x^2]-[E[x]]^2\\\\=316792-556.4^2\\\\=316792-309580.96=7211.04$

Standard deviation: $\sigma=\sqrt{7211.04}=84.9178426481\approx84.92$

Hence, The standard deviation is<u> 84.92</u> ppm.

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Step-by-step explanation:

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We are given $N_0=800$ at $t=0$ .

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Put into calculator:

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Divide both sides by -0.069:

$\frac{\ln(\frac{1}{2})}{-0.069}=t$

Input into calculator:

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$t=10.0$

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