Question

Assume that trees are subjected to different levels of carbon dioxide atmosphere with 4% of the trees in a minimal growth condition at 360 parts per million (ppm), 10% at 470 ppm (slow growth), 49% at 540 ppm (moderate growth), and 37% at 650 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

Answer 1

Answer: The mean is 566.5 ppm.

The standard deviation is 74.57 ppm.

Explanation:

Let X represents the concentration of carbon dioxide , then we have the table below

X p(x)

360 0.04

470 0.10

540 0.49

650 0.37

The mean for given Probability mass function:-

`E[x]=\sum_(i=1)^(n) x_i p(x_i)`

i.e.
`E[x]=360\cdot 0.04+470\cdot0.10+540\cdot0.49+650\cdot0.37`

`E[x]=566.5`

∴ The mean is 566.5 ppm.

`E[x^2]=\sum_(i=1)^(n) x_i^2 p(x_i)\\\\=360^2\cdot 0.04+470^2\cdot0.10+540^2\cdot0.49+650^2\cdot0.37\\\\=326483`

Standard deviation:-

`\sigma=√(E[x^2]-E[x]^2)\\\\=√(326483-566.5^2)\\\\=√(326483-320922.25)\\\\=√(5560.75)=74.5704365013\approx74.57`

∴ The standard deviation is 74.57 ppm.