Assume that trees are subjected to different levels of carbon dioxide atmosphere with 4% of the trees in a minimal growth condition at 340 parts per million (ppm), 11% at 430 ppm (slow growth), 49% at 530 ppm (moderate growth), and 36% at 650 ppm (rapid growth). What is the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees

Mean Σ(X) = 554.6

Standard Deviation = 84.397

Explanation:

The probability mass function of the random variable is provided in the text question. We are to calculate the mean, variance and standard deviation.

Mean Σ(X)

= (340 × 0.04) + (430 × 0.11) + (530 × 0.49)+ (650 × 0.36)

Mean Σ(X) = 554.6

Σ(X^2)

= (340^2 × 0.04) + (430^2 × 0.11) + (530^2 × 0.49)+ (650^2 × 0.36)

= 314,704

Var (X) = Σ(X^2) - Σ(X)^2

= 314,704 - (554.6)^2

= 314,704 - 307,581.16

Var (X) = 7,122.84

Standard Deviation = √(var(X))

= √(7,122.84)

Standard Deviation = 84.397

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