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

Question

Martin Mccall

Biology

2 November, 20:39

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

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Answers (1)

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Jayla Lane · Answer 1

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