Trees are subjected to different levels of carbon dioxide atmosphere with 6% of them in a minimal growth condition at 350 parts per million (ppm), 10% at 450 ppm (slow growth), 47% at 550 ppm (moderate growth), and 37% at 650 ppm (rapid growth). What are the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees in ppm?

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Mathematics

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5 April, 05:43

Trees are subjected to different levels of carbon dioxide atmosphere with 6% of them in a minimal growth condition at 350 parts per million (ppm), 10% at 450 ppm (slow growth), 47% at 550 ppm (moderate growth), and 37% at 650 ppm (rapid growth). What are the mean and standard deviation of the carbon dioxide atmosphere (in ppm) for these trees in ppm?

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Cash · Answer 1

Mean = 528 ppm

Standard deviation = 90.8 ppm

Step-by-step explanation:

Assuming a basis of 100 trees

6 trees with 350 ppm (minimal growth)

10 trees with 450 ppm (slow growth)

47 trees with 550 ppm (moderate growth)

37 trees with 650 ppm (rapid growth)

Mean = xbar = Σx/N

x = each variable

xbar = mean

N = number of variables = 100

Σx = sum of all variables = sum of all the ppm = (6*350) + (10*450) + (47*550) + (37*550) = 52800

xbar = 52800/100 = 528 ppm

Standard deviation = σ = √[Σ (x - xbar) ²/N]

x = each variable

xbar = mean = 528

N = number of variables = 12

Σ (x - xbar) ² = [6 (350 - 528) ²] + [10 (450 - 528) ²] + [47 (550 - 528) ²] + [37 (650 - 528) ²] = 824400

σ = √[Σ (x - xbar) ²/N] = √ (824400/100) = 90.8 ppm