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17 November, 08:17

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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  1. 17 November, 08:28
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    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
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