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?

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

the slope of the first function is -3. (3/1)

the slope of the second function is 1. (13-10/6-3).

since |-3| is greater than |1|, the function with the steepest slope is the first.

2 is the final ! : )