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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interphase   is the cell cycle stage in which the cell acquires nutrients, creates and uses proteins and other molecules and starts the process of cell division by replicating the dna. it is the longest stage in the eukaryote cell cycle.

the following   can go wrong during interphase:   a base pair could be mismatched during the duplication process causing a mutation.   correct answer: c

explanation: