This Question: 3 pts 760
The logistic growth function f(t) =
1+18 2 -0.131 describes the population of a species of butterflies t months after they are introduced to a non-threatening nabitat. How many butterflies are expected in the habitat after 14 months?
nearest whole number.
O A. 294 butterflies
OB. 194 butterflies
O C. 10,640 butterflies
OD. 94 butterflies

c(x)= 6x + 3

step-by-step explanation:

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

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