G(x) is f(x) rotated about the x-axis and then compressed vertically by a factor of 4/5.
Answer:
a) P=0.558
b) P=0.021
Step-by-step explanation:
We can model this random variable as a Poisson distribution with parameter λ=1/500*2000=4.
The approximate distribution of the number who carry this gene in a sample of 2000 individuals is:

a) We can calculate that the approximate probability that between 4 and 9 (inclusive) as:


b) The approximate probability that at least 9 carry the gene is:



Answer:
The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.
Step-by-step explanation:
The given equation is

Where Y is the annual growth rate of a certain country's population and x is the number of years after 1900.
Difference between 1950 and 1900 is 50.
Put x=50 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.
Difference between 1988 and 1900 is 88.
Put x=88 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.
Difference between 2010 and 1900 is 110.
Put x=110 in the given equation.


Therefore the estimated average annual growth rate of the country's population for 2010 is 0.2236.