Respuesta :
Answer:
a) 56.97% probability that fewer than 3 field mice are found on a given acre.
b) 41.90% probability that in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.
Step-by-step explanation:
To solve this question, we use the Poisson probability distribution and the binomial probability distribution.
Poisson distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
Binomial distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
If the average number of field mice in a 5-acre wheat field is estimated to be 12.
Find the probability that
(a) fewer than 3 field mice are found on a given acre.
5 acres have a mean of 12. For one acre
[tex]\mu = \frac{12}{5} = 2.4[/tex]
This probability is
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2.4}*(2.4)^{0}}{(0)!} = 0.0907[/tex]
[tex]P(X = 1) = \frac{e^{-2.4}*(2.4)^{1}}{(1)!} = 0.2177[/tex]
[tex]P(X = 2) = \frac{e^{-2.4}*(2.4)^{2}}{(2)!} = 0.2613[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0907 + 0.2177 + 0.2613 = 0.5697[/tex]
56.97% probability that fewer than 3 field mice are found on a given acre.
(b) in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.
Here we use the binomial probability distribution.
56.97% probability that fewer than 3 field mice are found on a given acre, which means that [tex]p = 0.5697[/tex]
This probability is P(X = 2) when n = 3.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{3,2}.(0.5697)^{2}.(0.4303)^{1} = 0.4190[/tex]
41.90% probability that in 2 of the next 3 acres inspected, fewer than 3 field mice are found on each acre.
