Answer:
a) 0.482 , 0.099 and 0.231 respectively
b) 0.609 and 0.053 respectively
Step-by-step explanation:
We can modeled the number of wolves in New Mexico and Arizona as a binomial random variable.
The binomial random variable has the following probability distribution :
[tex]P(X=k)=(nCk)p^{k}(1-p)^{n-k}[/tex]
Where P(X=k) is the probability of the variable X to assume the value k
p is the success probability
n is the sample
a) Before 1918.
We define the random variables :
X : ''number of male wolves''
Y : ''number of female wolves''
X ~ Bi (9,0.6)
Y ~ Bi (9,0.4)
[tex]P(X\geq 6)=P(X=6)+P(X=7)+P(X=8)+P(X=9)[/tex]
[tex]P(X\geq 6)=(9C6)0.6^{6}0.4^{3}+(9C7)0.6^{7}0.4^{2}+(9C8)0.6^{8}0.4^{1}+(9C9)0.6^{9}0.4^{0}[/tex]
[tex]P(X\geq 6)=0.251+0.161+0.060+0.010=0.482[/tex]
[tex]P(Y\geq 6)=P(Y=6)+P(Y=7)+P(Y=8)+P(Y=9)[/tex]
[tex]P(Y\geq 6)=(9C6)0.4^{6}0.6^{3}+(9C7)0.4^{7}0.6^{2}+(9C8)0.4^{8}0.6^{1}+(9C9)0.4^{9}0.6^{0}[/tex]
[tex]P(Y\geq 6)=0.074+0.021+(3.54)10^{-3}+(2.62)10^{-4}=0.099[/tex]
[tex]P(Y<3)=P(Y=0)+P(Y=1)+P(Y=2)[/tex]
[tex]P(Y<3)=(9C0)0.4^{0}0.6^{9}+(9C1)0.4^{1}0.6^{8}+(9C2)0.4^{2}0.6^{7}[/tex]
[tex]P(Y<3)=0.010+0.060+0.161=0.231[/tex]
b) For the period from 1918 to the present.
X and Y are the same random variables but with a different distribution :
X ~ Bi(9,0.65)
Y ~ Bi(9,0.35)
[tex]P(X\geq 6)=P(X=6)+P(X=7)+P(X=8)+P(X=9)[/tex]
[tex]P(X\geq 6)=(9C6)0.65^{6}0.35^{3}+(9C7)0.65^{7}0.35^{2}+(9C8)0.65^{8}0.35^{1}+(9C9)0.65^{9}0.35^{0}[/tex]
[tex]P(X\geq 6)=0.272+0.216+0.100+0.021=0.609[/tex]
[tex]P(Y\geq 6)=P(Y=6)+P(Y=7)+P(Y=8)+P(Y=9)[/tex]
[tex]P(Y\geq 6)=(9C6)0.35^{6}0.65^{3}+(9C7)0.35^{7}0.65^{2}+(9C8)0.35^{8}0.65^{1}+(9C9)0.35^{9}0.65^{0}[/tex]
[tex]P(Y\geq 6)=0.042+(9.79)10^{-3}+(1.32)10^{-3}+(7.89)10^{-5}=0.053[/tex]
