Respuesta :

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a.
[tex]\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)[/tex]
[tex]=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}[/tex]
[tex]\approx0.025[/tex]

b.
[tex]\mathbb P(X\ge2)=1-\mathbb P(X<2)=1-\mathbb P(X=0)-\mathbb P(X=1)[/tex]
[tex]=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}[/tex]
[tex]\approx0.929[/tex]

c.
[tex]\mathbb P(2\le X<5)=\mathbb P(X=2)+\mathbb P(X=3)+\mathbb P(X=4)[/tex]
[tex]=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}[/tex]
[tex]\approx0.663[/tex]

d.
[tex]\mathbb P(2<X\le5)=\mathbb P(X=3)+\mathbb P(X=4)+\mathbb P(X=5)[/tex]
[tex]=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}[/tex]
[tex]\approx0.669[/tex]

e.
[tex]\mathbb P(X=0)=\dbinom90(0.4)^0(1-0.4)^{9-0}\approx0.01[/tex]

f.
[tex]\mathbb P(X=7)=\dbinom97(0.4)^7(1-0.4)^{9-7}\approx0.021[/tex]

g, h.
For [tex]X\sim\mathcal B(n,p)[/tex], recall that [tex]\mathbb E[X]=\mu_X=np[/tex] and [tex]\mathbb V[X]={\sigma_X}^2=np(1-p)[/tex]. So

[tex]\mu_X=9(0.4)=3.6[/tex]
[tex]{\sigma_X}^2=9(0.4)(0.6)=2.16[/tex]

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