Question

Let x ∼ bin(9, 0.4). find

Answer 1

A.
$\mathbb P(X>6)=\mathbb P(X=7)+\mathbb P(X=8)+\mathbb P(X=9)$
$=\dbinom97(0.4)^7(1-0.4)^{9-7}+\dbinom98(0.4)^8(1-0.4)^{9-8}+\dbinom99(0.4)^9(1-0.4)^{9-9}$
$\approx0.025$

b.
$\mathbb P(X\ge2)=1-\mathbb P(X$
$=1-\dbinom90(0.4)^0(1-0.4)^{9-0}-\dbinom91(0.4)^1(1-0.4)^{9-1}$
$\approx0.929$

c.
$\mathbb P(2\le X$
$=\dbinom92(0.4)^2(1-0.4)^{9-2}+\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}$
$\approx0.663$

d.
$\mathbb P(2$
$=\dbinom93(0.4)^3(1-0.4)^{9-3}+\dbinom94(0.4)^4(1-0.4)^{9-4}+\dbinom95(0.4)^5(1-0.4)^{9-5}$
$\approx0.669$

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

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

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

$\mu_X=9(0.4)=3.6$
${\sigma_X}^2=9(0.4)(0.6)=2.16$