a) This is a geometric distribution wit p=0.2.
P(X=3)=(1−p)2p=0.82∗0.2=0.128.P(X=3)=(1-p)^2p=0.8^2*0.2=0.128.P(X=3)=(1−p)2p=0.82∗0.2=0.128.
b) There is a negative binomial distribution with p=0.2, r=3.
P(X=3)=C7−13−1(1−p)7−3p3=C62∗0.820.23=0.049.P(X=3)=C^{3-1}_{7-1}(1-p)^{7-3}p^3=C^2_6*0.8^20.2^3=0.049.P(X=3)=C7−13−1(1−p)7−3p3=C62∗0.820.23=0.049.
c) Mean: μ=rp=30.2=15.\mu=\frac{r}{p}=\frac{3}{0.2}=15.μ=pr=0.23=15.
Variance: σ2=r(1−p)r2=3∗0.80.22=60.\sigma^2=\frac{r(1-p)}{r^2}=\frac{3*0.8}{0.2^2}=60.σ2=r2r(1−p)=0.223∗0.8=60.
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