a) This is a geometric distribution wit p=0.2.
"P(X=3)=(1-p)^2p=0.8^2*0.2=0.128."
b) There is a negative binomial distribution with p=0.2, r=3.
"P(X=3)=C^{3-1}_{7-1}(1-p)^{7-3}p^3=C^2_6*0.8^20.2^3=0.049."
c) Mean: "\\mu=\\frac{r}{p}=\\frac{3}{0.2}=15."
Variance: "\\sigma^2=\\frac{r(1-p)}{r^2}=\\frac{3*0.8}{0.2^2}=60."
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