Question

5. The World Series determines the national championship team for Major League Baseball. In the World Series, two teams (A and B) play a series of games until one wins 4 games. Assume that the games are played independently and the probability that Team A wins is 0.40 and this does not change from game to game. Let the random variable X represent the total number of games required in a randomly selected World Series. The CDF for X is: 0 0.15524<x<5 etグ 2. 0.72352 6 S<7 al r27 (a) In this example, is X discrete or contimous? Explain. Be sure to reference both the (b) SHOW that k (c) Find the moment generating function (MGF) for X. Show and explain all work. te or continuous? Explain. Be sure to reference both the CDF and the definition of X in your explanation. 0.424 for 2-5. You may assume without proof that the CDF is 0.1552 for 4 K< 5. ethe NCP nf tut ck

Answer 1

$(a)~Since~X~represents~the~total~no.~of~games~required~in~a~randomly\\ selected~World~Series~hence~X~is~discrete~random~variable.~Moreover~F(x)\\ is~not~continuous~since~for~example\\ F(4-0)=0,~F(4)=0.1552.\\ (b)~The~pmf~of~X\\ P(X=4)=F(4)-F(4-0)=0.1552\\ ~P(X=5)=F(5)-F(5-0)=k-0.1552\\ P(X=6)=F(6)-F(6-0)=0.72352-k\\ P(X=7)=F(7)-F(7-0)=1-0.72352=0.27648 \\ P(X=x)=0~when~x\neq 4,5,6,7\\ P(X=5)=Probability~that~either~A~wins~4~games~or~B~wins~4~games\\ =\binom{4}{3}(0.6)*(0.4)^4&plus;\binom{4}{3}(0.4)*(0.6)^4=0.2688\\ Again~P(X=5)=k-0.1552\\ or,~0.2688=k-0.1552~or,~k=0.2688&plus;0.1552=0.424.\\ (c)~MGF~of~X=M_X(t)=E(e^{tX})\\ =0.1552e^{4t}&plus;0.2688e^{5t}&plus;0.29952e^{6t}&plus;0.27648e^{7t}\\ where ~P(X=7)=0.29952\\ (d)~E(X)=\left.\begin{matrix} \frac{\mathrm{d} M_X(t)}{\mathrm{d} t} \end{matrix}\right|_{t=0}=0.1552*4&plus;0.2688*5&plus;0.29952*6&plus;0.27648*7\\ =5.69728$