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Precalculus Mathematics Homework Help
Solve Probability Question: Find r & E[X] = 7.3
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[QUOTE="LCKurtz, post: 4749689, member: 198114"] Using ##p(k)=P(X=k)## you have shown ##p(k+1)>p(k)## if ##k<6.3##. In particular, with ##k=6## this says ##p(7) > p(6)##. And you know ##p(k+1) < p(k)## if ##k\ge 7##. So wouldn't that make ##p(7)## the largest value of any ##p(k)##? You have $$\sum_{k=0}^\infty ke^{-7.3}\frac{7.3^k}{k!}$$Notice that the first term when ##{k=0}## is zero so you have$$ E(X) = \sum_{k=1}^\infty ke^{-7.3}\frac{7.3^k}{k!}$$Also, since ##\frac k {k!} = \frac 1 {(k-1)!}##we have$$ E(X) = \sum_{k=1}^\infty e^{-7.3}\frac{7.3^k}{(k-1)!}= e^{-7.3}(7.3)\sum_{k=1}^\infty \frac{7.3^{k-1}}{(k-1)!}$$The two constants can be removed from the sum and it is now ready for your next step. [/QUOTE]
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Precalculus Mathematics Homework Help
Solve Probability Question: Find r & E[X] = 7.3
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