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anonymous
 3 years ago
I need help with problem 4. I'm supposed to use the Poisson Distribution, but I'm confused because part A says 6 or MORE. (I obviously can't go to infinity, so what do I do? If it was 6 or LESS, I would plug in 6, 5, 4, 3, 2, 1... in for P(x).)
anonymous
 3 years ago
I need help with problem 4. I'm supposed to use the Poisson Distribution, but I'm confused because part A says 6 or MORE. (I obviously can't go to infinity, so what do I do? If it was 6 or LESS, I would plug in 6, 5, 4, 3, 2, 1... in for P(x).)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 @phi @LoveYou*69 @karatechopper @jim_thompson5910 @Jemurray3 @Hero @AccessDenied

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you could find the probability of five or less, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Consider the fact that the only two possible outcomes are that it occurs five or fewer times, or six or more times.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, because it's not continuous. it's a discrete number. i have an example like that in my notes.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"Consider the fact that the only two possible outcomes are that it occurs five or fewer times, or six or more times." Ok, continue...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha well, that was the main hint... If there are only two possible outcomes, and you calculate the probability for one, then you immediately know the probability of the other, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because you can do 1the other

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so should i find plug in 6, 5, 4, 3, 2, 1 into the equation and then subtract that answer from 1??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Good idea. Though probably only from 5 down, if you want to include 6 in "6 or more"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i should do: p(5)= e^(2.5) * 2.5^(5) / 5! + p(4)= e^(2.5) * 2.5^(4) / 4! + ........

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I would calculate the prob of 0,1,2,3,4 or 5 events then 1  sum = pr(k≥6)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so what i said above ^^^ ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah and included 6. i needa include 0 and exclude 6. everything else seems good though, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, and for part b do i plug in 15 through 20 for p(x) and then add them up?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, what? do i place that 20 before e and before the ^ (x) ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(\lambda) ^  \lambda x) \

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm gonna log off but please continue helping (with #s 58 on the wkst) everyone!! i'll be back tomorrow to continue this madness. haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@phi @amistre64 can you clarify what i do with the lambda?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0for part b \[ Pr(k)= \frac{\lambda^k e^{\lambda}}{k!} \] λ= 20, the number of events in 8 hours

phi
 3 years ago
Best ResponseYou've already chosen the best response.0If I did it right, I got Pr(15 to 20) = 0.4542

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok what formula is that? i thought you said we use the poisson one for #4?
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