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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).)
 one year ago
 one year 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).)
 one year ago
 one year ago

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mariomintchevBest ResponseYou've already chosen the best response.0
@satellite73 @phi @LoveYou*69 @karatechopper @jim_thompson5910 @Jemurray3 @Hero @AccessDenied
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.0
So you could find the probability of five or less, right?
 one year ago

Jemurray3Best 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.
 one year ago

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

mariomintchevBest 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...
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.0
haha 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?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
because you can do 1the other
 one year ago

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

Jemurray3Best ResponseYou've already chosen the best response.0
Good idea. Though probably only from 5 down, if you want to include 6 in "6 or more"
 one year ago

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

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

mariomintchevBest ResponseYou've already chosen the best response.0
so what i said above ^^^ ???
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
yeah and included 6. i needa include 0 and exclude 6. everything else seems good though, right?
 one year ago

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

mariomintchevBest ResponseYou've already chosen the best response.0
wait, what? do i place that 20 before e and before the ^ (x) ??
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
\[(\lambda) ^  \lambda x) \
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i'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
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@phi @amistre64 can you clarify what i do with the lambda?
 one year ago

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

phiBest ResponseYou've already chosen the best response.0
If I did it right, I got Pr(15 to 20) = 0.4542
 one year ago

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