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mariomintchev
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).)
@satellite73 @phi @LoveYou*69 @karatechopper @jim_thompson5910 @Jemurray3 @Hero @AccessDenied
So you could find the probability of five or less, right?
Consider the fact that the only two possible outcomes are that it occurs five or fewer times, or six or more times.
yes, because it's not continuous. it's a discrete number. i have an example like that in my notes.
"Consider the fact that the only two possible outcomes are that it occurs five or fewer times, or six or more times." Ok, continue...
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?
because you can do 1-the other
so should i find plug in 6, 5, 4, 3, 2, 1 into the equation and then subtract that answer from 1??
Good idea. Though probably only from 5 down, if you want to include 6 in "6 or more"
so i should do: p(5)= e^(-2.5) * 2.5^(5) / 5! + p(4)= e^(-2.5) * 2.5^(4) / 4! + ........
I would calculate the prob of 0,1,2,3,4 or 5 events then 1 - sum = pr(k≥6)
so what i said above ^^^ ???
yeah and included 6. i needa include 0 and exclude 6. everything else seems good though, right?
ok, and for part b do i plug in 15 through 20 for p(x) and then add them up?
wait, what? do i place that 20 before e and before the ^ (-x) ??
\[(\lambda) ^ - \lambda x) \
i'm gonna log off but please continue helping (with #s 5-8 on the wkst) everyone!! i'll be back tomorrow to continue this madness. haha
@phi @amistre64 can you clarify what i do with the lambda?
for part b \[ Pr(k)= \frac{\lambda^k e^{-\lambda}}{k!} \] λ= 20, the number of events in 8 hours
If I did it right, I got Pr(15 to 20) = 0.4542
ok what formula is that? i thought you said we use the poisson one for #4?