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Mathematics 10 Online
OpenStudy (tomfoolery1):

When studying radioactive​ material, a nuclear engineer found that over 365​ days, 1,000,000 radioactive atoms decayed to 971,984 radioactive​ atoms, so 28,016 atoms decayed during 365 days. a. Find the mean number of radioactive atoms that decayed in a day. b. Find the probability that on a given​ day, 50 radioactive atoms decayed.

OpenStudy (threshsupport):

do you know that mean is

OpenStudy (threshsupport):

the mean teacher averaged their grades??

OpenStudy (threshsupport):

so you subtract 971,984 from 1 mil to get the change

OpenStudy (threshsupport):

then you divide that number by 365

OpenStudy (threshsupport):

to find the mean

OpenStudy (tomfoolery1):

Yes its the second half that frustrates me

OpenStudy (threshsupport):

idk about the second part

OpenStudy (tomfoolery1):

76.756

OpenStudy (threshsupport):

im not good at probability unless its like a bag of marbles

OpenStudy (tomfoolery1):

its called a poisson distribution

OpenStudy (anonymous):

Do you know why Poisson distribution should be used here?

OpenStudy (tomfoolery1):

i do not, why?

OpenStudy (anonymous):

Well, Poisson is used when: 1. Occurrences are all independent 2. We know the average number of occurrences in a time span. 3. We want to know the probability of x occurences

OpenStudy (tomfoolery1):

Okay

OpenStudy (anonymous):

So are you familiar with this: \[ \Pr(X=x) = \frac{\mu^xe^{-\mu}}{x!} \]

OpenStudy (anonymous):

Part one lets you find \(\mu\), and part two tells you to use \(x=50\).

OpenStudy (tomfoolery1):

yes i have seen this before! e = 2.71828 i believe!

OpenStudy (anonymous):

Anyway, can you plug it in then?

OpenStudy (anonymous):

\[ \Pr(X=50) = \frac{\left(\frac{28016}{365}\right)^{50}e^{-28016/365}}{50!} \]

OpenStudy (anonymous):

It's going to be a very small number, because getting exactly \(50\) is not a lot

OpenStudy (tomfoolery1):

what is e?

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