Mathematics
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OpenStudy (bloomlocke367):
Can someone help with probability?
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OpenStudy (bloomlocke367):
OpenStudy (bloomlocke367):
@TheSmartOne
OpenStudy (bloomlocke367):
can you help @xapproachesinfinity
OpenStudy (xapproachesinfinity):
i don't remember the prob equation for standard deviation and mean
i believe this is just a direct use of that equation
OpenStudy (bloomlocke367):
oh...
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OpenStudy (xapproachesinfinity):
hm how about reading what that graph is saying
OpenStudy (xapproachesinfinity):
taller than 66.7 would be 13.5+2.35=15.85%
OpenStudy (xapproachesinfinity):
my guess
OpenStudy (bloomlocke367):
I dunno....
OpenStudy (xapproachesinfinity):
well i don't remember any of this stuff
just think that's what the graph is showing hehehe
don't believe me though haha
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OpenStudy (zarkon):
15.85% is correct
OpenStudy (bloomlocke367):
can you explain @Zarkon ?
OpenStudy (zarkon):
you are just adding the percentages that lie above 66.7in
OpenStudy (zarkon):
the probability is the area under the curve from 66.7 to infinity
OpenStudy (bloomlocke367):
why do you add those 2 numbers though?
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OpenStudy (zarkon):
those are the ones above 66.7
OpenStudy (xapproachesinfinity):
is this the integral \[\int e^{-x^2}dx\]
OpenStudy (bloomlocke367):
how do you do part b?
OpenStudy (zarkon):
add up the probabilities and then multiply by 300
OpenStudy (zarkon):
Let \(X\sim N(\mu,\sigma^2)\)
then \[P(X\le x)=\int\limits_{-\infty}^{x}\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(t-\mu)^2}{2\sigma^2}}dt\]
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OpenStudy (bloomlocke367):
woah, I'm sorry, but that's Greek to me XD
OpenStudy (zarkon):
though you don't need to use that here since the probabilities are already given
OpenStudy (xapproachesinfinity):
i completely forgot this thing!
so mu is the mean?
OpenStudy (zarkon):
yes
OpenStudy (xapproachesinfinity):
i see, thanks :)
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OpenStudy (bloomlocke367):
what do you mean the prob. are given?
OpenStudy (xapproachesinfinity):
so in generality in \[\int e^{-x^2}dx\] where x is the random variable
i remember this integral from calc one of those special integrals
OpenStudy (xapproachesinfinity):
the probabilities are given to u
the percentages are the chances
OpenStudy (xapproachesinfinity):
just add them up and multiply by 300
OpenStudy (bloomlocke367):
wait, so 13.5+34+34+13.5
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OpenStudy (xapproachesinfinity):
yes
OpenStudy (bloomlocke367):
=94.9
OpenStudy (xapproachesinfinity):
multiply by 300
OpenStudy (bloomlocke367):
94.9*300=28470
OpenStudy (xapproachesinfinity):
sounds good :)
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OpenStudy (bloomlocke367):
that doesn't make sense, pertaining to the question...
OpenStudy (xapproachesinfinity):
why is not reasonable?
OpenStudy (bloomlocke367):
if the data's based on 300 students, how is the number greater than 300?
OpenStudy (xapproachesinfinity):
oh you should divide the percentages by 100
OpenStudy (xapproachesinfinity):
add those values and divide by 100 before you multiply
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OpenStudy (bloomlocke367):
so divide 94.95 by 100? why?
OpenStudy (xapproachesinfinity):
we are looking for prob not percentages
OpenStudy (bloomlocke367):
oh, okay
OpenStudy (xapproachesinfinity):
should make since now :)
OpenStudy (bloomlocke367):
284.85... and since we're talking about people, I would need to round. would I round up?
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OpenStudy (xapproachesinfinity):
yeah
OpenStudy (bloomlocke367):
oh, okay, thanks!
OpenStudy (xapproachesinfinity):
np!