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OpenStudy (bloomlocke367):

Can someone help with probability?

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...

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

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?

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 \[P(X\le x)=\int\limits_{-\infty}^{x}\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(t-\mu)^2}{2\sigma^2}}dt\]

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 :)

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

OpenStudy (xapproachesinfinity):

yes

OpenStudy (bloomlocke367):

=94.9

OpenStudy (xapproachesinfinity):

multiply by 300

OpenStudy (bloomlocke367):

94.9*300=28470

OpenStudy (xapproachesinfinity):

sounds good :)

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

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?

OpenStudy (xapproachesinfinity):

yeah

OpenStudy (bloomlocke367):

oh, okay, thanks!

OpenStudy (xapproachesinfinity):

np!

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