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anonymous
 3 years ago
Help with #6 on the worksheet....
anonymous
 3 years ago
Help with #6 on the worksheet....

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is what i have: dw:1360104752412:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0b) Expected Value (Height) = ? c) Pr(X>160) d) e) Pr(165<x<185) f) Pr(X>185) Help????

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok b) is 1/(ba) so 1/(210150)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What you said above, 1/(ba), is the probability density function: P(x) = 1/(ba) for all x between a and b, and zero elsewhere. Because it's constant, you can multiply it by the length of any interval to find the probability that someone's height falls in said interval. So, (c) is (160150) / (210150) = 10/60 = 1/6.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0o ok i see... please continue.... my notes have what you just said but are not very clear....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for D would it be 2 times the numerator??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The expectation value is fairly clearly the average height. The rest are just multiplying your probability density function by intervals of different lengths.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, it is not. If A and B are uncorrelated random events, then P(A and B) = P(A) * P(B)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for part E would i do: (185165)/(210150) ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and F would be (210200)/(210150) ?? lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Jemurray3 , you still alive? haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You seem to understand it well enough without me at this point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok. im just making sure.
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