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mariomintchevBest ResponseYou've already chosen the best response.0
this is what i have: dw:1360104752412:dw
 one year ago

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

mariomintchevBest ResponseYou've already chosen the best response.0
ok b) is 1/(ba) so 1/(210150)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
hi @Jemurray3 ! :D
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
**** c) Pr(X<160)
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
What 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.
 one year ago

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

mariomintchevBest ResponseYou've already chosen the best response.0
for D would it be 2 times the numerator??
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
The expectation value is fairly clearly the average height. The rest are just multiplying your probability density function by intervals of different lengths.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
No, it is not. If A and B are uncorrelated random events, then P(A and B) = P(A) * P(B)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so 1/6 times 1/6 ??
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
for part E would i do: (185165)/(210150) ??
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
and F would be (210200)/(210150) ?? lol
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@Jemurray3 , you still alive? haha
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
You seem to understand it well enough without me at this point
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok. im just making sure.
 one year ago
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