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JerJason
Group Title
If the probability density of a random variable is given by:
f(x)= Kx^2 for 0<x<1
0 elsewhere
find the value k and the probability that the random variable takes on a value between
(a) 1/4 and 3/4
(b) greater than 2/3
 one year ago
 one year ago
JerJason Group Title
If the probability density of a random variable is given by: f(x)= Kx^2 for 0<x<1 0 elsewhere find the value k and the probability that the random variable takes on a value between (a) 1/4 and 3/4 (b) greater than 2/3
 one year ago
 one year ago

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JerJason Group TitleBest ResponseYou've already chosen the best response.1
So that I'm understanding this correctly. I'm finding the value of "k" in order to verify that the function f(x)=kx^2 is a probability density function based on these conditions: f(x)>=0 and f(x) dx = 1 when x is between infinity and infinity.
 one year ago

JerJason Group TitleBest ResponseYou've already chosen the best response.1
Therefore, \[f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1\] so k=3
 one year ago

JerJason Group TitleBest ResponseYou've already chosen the best response.1
From here I'm plugging in the limits between 1/4 (.25) and 3/4 (.75). \[\int\limits_{.25}^{.75}kx^2dx=k \frac{ x^3 }{ 3 }_{.25}^{.75}=k(\frac{ .4218.0156 }{ 3 })=.4062/3=.1354k\]
 one year ago

JerJason Group TitleBest ResponseYou've already chosen the best response.1
I have a filling I did that last step wrong.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
thats correct, 0.1354*3=0.4062 is the correct probability for a)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
you know how to do part b), right ?
 one year ago

JerJason Group TitleBest ResponseYou've already chosen the best response.1
^ alright good. For (b) do I convert f(x) into F(x), which is the distribution function then plug in the limit .67 (or 2/3) for F(x) and then minus that from 1 like so? Find: \[P(x>.67)\] \[F(x)=\int\limits_{0}^{x}kx^2dx=kx^3/3\] \[F(.67)=3*.67^3/3=.902289/3=.3008\] \[1.3008=.70\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
that is correct, but there is also another way. \(\huge \int \limits_{(2/3)}^1 kx^2dx=....\) and u get same result 0.703
 one year ago

JerJason Group TitleBest ResponseYou've already chosen the best response.1
yes that makes sense as well. Thanks.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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