anonymous
  • anonymous
If the probability density of a random variable is given by: f(x)= Kx^2 for 0
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Therefore, \[f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1\] so k=3
anonymous
  • anonymous
From here I'm plugging in the limits between 1/4 (.25) and 3/4 (.75). \[\int\limits_{.25}^{.75}kx^2dx=k \frac{ x^3 }{ 3 }_{.25}^{.75}=k(\frac{ .4218-.0156 }{ 3 })=.4062/3=.1354k\]

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anonymous
  • anonymous
I have a filling I did that last step wrong.
hartnn
  • hartnn
thats correct, 0.1354*3=0.4062 is the correct probability for a)
hartnn
  • hartnn
you know how to do part b), right ?
anonymous
  • anonymous
^ 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: \[P(x>.67)\] \[F(x)=\int\limits_{0}^{x}kx^2dx=kx^3/3\] \[F(.67)=3*.67^3/3=.902289/3=.3008\] \[1-.3008=.70\]
hartnn
  • hartnn
that is correct, but there is also another way. \(\huge \int \limits_{(2/3)}^1 kx^2dx=....\) and u get same result 0.703
anonymous
  • anonymous
yes that makes sense as well. Thanks.
hartnn
  • hartnn
welcome ^_^

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