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JerJason

  • 3 years ago

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

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  1. JerJason
    • 3 years ago
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    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.

  2. JerJason
    • 3 years ago
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    Therefore, \[f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1\] so k=3

  3. JerJason
    • 3 years ago
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    From here I'm plugging in the limits between 1/4 (.25) and 3/4 (.75). \[\int\limits_{.25}^{.75}kx^2dx=k \frac{ x^3 }{ 3 }_{.25}^{.75}=k(\frac{ .4218-.0156 }{ 3 })=.4062/3=.1354k\]

  4. JerJason
    • 3 years ago
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    I have a filling I did that last step wrong.

  5. hartnn
    • 3 years ago
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    thats correct, 0.1354*3=0.4062 is the correct probability for a)

  6. hartnn
    • 3 years ago
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    you know how to do part b), right ?

  7. JerJason
    • 3 years ago
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    ^ 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: \[P(x>.67)\] \[F(x)=\int\limits_{0}^{x}kx^2dx=kx^3/3\] \[F(.67)=3*.67^3/3=.902289/3=.3008\] \[1-.3008=.70\]

  8. hartnn
    • 3 years ago
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    that is correct, but there is also another way. \(\huge \int \limits_{(2/3)}^1 kx^2dx=....\) and u get same result 0.703

  9. JerJason
    • 3 years ago
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    yes that makes sense as well. Thanks.

  10. hartnn
    • 3 years ago
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    welcome ^_^

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