The probability density function of a random variable is given by $f_X(x) = frac3208x^2$ Find $E(X)$

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The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



$(a)$ Compute $E(X)$



$(b)$ Compute $E[X(X+1)]$



My attempt



$(a)$



$$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



$(b)$



$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



$$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$



Would this be right?




probability expectation




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  • 4




    Yes, you are right.
    – Kavi Rama Murthy
    Jul 25 at 23:34






  • 2




    Yes. Place it in the answer and accept it.
    – Graham Kemp
    Jul 25 at 23:58














up vote
2
down vote

favorite












The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



$(a)$ Compute $E(X)$



$(b)$ Compute $E[X(X+1)]$



My attempt



$(a)$



$$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



$(b)$



$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



$$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$



Would this be right?




probability expectation




share|cite|improve this question















  • 4




    Yes, you are right.
    – Kavi Rama Murthy
    Jul 25 at 23:34






  • 2




    Yes. Place it in the answer and accept it.
    – Graham Kemp
    Jul 25 at 23:58












up vote
2
down vote

favorite









up vote
2
down vote

favorite











The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



$(a)$ Compute $E(X)$



$(b)$ Compute $E[X(X+1)]$



My attempt



$(a)$



$$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



$(b)$



$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



$$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$



Would this be right?




probability expectation




share|cite|improve this question











The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



$(a)$ Compute $E(X)$



$(b)$ Compute $E[X(X+1)]$



My attempt



$(a)$



$$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



$(b)$



$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



$$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$



Would this be right?





probability expectation





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 25 at 23:32









Bas bas

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  • 4




    Yes, you are right.
    – Kavi Rama Murthy
    Jul 25 at 23:34






  • 2




    Yes. Place it in the answer and accept it.
    – Graham Kemp
    Jul 25 at 23:58












  • 4




    Yes, you are right.
    – Kavi Rama Murthy
    Jul 25 at 23:34






  • 2




    Yes. Place it in the answer and accept it.
    – Graham Kemp
    Jul 25 at 23:58







4




4




Yes, you are right.
– Kavi Rama Murthy
Jul 25 at 23:34




Yes, you are right.
– Kavi Rama Murthy
Jul 25 at 23:34




2




2




Yes. Place it in the answer and accept it.
– Graham Kemp
Jul 25 at 23:58




Yes. Place it in the answer and accept it.
– Graham Kemp
Jul 25 at 23:58










1 Answer
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The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



$(a)$ Compute $E(X)$



$(b)$ Compute $E[X(X+1)]$



$(a)$



$$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



$(b)$



$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



$$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$






share|cite|improve this answer





















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    The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



    $(a)$ Compute $E(X)$



    $(b)$ Compute $E[X(X+1)]$



    $(a)$



    $$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



    $(b)$



    $$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



    $$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



    Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$






    share|cite|improve this answer

























      up vote
      1
      down vote













      The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



      $(a)$ Compute $E(X)$



      $(b)$ Compute $E[X(X+1)]$



      $(a)$



      $$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



      $(b)$



      $$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



      $$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



      Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



        $(a)$ Compute $E(X)$



        $(b)$ Compute $E[X(X+1)]$



        $(a)$



        $$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



        $(b)$



        $$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



        $$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



        Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$






        share|cite|improve this answer













        The probability density function of a random variable is given by $$f_X(x) = frac3208x^2$$ whenever $2 < x < 6$. $0$ otherwise



        $(a)$ Compute $E(X)$



        $(b)$ Compute $E[X(X+1)]$



        $(a)$



        $$E(X) = int_2^6 xfrac3208x^2dx = frac3208cdotfrac14(x^4)bigg|_2^6 = 60/13$$



        $(b)$



        $$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$



        $$E(X^2) = int_2^6x^2 frac3208x^2 dx = frac3208frac15(x^5)bigg|_2^6 = 1452/65$$



        Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 approx 26.9538 $$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 26 at 7:20









        Bas bas

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