Question on finding the median of a discrete random variable












2












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enter image description here



I have been trying to solve this math problem, but I end up with a different answer, I suspect the publisher probably used a different approach but I am not sure which one.



Here is my attempt:



My cumulative distribution function, $F(x) = P(X leq x)$ is same as what they have.



To find the median I used this definition: The median is the value of $X$ for which $P(X leq x)$ is greater than or equal to 0.5 and $P(X geq x)$ is greater than or equal to 0.5.



So the median is 2 because $P(X leq 2) = 0.7$ and $P(X geq 2) = 0.7$



I will appreciate any help.






probability-distributions







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  • 1




    $begingroup$
    Bad textbook. You were right to stick to the definition.
    $endgroup$
    – Did
    Jan 1 at 18:24
















2












$begingroup$


enter image description here



I have been trying to solve this math problem, but I end up with a different answer, I suspect the publisher probably used a different approach but I am not sure which one.



Here is my attempt:



My cumulative distribution function, $F(x) = P(X leq x)$ is same as what they have.



To find the median I used this definition: The median is the value of $X$ for which $P(X leq x)$ is greater than or equal to 0.5 and $P(X geq x)$ is greater than or equal to 0.5.



So the median is 2 because $P(X leq 2) = 0.7$ and $P(X geq 2) = 0.7$



I will appreciate any help.






probability-distributions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Bad textbook. You were right to stick to the definition.
    $endgroup$
    – Did
    Jan 1 at 18:24














2












2








2





$begingroup$


enter image description here



I have been trying to solve this math problem, but I end up with a different answer, I suspect the publisher probably used a different approach but I am not sure which one.



Here is my attempt:



My cumulative distribution function, $F(x) = P(X leq x)$ is same as what they have.



To find the median I used this definition: The median is the value of $X$ for which $P(X leq x)$ is greater than or equal to 0.5 and $P(X geq x)$ is greater than or equal to 0.5.



So the median is 2 because $P(X leq 2) = 0.7$ and $P(X geq 2) = 0.7$



I will appreciate any help.






probability-distributions







share|cite|improve this question











$endgroup$




enter image description here



I have been trying to solve this math problem, but I end up with a different answer, I suspect the publisher probably used a different approach but I am not sure which one.



Here is my attempt:



My cumulative distribution function, $F(x) = P(X leq x)$ is same as what they have.



To find the median I used this definition: The median is the value of $X$ for which $P(X leq x)$ is greater than or equal to 0.5 and $P(X geq x)$ is greater than or equal to 0.5.



So the median is 2 because $P(X leq 2) = 0.7$ and $P(X geq 2) = 0.7$



I will appreciate any help.






probability-distributions




probability-distributions






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 10:55









twnly

1,2341214




1,2341214










asked Jan 1 at 10:37









Meghan CMeghan C

21328




21328








  • 1




    $begingroup$
    Bad textbook. You were right to stick to the definition.
    $endgroup$
    – Did
    Jan 1 at 18:24














  • 1




    $begingroup$
    Bad textbook. You were right to stick to the definition.
    $endgroup$
    – Did
    Jan 1 at 18:24








1




1




$begingroup$
Bad textbook. You were right to stick to the definition.
$endgroup$
– Did
Jan 1 at 18:24




$begingroup$
Bad textbook. You were right to stick to the definition.
$endgroup$
– Did
Jan 1 at 18:24










1 Answer
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2












$begingroup$

I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $Xsim Bin(4,0.5)$



Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:




  1. $P(Xleq x_m)geq 0.5$


  2. $P(Xgeq x_m)geq 0.5$



For $x_m=1$ we get:




  1. $sum_{k=0}^1 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}=frac{5}{16} not geq 0.5 quad quad large{color{red}{times}}$


  2. $sum_{k=1}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{4}{16}+frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{15}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$




  1. $sum_{k=0}^2 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}+frac{6}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$


  2. $sum_{k=2}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.






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    2












    $begingroup$

    I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $Xsim Bin(4,0.5)$



    Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:




    1. $P(Xleq x_m)geq 0.5$


    2. $P(Xgeq x_m)geq 0.5$



    For $x_m=1$ we get:




    1. $sum_{k=0}^1 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}=frac{5}{16} not geq 0.5 quad quad large{color{red}{times}}$


    2. $sum_{k=1}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{4}{16}+frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{15}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



    We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$




    1. $sum_{k=0}^2 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}+frac{6}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$


    2. $sum_{k=2}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



    Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $Xsim Bin(4,0.5)$



      Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:




      1. $P(Xleq x_m)geq 0.5$


      2. $P(Xgeq x_m)geq 0.5$



      For $x_m=1$ we get:




      1. $sum_{k=0}^1 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}=frac{5}{16} not geq 0.5 quad quad large{color{red}{times}}$


      2. $sum_{k=1}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{4}{16}+frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{15}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



      We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$




      1. $sum_{k=0}^2 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}+frac{6}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$


      2. $sum_{k=2}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



      Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $Xsim Bin(4,0.5)$



        Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:




        1. $P(Xleq x_m)geq 0.5$


        2. $P(Xgeq x_m)geq 0.5$



        For $x_m=1$ we get:




        1. $sum_{k=0}^1 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}=frac{5}{16} not geq 0.5 quad quad large{color{red}{times}}$


        2. $sum_{k=1}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{4}{16}+frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{15}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



        We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$




        1. $sum_{k=0}^2 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}+frac{6}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$


        2. $sum_{k=2}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



        Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.






        share|cite|improve this answer









        $endgroup$



        I agree that the median $x_m$ is $2$ and must be an integer. The random variable is distributed as $Xsim Bin(4,0.5)$



        Then we have to find the integer of $X$ which satisfies the following two conditions at the same time:




        1. $P(Xleq x_m)geq 0.5$


        2. $P(Xgeq x_m)geq 0.5$



        For $x_m=1$ we get:




        1. $sum_{k=0}^1 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}=frac{5}{16} not geq 0.5 quad quad large{color{red}{times}}$


        2. $sum_{k=1}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{4}{16}+frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{15}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



        We have to check the next higher integer since the first condition is not fulfilled: $x_m=2$




        1. $sum_{k=0}^2 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{1}{16}+frac{4}{16}+frac{6}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$


        2. $sum_{k=2}^4 binom{4}{k} cdot 0.5^kcdot 0.5^{4-k}=frac{6}{16}+frac{4}{16}+frac{1}{16}=frac{11}{16} geq 0.5 quad quad large{color{green}{checkmark}}$



        Both conditions are fulfilled. I´ve never got the number $0.7$ but the median is indeed $2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 18:09









        callculuscallculus

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        18.9k31729






























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