Median of a fair $6$ sided die, rolling once












1












$begingroup$


If you roll a $6$ sided die once. What would be the median outcome?



I made a table:



begin{array}{|c|c|c|}
hline
x & P(X= x) & F(x) = P(X le x) \ hline
1 & 1/6 & 1/6 \ hline
2 & 1/6 & 2/6 \ hline
3 & 1/6 & 3/6 = 1/2 \ hline
4 & 1/6 & 4/6 \ hline
5 & 1/6 & 5/6 \ hline
6 & 1/6 & 1 \ hline
end{array}



So by definition, whatever value the first time $F(x) ge 1/2$ will be the median. I thought the median would be $3$ then. But $F(x) = 1/2$ for any $x$ s.t $3 le x < 4$ and if you roll the die many times it will converge to $3.5$. So then there would be infinitely many medians between $3$ and $4$? Can anyone explain more on this for me?






probability







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  • 1




    $begingroup$
    Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
    $endgroup$
    – Alex
    Jan 25 at 15:30
















1












$begingroup$


If you roll a $6$ sided die once. What would be the median outcome?



I made a table:



begin{array}{|c|c|c|}
hline
x & P(X= x) & F(x) = P(X le x) \ hline
1 & 1/6 & 1/6 \ hline
2 & 1/6 & 2/6 \ hline
3 & 1/6 & 3/6 = 1/2 \ hline
4 & 1/6 & 4/6 \ hline
5 & 1/6 & 5/6 \ hline
6 & 1/6 & 1 \ hline
end{array}



So by definition, whatever value the first time $F(x) ge 1/2$ will be the median. I thought the median would be $3$ then. But $F(x) = 1/2$ for any $x$ s.t $3 le x < 4$ and if you roll the die many times it will converge to $3.5$. So then there would be infinitely many medians between $3$ and $4$? Can anyone explain more on this for me?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
    $endgroup$
    – Alex
    Jan 25 at 15:30














1












1








1





$begingroup$


If you roll a $6$ sided die once. What would be the median outcome?



I made a table:



begin{array}{|c|c|c|}
hline
x & P(X= x) & F(x) = P(X le x) \ hline
1 & 1/6 & 1/6 \ hline
2 & 1/6 & 2/6 \ hline
3 & 1/6 & 3/6 = 1/2 \ hline
4 & 1/6 & 4/6 \ hline
5 & 1/6 & 5/6 \ hline
6 & 1/6 & 1 \ hline
end{array}



So by definition, whatever value the first time $F(x) ge 1/2$ will be the median. I thought the median would be $3$ then. But $F(x) = 1/2$ for any $x$ s.t $3 le x < 4$ and if you roll the die many times it will converge to $3.5$. So then there would be infinitely many medians between $3$ and $4$? Can anyone explain more on this for me?






probability







share|cite|improve this question









$endgroup$




If you roll a $6$ sided die once. What would be the median outcome?



I made a table:



begin{array}{|c|c|c|}
hline
x & P(X= x) & F(x) = P(X le x) \ hline
1 & 1/6 & 1/6 \ hline
2 & 1/6 & 2/6 \ hline
3 & 1/6 & 3/6 = 1/2 \ hline
4 & 1/6 & 4/6 \ hline
5 & 1/6 & 5/6 \ hline
6 & 1/6 & 1 \ hline
end{array}



So by definition, whatever value the first time $F(x) ge 1/2$ will be the median. I thought the median would be $3$ then. But $F(x) = 1/2$ for any $x$ s.t $3 le x < 4$ and if you roll the die many times it will converge to $3.5$. So then there would be infinitely many medians between $3$ and $4$? Can anyone explain more on this for me?






probability




probability






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asked Jan 25 at 15:16









ItsnhantransitiveItsnhantransitive

789315




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  • 1




    $begingroup$
    Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
    $endgroup$
    – Alex
    Jan 25 at 15:30














  • 1




    $begingroup$
    Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
    $endgroup$
    – Alex
    Jan 25 at 15:30








1




1




$begingroup$
Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
$endgroup$
– Alex
Jan 25 at 15:30




$begingroup$
Yep, there is no unique median here, unless you restrict the possible values for the median to ${1, 2, 3, 4, 5, 6}$.
$endgroup$
– Alex
Jan 25 at 15:30










1 Answer
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By definition a median connected with a random variable $X$ (or if you like its distribution) is a real number $m$ that satisfies the inequalities:$$P(Xgeq m)geq0.5text{ and }P(Xleq m)geq0.5$$ So in your case every element of $[3,4]$ is a median.






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    $begingroup$

    By definition a median connected with a random variable $X$ (or if you like its distribution) is a real number $m$ that satisfies the inequalities:$$P(Xgeq m)geq0.5text{ and }P(Xleq m)geq0.5$$ So in your case every element of $[3,4]$ is a median.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      By definition a median connected with a random variable $X$ (or if you like its distribution) is a real number $m$ that satisfies the inequalities:$$P(Xgeq m)geq0.5text{ and }P(Xleq m)geq0.5$$ So in your case every element of $[3,4]$ is a median.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        By definition a median connected with a random variable $X$ (or if you like its distribution) is a real number $m$ that satisfies the inequalities:$$P(Xgeq m)geq0.5text{ and }P(Xleq m)geq0.5$$ So in your case every element of $[3,4]$ is a median.






        share|cite|improve this answer









        $endgroup$



        By definition a median connected with a random variable $X$ (or if you like its distribution) is a real number $m$ that satisfies the inequalities:$$P(Xgeq m)geq0.5text{ and }P(Xleq m)geq0.5$$ So in your case every element of $[3,4]$ is a median.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 16:13









        drhabdrhab

        103k545136




        103k545136






























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