Question on finding the median of a discrete random variable












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












$begingroup$


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






share|cite|improve this question















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
1






active

oldest

votes


















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









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058368%2fquestion-on-finding-the-median-of-a-discrete-random-variable%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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

        18.9k31729




        18.9k31729






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058368%2fquestion-on-finding-the-median-of-a-discrete-random-variable%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

            Image
            0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
            Read more

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more