Help With Statistics and Distributions!!











up vote
2
down vote

favorite












So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!










share|cite|improve this question






















  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    Nov 19 at 19:09

















up vote
2
down vote

favorite












So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!










share|cite|improve this question






















  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    Nov 19 at 19:09















up vote
2
down vote

favorite









up vote
2
down vote

favorite











So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!










share|cite|improve this question













So the question asks:
Airlines find that each passenger who reserves a seat fails to turn up with probability 0.01 independently of other passengers. Consequently, Bryanair always sell 100 tickets for their 99 seat aeroplane, and DifficultJet always sell 200 tickets for their 198 seat aeroplane. Which is more over-booked?



At first, I thought I should use the Bernoulli distribution because there are no set trials, with p=0.01, but x has to been either 0 or 1 for this distribution and surely x=100 for Bryanair and y=199,200 for DifficultJet. So, then I tried Binomial, but then surely n=1, so x>n and this is just confusing.
Please help!







probability statistics probability-distributions binomial-distribution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 19 at 19:03









mathsatbath

162




162












  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    Nov 19 at 19:09




















  • What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
    – lulu
    Nov 19 at 19:09


















What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
Nov 19 at 19:09






What's wrong with Bernoulli here? For each airline (separately) you are asking to compute the probability that the flight will be over booked. For $B$ that just means everybody shows up, so $.99^{100}=.366$. Now compare that to airline $D$.
– lulu
Nov 19 at 19:09












1 Answer
1






active

oldest

votes

















up vote
3
down vote













First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




  1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


  2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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',
    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%2f3005364%2fhelp-with-statistics-and-distributions%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








    up vote
    3
    down vote













    First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



    Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




    1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


    2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



    Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






    share|cite|improve this answer

























      up vote
      3
      down vote













      First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



      Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




      1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


      2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



      Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



        Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




        1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


        2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



        Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?






        share|cite|improve this answer












        First of all, you may benefit from being a bit more precise in your thinking. It doesn't mean much to say you're going to "use" a certain distribution. Use it as what, a paperweight? A random variable, that is, a random number, is something that has a distribution. So you should articulate carefully which random number it is which you think has a Bernoulli or a Binomial distribution. For example you could say "the number of passengers who don't cancel on a Bryanair flight follows a Bernoulli distribution", or something like that.



        Once you've figured out exactly what it is you're trying to say, you need to start from a conceptual understanding of what the different distributions mean. So:




        1. If you flip a coin (or do any other experiment which can have two different results) the result of the coin flip follows a Bernoulli distribution of parameters $p$ (normally we call one of the results $1$ and the other $0$, and say that $p$ is the probability of getting $1$).


        2. If you flip a coin with probability $p$ of coming up heads (or what-have-you) $n$ times, the number of heads you get follows a Binomial distribution of parameters $n$, $p$.



        Can you find a random variable in the problem which follows a Bernoulli distribution? Can you find one which follows a Binomial?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 19:35









        Jack M

        18.4k33778




        18.4k33778






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3005364%2fhelp-with-statistics-and-distributions%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

            Plaza Victoria

            Puebla de Zaragoza

            Musa