Exponential and Poisson distribution, machine












0












$begingroup$


I have this question where i am unsure how to solve it.



X...how often a machine does not work
E(X)= 3 per day= 1/8 per hour



X-Poisson distributed



What is the probability that no machine breaks down for more than 5hours.



I know that the time between poisson distributed events is exponentially distributed.



So P(Y>5)? With Y being exp. distributed. That would be the same as 1-P(Y<5)



But what is lambda, i know that if Y($lambda_1$) is exponentially distributed than X is distributed with ($lambda_2*(t_2-t_1))$



Can I know say that $lambda_1/(t_2-t_1)$



$lambda_2=1/8$ than
$lambda_1=1/40$



And we would get $e^{(-5/40)}$ as solution but this just does not seem correct to me....
I don’t feel like i really get what i am doing.....






probability poisson-distribution exponential-distribution







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I have this question where i am unsure how to solve it.



    X...how often a machine does not work
    E(X)= 3 per day= 1/8 per hour



    X-Poisson distributed



    What is the probability that no machine breaks down for more than 5hours.



    I know that the time between poisson distributed events is exponentially distributed.



    So P(Y>5)? With Y being exp. distributed. That would be the same as 1-P(Y<5)



    But what is lambda, i know that if Y($lambda_1$) is exponentially distributed than X is distributed with ($lambda_2*(t_2-t_1))$



    Can I know say that $lambda_1/(t_2-t_1)$



    $lambda_2=1/8$ than
    $lambda_1=1/40$



    And we would get $e^{(-5/40)}$ as solution but this just does not seem correct to me....
    I don’t feel like i really get what i am doing.....






    probability poisson-distribution exponential-distribution







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have this question where i am unsure how to solve it.



      X...how often a machine does not work
      E(X)= 3 per day= 1/8 per hour



      X-Poisson distributed



      What is the probability that no machine breaks down for more than 5hours.



      I know that the time between poisson distributed events is exponentially distributed.



      So P(Y>5)? With Y being exp. distributed. That would be the same as 1-P(Y<5)



      But what is lambda, i know that if Y($lambda_1$) is exponentially distributed than X is distributed with ($lambda_2*(t_2-t_1))$



      Can I know say that $lambda_1/(t_2-t_1)$



      $lambda_2=1/8$ than
      $lambda_1=1/40$



      And we would get $e^{(-5/40)}$ as solution but this just does not seem correct to me....
      I don’t feel like i really get what i am doing.....






      probability poisson-distribution exponential-distribution







      share|cite|improve this question









      $endgroup$




      I have this question where i am unsure how to solve it.



      X...how often a machine does not work
      E(X)= 3 per day= 1/8 per hour



      X-Poisson distributed



      What is the probability that no machine breaks down for more than 5hours.



      I know that the time between poisson distributed events is exponentially distributed.



      So P(Y>5)? With Y being exp. distributed. That would be the same as 1-P(Y<5)



      But what is lambda, i know that if Y($lambda_1$) is exponentially distributed than X is distributed with ($lambda_2*(t_2-t_1))$



      Can I know say that $lambda_1/(t_2-t_1)$



      $lambda_2=1/8$ than
      $lambda_1=1/40$



      And we would get $e^{(-5/40)}$ as solution but this just does not seem correct to me....
      I don’t feel like i really get what i am doing.....






      probability poisson-distribution exponential-distribution




      probability poisson-distribution exponential-distribution






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 6 at 9:11









      LillysLillys

      778




      778






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.



          Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $lambda t=frac18t$.



          (Note that e.g. $mathbb EN_1=lambda1=frac18$ matching with the info in your question.)



          So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $lambda 5=frac58$.



          We find:$$P(N_5=0)=e^{-frac58}$$




          For clarity if we define $X_1$ as the time that the first break down takes place then the events ${N_5=0}$ and ${X_1>5}$ are exactly the same.



          So you are indeed justified to find the answer by calculation of $P(X_1>5)$.



          Indeed $X_1$ has exponential distribution, and this with parameter $lambda=frac18$, so that:$$P(X_1>5)=e^{-frac58}$$






          share|cite|improve this answer









          $endgroup$













            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',
            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%2f3063631%2fexponential-and-poisson-distribution-machine%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









            1












            $begingroup$

            It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.



            Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $lambda t=frac18t$.



            (Note that e.g. $mathbb EN_1=lambda1=frac18$ matching with the info in your question.)



            So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $lambda 5=frac58$.



            We find:$$P(N_5=0)=e^{-frac58}$$




            For clarity if we define $X_1$ as the time that the first break down takes place then the events ${N_5=0}$ and ${X_1>5}$ are exactly the same.



            So you are indeed justified to find the answer by calculation of $P(X_1>5)$.



            Indeed $X_1$ has exponential distribution, and this with parameter $lambda=frac18$, so that:$$P(X_1>5)=e^{-frac58}$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.



              Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $lambda t=frac18t$.



              (Note that e.g. $mathbb EN_1=lambda1=frac18$ matching with the info in your question.)



              So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $lambda 5=frac58$.



              We find:$$P(N_5=0)=e^{-frac58}$$




              For clarity if we define $X_1$ as the time that the first break down takes place then the events ${N_5=0}$ and ${X_1>5}$ are exactly the same.



              So you are indeed justified to find the answer by calculation of $P(X_1>5)$.



              Indeed $X_1$ has exponential distribution, and this with parameter $lambda=frac18$, so that:$$P(X_1>5)=e^{-frac58}$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.



                Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $lambda t=frac18t$.



                (Note that e.g. $mathbb EN_1=lambda1=frac18$ matching with the info in your question.)



                So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $lambda 5=frac58$.



                We find:$$P(N_5=0)=e^{-frac58}$$




                For clarity if we define $X_1$ as the time that the first break down takes place then the events ${N_5=0}$ and ${X_1>5}$ are exactly the same.



                So you are indeed justified to find the answer by calculation of $P(X_1>5)$.



                Indeed $X_1$ has exponential distribution, and this with parameter $lambda=frac18$, so that:$$P(X_1>5)=e^{-frac58}$$






                share|cite|improve this answer









                $endgroup$



                It is handsome to define $N_t$ as the number of break downs that take place within $t$ hours.



                Then for every $t>0$ random variable $N_t$ has Poisson distribution with parameter $lambda t=frac18t$.



                (Note that e.g. $mathbb EN_1=lambda1=frac18$ matching with the info in your question.)



                So you are asked to find $P(N_5=0)$ where $N_5$ has Poisson distribution with parameter $lambda 5=frac58$.



                We find:$$P(N_5=0)=e^{-frac58}$$




                For clarity if we define $X_1$ as the time that the first break down takes place then the events ${N_5=0}$ and ${X_1>5}$ are exactly the same.



                So you are indeed justified to find the answer by calculation of $P(X_1>5)$.



                Indeed $X_1$ has exponential distribution, and this with parameter $lambda=frac18$, so that:$$P(X_1>5)=e^{-frac58}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 11:33









                drhabdrhab

                103k545136




                103k545136






























                    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%2f3063631%2fexponential-and-poisson-distribution-machine%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

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
                    Read more

                    build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

                    Image
                    0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
                    Read more