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






















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          $begingroup$

          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









          $endgroup$













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            $begingroup$

            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






























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