life expectancy of new computer with exponential distribution












0












$begingroup$


I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.



Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?



what I have:
f(x) = $lambda cdot e^{-lambda cdot x}$
given: $lambda = 3, x = 8$
so simply
$= 3 cdot e^{-3 cdot 8}
= 3cdot e^{-24}$?



sorry if this is too low level question. But I'm a bit confused.






probability statistics probability-distributions exponential-distribution







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.



    Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?



    what I have:
    f(x) = $lambda cdot e^{-lambda cdot x}$
    given: $lambda = 3, x = 8$
    so simply
    $= 3 cdot e^{-3 cdot 8}
    = 3cdot e^{-24}$    ?



    sorry if this is too low level question. But I'm a bit confused.






    probability statistics probability-distributions exponential-distribution







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.



      Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?



      what I have:
      f(x) = $lambda cdot e^{-lambda cdot x}$
      given: $lambda = 3, x = 8$
      so simply
      $= 3 cdot e^{-3 cdot 8}
      = 3cdot e^{-24}$      ?



      sorry if this is too low level question. But I'm a bit confused.






      probability statistics probability-distributions exponential-distribution







      share|cite|improve this question











      $endgroup$




      I will appreciate someone to verify my answer for exponential distribution question as I am teaching myself and don't have much confidence.



      Let X, the number of years a computer works, be a random variable that follows an exponential distribution with a lambda of 3 years. You just bought a computer, what is the probability that the computer will work in 8 years?



      what I have:
      f(x) = $lambda cdot e^{-lambda cdot x}$
      given: $lambda = 3, x = 8$
      so simply
      $= 3 cdot e^{-3 cdot 8}
      = 3cdot e^{-24}$      ?



      sorry if this is too low level question. But I'm a bit confused.






      probability statistics probability-distributions exponential-distribution




      probability statistics probability-distributions exponential-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 18:59









      twnly

      1,1911214




      1,1911214










      asked Dec 19 '18 at 6:39









      Daniel KimDaniel Kim

      111




      111






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.



          By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.



          You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).



          Hope this helps.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
            $endgroup$
            – Daniel Kim
            Dec 19 '18 at 7:05












          • $begingroup$
            @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
            $endgroup$
            – Aditya Dua
            Dec 19 '18 at 7:13





















          0












          $begingroup$

          As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$



          Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
          begin{align*}
          P(X > 8) &= 1 - P(X leq 8) \
          & = 1 - F_X(8) \
          & = 1 - (1-e^{-frac{1}{3}x}) \
          & = e^{-frac{8}{3}}
          end{align*}



          This says that there is approximately 7% chance that the computer is still working after 8 years.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            2












            $begingroup$

            If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.



            By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.



            You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).



            Hope this helps.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
              $endgroup$
              – Daniel Kim
              Dec 19 '18 at 7:05












            • $begingroup$
              @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
              $endgroup$
              – Aditya Dua
              Dec 19 '18 at 7:13


















            2












            $begingroup$

            If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.



            By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.



            You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).



            Hope this helps.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
              $endgroup$
              – Daniel Kim
              Dec 19 '18 at 7:05












            • $begingroup$
              @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
              $endgroup$
              – Aditya Dua
              Dec 19 '18 at 7:13
















            2












            2








            2





            $begingroup$

            If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.



            By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.



            You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).



            Hope this helps.






            share|cite|improve this answer









            $endgroup$



            If $X$ is exponentially distributed with parameter $lambda$, the probability $P(X>x)=e^{-lambda x}$. You can substitute $lambda = 3$ and $x=8$ to get the probability that the computer is still working after 8 years.



            By the way, some clarity is required on what $lambda$ represents. Typically, $lambda$ is the parameter of the distribution and $1/lambda$ is the mean. In your example, if the mean life is 3 years (seems reasonable), $lambda = 1/3$.



            You substituted $x=8$ in the probability density function or PDF. Instead, you need to use the cumulative distribution function or CDF. For a continuous distribution (like the exponential), it is not meaningful to compute the probability $P(X=x)$. However, you can compute the probability $P(X>x)$ or $P(X leq x)$ using the CDF (which is obtained by integrating the PDF).



            Hope this helps.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 19 '18 at 6:48









            Aditya DuaAditya Dua

            1,15418




            1,15418












            • $begingroup$
              Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
              $endgroup$
              – Daniel Kim
              Dec 19 '18 at 7:05












            • $begingroup$
              @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
              $endgroup$
              – Aditya Dua
              Dec 19 '18 at 7:13




















            • $begingroup$
              Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
              $endgroup$
              – Daniel Kim
              Dec 19 '18 at 7:05












            • $begingroup$
              @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
              $endgroup$
              – Aditya Dua
              Dec 19 '18 at 7:13


















            $begingroup$
            Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
            $endgroup$
            – Daniel Kim
            Dec 19 '18 at 7:05






            $begingroup$
            Hi Aditya, I want to ask some clarification. The equation you used is CDF (P(X>x) = e^(-lambda*x)), it uses the lambda value 3 and x value 8. Therefore, the answer for my question would be e^(-24) ?
            $endgroup$
            – Daniel Kim
            Dec 19 '18 at 7:05














            $begingroup$
            @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
            $endgroup$
            – Aditya Dua
            Dec 19 '18 at 7:13






            $begingroup$
            @DanielKim Actually, as I explained above, if the mean life of the computer is 3 years, your $lambda = 1/3$ (since the mean of an exponential distribution is $1/lambda$). So your answer is $e^{-8/3}$.
            $endgroup$
            – Aditya Dua
            Dec 19 '18 at 7:13













            0












            $begingroup$

            As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$



            Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
            begin{align*}
            P(X > 8) &= 1 - P(X leq 8) \
            & = 1 - F_X(8) \
            & = 1 - (1-e^{-frac{1}{3}x}) \
            & = e^{-frac{8}{3}}
            end{align*}



            This says that there is approximately 7% chance that the computer is still working after 8 years.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$



              Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
              begin{align*}
              P(X > 8) &= 1 - P(X leq 8) \
              & = 1 - F_X(8) \
              & = 1 - (1-e^{-frac{1}{3}x}) \
              & = e^{-frac{8}{3}}
              end{align*}



              This says that there is approximately 7% chance that the computer is still working after 8 years.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$



                Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
                begin{align*}
                P(X > 8) &= 1 - P(X leq 8) \
                & = 1 - F_X(8) \
                & = 1 - (1-e^{-frac{1}{3}x}) \
                & = e^{-frac{8}{3}}
                end{align*}



                This says that there is approximately 7% chance that the computer is still working after 8 years.






                share|cite|improve this answer









                $endgroup$



                As Aditya correctly pointed out, if your exponential distribution has a mean of 3 years, this means that $frac{1}{lambda} = 3$ and so $lambda = frac{1}{3}$ simply by definition. Let $X sim exp(-frac{1}{3})$



                Since the exponential distribution is continuous, the $P(X = 8) = 0$ and so the closest you can look is the probability that your computer will still be working after 8 years, or $P(X>8)$. we have that
                begin{align*}
                P(X > 8) &= 1 - P(X leq 8) \
                & = 1 - F_X(8) \
                & = 1 - (1-e^{-frac{1}{3}x}) \
                & = e^{-frac{8}{3}}
                end{align*}



                This says that there is approximately 7% chance that the computer is still working after 8 years.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 17:27









                Amin SammaraAmin Sammara

                186




                186






























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