Exponential Distruproblem












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Four tigers in a reserve forest are monitored using geo tags. The waiting times for responses from 4 tigers in the reserve follow an iid exponential distribution with mean 3. If the system has to locate all 4 tigers within 5 minutes, it has to reduce the expected response time of each geo tag. What is the maximum expected response time that will produce a location for all four tigers within 5 minutes or less with at least $90%$ probability.



I don’t understand this question, can someone explain what it is asking for and provide the solution?










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  • Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
    – Tki Deneb
    Dec 4 '18 at 9:37


















0














Four tigers in a reserve forest are monitored using geo tags. The waiting times for responses from 4 tigers in the reserve follow an iid exponential distribution with mean 3. If the system has to locate all 4 tigers within 5 minutes, it has to reduce the expected response time of each geo tag. What is the maximum expected response time that will produce a location for all four tigers within 5 minutes or less with at least $90%$ probability.



I don’t understand this question, can someone explain what it is asking for and provide the solution?










share|cite|improve this question
























  • Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
    – Tki Deneb
    Dec 4 '18 at 9:37
















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Four tigers in a reserve forest are monitored using geo tags. The waiting times for responses from 4 tigers in the reserve follow an iid exponential distribution with mean 3. If the system has to locate all 4 tigers within 5 minutes, it has to reduce the expected response time of each geo tag. What is the maximum expected response time that will produce a location for all four tigers within 5 minutes or less with at least $90%$ probability.



I don’t understand this question, can someone explain what it is asking for and provide the solution?










share|cite|improve this question















Four tigers in a reserve forest are monitored using geo tags. The waiting times for responses from 4 tigers in the reserve follow an iid exponential distribution with mean 3. If the system has to locate all 4 tigers within 5 minutes, it has to reduce the expected response time of each geo tag. What is the maximum expected response time that will produce a location for all four tigers within 5 minutes or less with at least $90%$ probability.



I don’t understand this question, can someone explain what it is asking for and provide the solution?







probability probability-distributions random-variables exponential-distribution






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edited Dec 4 '18 at 9:40









Tki Deneb

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asked Dec 3 '18 at 20:48









user601297

1226




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  • Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
    – Tki Deneb
    Dec 4 '18 at 9:37




















  • Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
    – Tki Deneb
    Dec 4 '18 at 9:37


















Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
– Tki Deneb
Dec 4 '18 at 9:37






Let $X_1,dots,X_4$ be the response time for the $4$ tigers. Those random variables are iid exponentially distributed with mean $3$. Now we have $$ P(max(X_1,dots,X_4) leq 5) < 0.9. $$ But you want that probability to be $geq 0.9$. To achive that, you can change the mean of the exponential distributions to any $mu in mathbb{R}$. How big can you choose $mu$ such that the above probability is $geq 0.9$?
– Tki Deneb
Dec 4 '18 at 9:37












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The first step is to find the distribution of the max of exponential RVs. Let $X_n sim exp(lambda)$ be i.i.d. RVs and let $Y = max_n {X_n}$.



$P(Y leq y)$



$=P(max_n {X_n} leq y)$



$=P(X_1 leq y, ; X_2 leq y, ; ..., ; X_n leq y)$



$=P(X_1 leq y)^n$ (using independence)



$=(1-e^{-lambda y})^n$ (using definition of exponential CDF)



In your example, $n=4$, so you can compute $lambda$ from the inequality $P(Y leq 5) geq 0.9$, i.e. $(1-e^{-5 lambda})^4 geq 0.9$.






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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    The first step is to find the distribution of the max of exponential RVs. Let $X_n sim exp(lambda)$ be i.i.d. RVs and let $Y = max_n {X_n}$.



    $P(Y leq y)$



    $=P(max_n {X_n} leq y)$



    $=P(X_1 leq y, ; X_2 leq y, ; ..., ; X_n leq y)$



    $=P(X_1 leq y)^n$ (using independence)



    $=(1-e^{-lambda y})^n$ (using definition of exponential CDF)



    In your example, $n=4$, so you can compute $lambda$ from the inequality $P(Y leq 5) geq 0.9$, i.e. $(1-e^{-5 lambda})^4 geq 0.9$.






    share|cite|improve this answer


























      0














      The first step is to find the distribution of the max of exponential RVs. Let $X_n sim exp(lambda)$ be i.i.d. RVs and let $Y = max_n {X_n}$.



      $P(Y leq y)$



      $=P(max_n {X_n} leq y)$



      $=P(X_1 leq y, ; X_2 leq y, ; ..., ; X_n leq y)$



      $=P(X_1 leq y)^n$ (using independence)



      $=(1-e^{-lambda y})^n$ (using definition of exponential CDF)



      In your example, $n=4$, so you can compute $lambda$ from the inequality $P(Y leq 5) geq 0.9$, i.e. $(1-e^{-5 lambda})^4 geq 0.9$.






      share|cite|improve this answer
























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        0






        The first step is to find the distribution of the max of exponential RVs. Let $X_n sim exp(lambda)$ be i.i.d. RVs and let $Y = max_n {X_n}$.



        $P(Y leq y)$



        $=P(max_n {X_n} leq y)$



        $=P(X_1 leq y, ; X_2 leq y, ; ..., ; X_n leq y)$



        $=P(X_1 leq y)^n$ (using independence)



        $=(1-e^{-lambda y})^n$ (using definition of exponential CDF)



        In your example, $n=4$, so you can compute $lambda$ from the inequality $P(Y leq 5) geq 0.9$, i.e. $(1-e^{-5 lambda})^4 geq 0.9$.






        share|cite|improve this answer












        The first step is to find the distribution of the max of exponential RVs. Let $X_n sim exp(lambda)$ be i.i.d. RVs and let $Y = max_n {X_n}$.



        $P(Y leq y)$



        $=P(max_n {X_n} leq y)$



        $=P(X_1 leq y, ; X_2 leq y, ; ..., ; X_n leq y)$



        $=P(X_1 leq y)^n$ (using independence)



        $=(1-e^{-lambda y})^n$ (using definition of exponential CDF)



        In your example, $n=4$, so you can compute $lambda$ from the inequality $P(Y leq 5) geq 0.9$, i.e. $(1-e^{-5 lambda})^4 geq 0.9$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 1:33









        Aditya Dua

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