Statistics Olympiad Problem












2












$begingroup$



Given that the mean, median, range and the only mode of 200 integers are also 200. If $A$ is the largest integer among those 200 integers, find the maximum value of $A$.




I have asked some of my friends and colleagues to solve this problem, but no one give me a light.



Attempt:



Assuming first that all the numbers are $200$.
To maximize $A$, but satisfies all the criterion given, we need to make $A$ ascending while descending the value of other numbers.
Logically,
$100, 200, 200, cdots, 300$
still satisfies.
Maybe we have$A_{text{max}} = 300$?
I don't know how to approach it clearly.






statistics







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Given that the mean, median, range and the only mode of 200 integers are also 200. If $A$ is the largest integer among those 200 integers, find the maximum value of $A$.




    I have asked some of my friends and colleagues to solve this problem, but no one give me a light.



    Attempt:



    Assuming first that all the numbers are $200$.
    To maximize $A$, but satisfies all the criterion given, we need to make $A$ ascending while descending the value of other numbers.
    Logically,
    $100, 200, 200, cdots, 300$
    still satisfies.
    Maybe we have$A_{text{max}} = 300$?
    I don't know how to approach it clearly.






    statistics







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Given that the mean, median, range and the only mode of 200 integers are also 200. If $A$ is the largest integer among those 200 integers, find the maximum value of $A$.




      I have asked some of my friends and colleagues to solve this problem, but no one give me a light.



      Attempt:



      Assuming first that all the numbers are $200$.
      To maximize $A$, but satisfies all the criterion given, we need to make $A$ ascending while descending the value of other numbers.
      Logically,
      $100, 200, 200, cdots, 300$
      still satisfies.
      Maybe we have$A_{text{max}} = 300$?
      I don't know how to approach it clearly.






      statistics







      share|cite|improve this question











      $endgroup$





      Given that the mean, median, range and the only mode of 200 integers are also 200. If $A$ is the largest integer among those 200 integers, find the maximum value of $A$.




      I have asked some of my friends and colleagues to solve this problem, but no one give me a light.



      Attempt:



      Assuming first that all the numbers are $200$.
      To maximize $A$, but satisfies all the criterion given, we need to make $A$ ascending while descending the value of other numbers.
      Logically,
      $100, 200, 200, cdots, 300$
      still satisfies.
      Maybe we have$A_{text{max}} = 300$?
      I don't know how to approach it clearly.






      statistics




      statistics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 7:47









      David G. Stork

      11.6k41534




      11.6k41534










      asked Jan 16 at 7:35









      Shane Dizzy SukardyShane Dizzy Sukardy

      60819




      60819






















          2 Answers
          2






          active

          oldest

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          6












          $begingroup$

          If $Agt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible



          If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible



          If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $frac{1}{200}(99 times 199 + 100 times 200 +1 times 399)=200.5$, making the mean exceed $200$, so that is not possible



          If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with





          • $198$ appearing $99$ times,


          • $200$ appearing $100$ times, and


          • $398$ appearing $1$ time,


          with mean $frac{1}{200}(99 times 198 + 100 times 200 +1 times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible



          So the maximum possible value of $A$ is $398$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            The distribution consisting of ${196, underbrace{198, ldots, 198}_{96}, underbrace{200, ldots 200}_{102}, 396 }$



            Has mean = mode = range = $200$ and $A_{max} = 396$.



            Here's the histogram, with the two "singletons" barely visible:



            enter image description here






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @JackMoody: I posted my solution within 10 seconds of Henry.
              $endgroup$
              – David G. Stork
              Jan 16 at 17:10










            • $begingroup$
              For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
              $endgroup$
              – Michael Lugo
              Jan 16 at 17:50










            • $begingroup$
              @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
              $endgroup$
              – Jack Moody
              Jan 16 at 17:55










            • $begingroup$
              @JackMoody: $398$ is just an upper bound.
              $endgroup$
              – David G. Stork
              Jan 16 at 17:56














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            2 Answers
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            active

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            2 Answers
            2






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            active

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            6












            $begingroup$

            If $Agt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible



            If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible



            If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $frac{1}{200}(99 times 199 + 100 times 200 +1 times 399)=200.5$, making the mean exceed $200$, so that is not possible



            If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with





            • $198$ appearing $99$ times,


            • $200$ appearing $100$ times, and


            • $398$ appearing $1$ time,


            with mean $frac{1}{200}(99 times 198 + 100 times 200 +1 times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible



            So the maximum possible value of $A$ is $398$






            share|cite|improve this answer











            $endgroup$


















              6












              $begingroup$

              If $Agt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible



              If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible



              If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $frac{1}{200}(99 times 199 + 100 times 200 +1 times 399)=200.5$, making the mean exceed $200$, so that is not possible



              If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with





              • $198$ appearing $99$ times,


              • $200$ appearing $100$ times, and


              • $398$ appearing $1$ time,


              with mean $frac{1}{200}(99 times 198 + 100 times 200 +1 times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible



              So the maximum possible value of $A$ is $398$






              share|cite|improve this answer











              $endgroup$
















                6












                6








                6





                $begingroup$

                If $Agt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible



                If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible



                If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $frac{1}{200}(99 times 199 + 100 times 200 +1 times 399)=200.5$, making the mean exceed $200$, so that is not possible



                If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with





                • $198$ appearing $99$ times,


                • $200$ appearing $100$ times, and


                • $398$ appearing $1$ time,


                with mean $frac{1}{200}(99 times 198 + 100 times 200 +1 times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible



                So the maximum possible value of $A$ is $398$






                share|cite|improve this answer











                $endgroup$



                If $Agt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible



                If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible



                If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $frac{1}{200}(99 times 199 + 100 times 200 +1 times 399)=200.5$, making the mean exceed $200$, so that is not possible



                If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with





                • $198$ appearing $99$ times,


                • $200$ appearing $100$ times, and


                • $398$ appearing $1$ time,


                with mean $frac{1}{200}(99 times 198 + 100 times 200 +1 times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible



                So the maximum possible value of $A$ is $398$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 16 at 9:29

























                answered Jan 16 at 8:35









                HenryHenry

                101k482169




                101k482169























                    0












                    $begingroup$

                    The distribution consisting of ${196, underbrace{198, ldots, 198}_{96}, underbrace{200, ldots 200}_{102}, 396 }$



                    Has mean = mode = range = $200$ and $A_{max} = 396$.



                    Here's the histogram, with the two "singletons" barely visible:



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @JackMoody: I posted my solution within 10 seconds of Henry.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:10










                    • $begingroup$
                      For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                      $endgroup$
                      – Michael Lugo
                      Jan 16 at 17:50










                    • $begingroup$
                      @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                      $endgroup$
                      – Jack Moody
                      Jan 16 at 17:55










                    • $begingroup$
                      @JackMoody: $398$ is just an upper bound.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:56


















                    0












                    $begingroup$

                    The distribution consisting of ${196, underbrace{198, ldots, 198}_{96}, underbrace{200, ldots 200}_{102}, 396 }$



                    Has mean = mode = range = $200$ and $A_{max} = 396$.



                    Here's the histogram, with the two "singletons" barely visible:



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      @JackMoody: I posted my solution within 10 seconds of Henry.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:10










                    • $begingroup$
                      For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                      $endgroup$
                      – Michael Lugo
                      Jan 16 at 17:50










                    • $begingroup$
                      @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                      $endgroup$
                      – Jack Moody
                      Jan 16 at 17:55










                    • $begingroup$
                      @JackMoody: $398$ is just an upper bound.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:56
















                    0












                    0








                    0





                    $begingroup$

                    The distribution consisting of ${196, underbrace{198, ldots, 198}_{96}, underbrace{200, ldots 200}_{102}, 396 }$



                    Has mean = mode = range = $200$ and $A_{max} = 396$.



                    Here's the histogram, with the two "singletons" barely visible:



                    enter image description here






                    share|cite|improve this answer











                    $endgroup$



                    The distribution consisting of ${196, underbrace{198, ldots, 198}_{96}, underbrace{200, ldots 200}_{102}, 396 }$



                    Has mean = mode = range = $200$ and $A_{max} = 396$.



                    Here's the histogram, with the two "singletons" barely visible:



                    enter image description here







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 16 at 17:54

























                    answered Jan 16 at 7:43









                    David G. StorkDavid G. Stork

                    11.6k41534




                    11.6k41534












                    • $begingroup$
                      @JackMoody: I posted my solution within 10 seconds of Henry.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:10










                    • $begingroup$
                      For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                      $endgroup$
                      – Michael Lugo
                      Jan 16 at 17:50










                    • $begingroup$
                      @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                      $endgroup$
                      – Jack Moody
                      Jan 16 at 17:55










                    • $begingroup$
                      @JackMoody: $398$ is just an upper bound.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:56




















                    • $begingroup$
                      @JackMoody: I posted my solution within 10 seconds of Henry.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:10










                    • $begingroup$
                      For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                      $endgroup$
                      – Michael Lugo
                      Jan 16 at 17:50










                    • $begingroup$
                      @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                      $endgroup$
                      – Jack Moody
                      Jan 16 at 17:55










                    • $begingroup$
                      @JackMoody: $398$ is just an upper bound.
                      $endgroup$
                      – David G. Stork
                      Jan 16 at 17:56


















                    $begingroup$
                    @JackMoody: I posted my solution within 10 seconds of Henry.
                    $endgroup$
                    – David G. Stork
                    Jan 16 at 17:10




                    $begingroup$
                    @JackMoody: I posted my solution within 10 seconds of Henry.
                    $endgroup$
                    – David G. Stork
                    Jan 16 at 17:10












                    $begingroup$
                    For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                    $endgroup$
                    – Michael Lugo
                    Jan 16 at 17:50




                    $begingroup$
                    For those coming along and seeing this later: the issue @JackMoody identified has been fixed.
                    $endgroup$
                    – Michael Lugo
                    Jan 16 at 17:50












                    $begingroup$
                    @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                    $endgroup$
                    – Jack Moody
                    Jan 16 at 17:55




                    $begingroup$
                    @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry?
                    $endgroup$
                    – Jack Moody
                    Jan 16 at 17:55












                    $begingroup$
                    @JackMoody: $398$ is just an upper bound.
                    $endgroup$
                    – David G. Stork
                    Jan 16 at 17:56






                    $begingroup$
                    @JackMoody: $398$ is just an upper bound.
                    $endgroup$
                    – David G. Stork
                    Jan 16 at 17:56




















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