Statistics Olympiad Problem
$begingroup$
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
edited Jan 16 at 7:47
David G. Stork
11.6k41534
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 7:47
David G. Stork
11.6k41534
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 7:47
David G. Stork
11.6k41534
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
$endgroup$
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
edited Jan 16 at 7:47
David G. Stork
11.6k41534
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
edited Jan 16 at 7:47
David G. Stork
11.6k41534
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
edited Jan 16 at 7:47
David G. Stork
11.6k41534
edited Jan 16 at 7:47
David G. Stork
11.6k41534
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
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
asked Jan 16 at 7:35
Shane Dizzy SukardyShane Dizzy Sukardy
60819
60819
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $398$
answered Jan 16 at 8:35
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
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:
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
@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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $398$
answered Jan 16 at 8:35
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
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 $398$
answered Jan 16 at 8:35
HenryHenry
101k482169
$endgroup$
add a comment |
$begingroup$
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 $398$
answered Jan 16 at 8:35
HenryHenry
101k482169
$endgroup$
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 $398$
answered Jan 16 at 8:35
HenryHenry
101k482169
edited Jan 16 at 9:29
answered Jan 16 at 8:35
HenryHenry
101k482169
answered Jan 16 at 8:35
HenryHenry
101k482169
answered Jan 16 at 8:35
HenryHenry
101k482169
101k482169
add a comment |
add a comment |
$begingroup$
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:
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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:
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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:
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
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:
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
edited Jan 16 at 17:54
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
answered Jan 16 at 7:43
David G. StorkDavid G. Stork
11.6k41534
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
add a comment |
$begingroup$
@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
add a comment |
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