Thorem stating how much of a population is within $n$ standard deviations from the mean
In a statistics class I took in college, I remember learning about a theorem stating that $50%$ of the population must be within $1sigma$ from $mu$, $75%$ within $2sigma$, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.
People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?
probability-distributions standard-deviation
asked Dec 28 '18 at 5:09
David Ehrmann
1284
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David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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In a statistics class I took in college, I remember learning about a theorem stating that $50%$ of the population must be within $1sigma$ from $mu$, $75%$ within $2sigma$, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.
People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?
probability-distributions standard-deviation
asked Dec 28 '18 at 5:09
David Ehrmann
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31
add a comment |
In a statistics class I took in college, I remember learning about a theorem stating that $50%$ of the population must be within $1sigma$ from $mu$, $75%$ within $2sigma$, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.
People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?
probability-distributions standard-deviation
asked Dec 28 '18 at 5:09
David Ehrmann
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In a statistics class I took in college, I remember learning about a theorem stating that $50%$ of the population must be within $1sigma$ from $mu$, $75%$ within $2sigma$, and so on, regardless of the distribution. The distribution can be tighter, but this much is a guarantee.
People don't seem to be familiar with this, I'm not sure I remembered the numbers right, and I'm not sure this even exists. Is this a thing?
probability-distributions standard-deviation
probability-distributions standard-deviation
asked Dec 28 '18 at 5:09
David Ehrmann
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 28 '18 at 5:09
David Ehrmann
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 28 '18 at 20:10
asked Dec 28 '18 at 5:09
David Ehrmann
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 28 '18 at 5:09
David Ehrmann
1284
asked Dec 28 '18 at 5:09
David Ehrmann
1284
1284
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
David Ehrmann is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31
add a comment |
For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31
For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31
add a comment |
2 Answers
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Yes, some of it is true, and comes from Tschebychev inequality. It says that
$$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.
This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
add a comment |
You're thinking of Chebyshev's inequality.
For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.
Most distributions, of course, are much tighter than this; the theorem is the worst case.
answered Dec 28 '18 at 5:20
jmerry
1,57718
add a comment |
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2 Answers
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Yes, some of it is true, and comes from Tschebychev inequality. It says that
$$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.
This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
add a comment |
Yes, some of it is true, and comes from Tschebychev inequality. It says that
$$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.
This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
add a comment |
Yes, some of it is true, and comes from Tschebychev inequality. It says that
$$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.
This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
Yes, some of it is true, and comes from Tschebychev inequality. It says that
$$P(|X-mu|le nsigma)ge 1-frac1{n^2}.$$
This gives the mentioned $0.75$ for $n=2$, but the $0.5$ actually appears if you take $n=sqrt2=1.41ldots$. It says nothing for $n=1$.
This is valid for any distribution, provided it has a well defined finite mean and a finite variance/s.d. For other distributions exact values can be determined, which are necessarily equal or greater (quite bigger, for most usual distributions) than those given by this theorem.
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
answered Dec 28 '18 at 5:20
Alejandro Nasif Salum
4,399118
4,399118
add a comment |
add a comment |
You're thinking of Chebyshev's inequality.
For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.
Most distributions, of course, are much tighter than this; the theorem is the worst case.
answered Dec 28 '18 at 5:20
jmerry
1,57718
add a comment |
You're thinking of Chebyshev's inequality.
For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.
Most distributions, of course, are much tighter than this; the theorem is the worst case.
answered Dec 28 '18 at 5:20
jmerry
1,57718
add a comment |
You're thinking of Chebyshev's inequality.
For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.
Most distributions, of course, are much tighter than this; the theorem is the worst case.
answered Dec 28 '18 at 5:20
jmerry
1,57718
You're thinking of Chebyshev's inequality.
For any $k$, in any distribution with a mean and standard deviation, the probability of being more than $k$ standard deviations away from the mean is no more than $frac1{k^2}$.
Most distributions, of course, are much tighter than this; the theorem is the worst case.
answered Dec 28 '18 at 5:20
jmerry
1,57718
answered Dec 28 '18 at 5:20
jmerry
1,57718
answered Dec 28 '18 at 5:20
jmerry
1,57718
answered Dec 28 '18 at 5:20
jmerry
1,57718
1,57718
add a comment |
add a comment |
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For the sake of completeness: note that there is also Markov's inequality, which is for nonegative random variables, but does not involve the standard deviation.
– Mehrdad
Dec 28 '18 at 10:04
You have $1sigma$ twice.
– Asaf Karagila♦
Dec 28 '18 at 11:31