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 2 days ago
David Ehrmann
1284
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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 2 days ago
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
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
asked 2 days ago
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 2 days ago
David Ehrmann
1284
asked 2 days ago
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
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago
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
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago
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
2 days ago
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
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago
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 2 days ago
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 2 days ago
jmerry
1,44817
add a comment |
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2 Answers
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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 2 days ago
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 2 days ago
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 2 days ago
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 2 days ago
Alejandro Nasif Salum
4,399118
answered 2 days ago
Alejandro Nasif Salum
4,399118
answered 2 days ago
Alejandro Nasif Salum
4,399118
answered 2 days ago
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 2 days ago
jmerry
1,44817
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 2 days ago
jmerry
1,44817
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 2 days ago
jmerry
1,44817
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 2 days ago
jmerry
1,44817
answered 2 days ago
jmerry
1,44817
answered 2 days ago
jmerry
1,44817
answered 2 days ago
jmerry
1,44817
1,44817
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
2 days ago
You have $1sigma$ twice.
– Asaf Karagila♦
2 days ago