Thorem stating how much of a population is within $n$ standard deviations from the mean
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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 EhrmannDavid Ehrmann
1334
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add a comment |
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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 EhrmannDavid Ehrmann
1334
$endgroup$
$begingroup$
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.
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– Mehrdad
Dec 28 '18 at 10:04
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You have $1sigma$ twice.
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– Asaf Karagila♦
Dec 28 '18 at 11:31
add a comment |
$begingroup$
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 EhrmannDavid Ehrmann
1334
$endgroup$
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 EhrmannDavid Ehrmann
1334
asked Dec 28 '18 at 5:09
David EhrmannDavid Ehrmann
1334
edited Dec 28 '18 at 20:10
David Ehrmann
asked Dec 28 '18 at 5:09
David EhrmannDavid Ehrmann
1334
asked Dec 28 '18 at 5:09
David EhrmannDavid Ehrmann
1334
asked Dec 28 '18 at 5:09
David EhrmannDavid Ehrmann
1334
1334
$begingroup$
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.
$endgroup$
– Mehrdad
Dec 28 '18 at 10:04
$begingroup$
You have $1sigma$ twice.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 11:31
add a comment |
$begingroup$
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.
$endgroup$
– Mehrdad
Dec 28 '18 at 10:04
$begingroup$
You have $1sigma$ twice.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 11:31
$begingroup$
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.
$endgroup$
– Mehrdad
Dec 28 '18 at 10:04
$begingroup$
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.
$endgroup$
– Mehrdad
Dec 28 '18 at 10:04
$begingroup$
You have $1sigma$ twice.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 11:31
$begingroup$
You have $1sigma$ twice.
$endgroup$
– 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 SalumAlejandro Nasif Salum
4,765118
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add a comment |
$begingroup$
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
jmerryjmerry
9,4881124
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
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 SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
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 SalumAlejandro Nasif Salum
4,765118
$endgroup$
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 SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:20
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:20
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Dec 28 '18 at 5:20
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
$begingroup$
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
jmerryjmerry
9,4881124
$endgroup$
add a comment |
$begingroup$
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
jmerryjmerry
9,4881124
$endgroup$
add a comment |
$begingroup$
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
jmerryjmerry
9,4881124
$endgroup$
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
jmerryjmerry
9,4881124
answered Dec 28 '18 at 5:20
jmerryjmerry
9,4881124
answered Dec 28 '18 at 5:20
jmerryjmerry
9,4881124
answered Dec 28 '18 at 5:20
jmerryjmerry
9,4881124
9,4881124
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– Mehrdad
Dec 28 '18 at 10:04
$begingroup$
You have $1sigma$ twice.
$endgroup$
– Asaf Karagila♦
Dec 28 '18 at 11:31