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 1σ from μ, 75% within 1σ, 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 hours ago
David Ehrmann
1133
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.
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 1σ from μ, 75% within 1σ, 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 hours ago
David Ehrmann
1133
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.
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 1σ from μ, 75% within 1σ, 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 hours ago
David Ehrmann
1133
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 1σ from μ, 75% within 1σ, 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 hours ago
David Ehrmann
1133
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 hours ago
David Ehrmann
1133
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 hours ago
David Ehrmann
1133
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 hours ago
David Ehrmann
1133
asked 2 hours ago
David Ehrmann
1133
1133
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.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 hours ago
Alejandro Nasif Salum
4,294118
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 hours ago
jmerry
1,26916
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
David Ehrmann is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054589%2fthorem-stating-how-much-of-a-population-is-within-n-standard-deviations-from-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 hours ago
Alejandro Nasif Salum
4,294118
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 hours ago
Alejandro Nasif Salum
4,294118
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 hours ago
Alejandro Nasif Salum
4,294118
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 hours ago
Alejandro Nasif Salum
4,294118
answered 2 hours ago
Alejandro Nasif Salum
4,294118
answered 2 hours ago
Alejandro Nasif Salum
4,294118
answered 2 hours ago
Alejandro Nasif Salum
4,294118
4,294118
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 hours ago
jmerry
1,26916
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 hours ago
jmerry
1,26916
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 hours ago
jmerry
1,26916
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 hours ago
jmerry
1,26916
answered 2 hours ago
jmerry
1,26916
answered 2 hours ago
jmerry
1,26916
answered 2 hours ago
jmerry
1,26916
1,26916
add a comment |
add a comment |
David Ehrmann is a new contributor. Be nice, and check out our Code of Conduct.
David Ehrmann is a new contributor. Be nice, and check out our Code of Conduct.
David Ehrmann is a new contributor. Be nice, and check out our Code of Conduct.
David Ehrmann is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054589%2fthorem-stating-how-much-of-a-population-is-within-n-standard-deviations-from-the%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
