Normal Distribution to Standard Normal Distribution
$begingroup$
Is there a way to to convert from a normal distribution to a standard normal distribution.
For example:
$ Y = N(-8,4) $
An answer to the question says that this is the same as writing:
$$frac{Y+8}{2} = N(0,1)$$
Can anyone explain to me how they made this transformation?
Thank you for any guidance.
probability probability-distributions normal-distribution
asked Dec 17 '18 at 3:42
SafderSafder
18410
$endgroup$
add a comment |
$begingroup$
Is there a way to to convert from a normal distribution to a standard normal distribution.
For example:
$ Y = N(-8,4) $
An answer to the question says that this is the same as writing:
$$frac{Y+8}{2} = N(0,1)$$
Can anyone explain to me how they made this transformation?
Thank you for any guidance.
probability probability-distributions normal-distribution
asked Dec 17 '18 at 3:42
SafderSafder
18410
$endgroup$
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12
add a comment |
$begingroup$
Is there a way to to convert from a normal distribution to a standard normal distribution.
For example:
$ Y = N(-8,4) $
An answer to the question says that this is the same as writing:
$$frac{Y+8}{2} = N(0,1)$$
Can anyone explain to me how they made this transformation?
Thank you for any guidance.
probability probability-distributions normal-distribution
asked Dec 17 '18 at 3:42
SafderSafder
18410
$endgroup$
Is there a way to to convert from a normal distribution to a standard normal distribution.
For example:
$ Y = N(-8,4) $
An answer to the question says that this is the same as writing:
$$frac{Y+8}{2} = N(0,1)$$
Can anyone explain to me how they made this transformation?
Thank you for any guidance.
probability probability-distributions normal-distribution
probability probability-distributions normal-distribution
asked Dec 17 '18 at 3:42
SafderSafder
18410
asked Dec 17 '18 at 3:42
SafderSafder
18410
asked Dec 17 '18 at 3:42
SafderSafder
18410
asked Dec 17 '18 at 3:42
SafderSafder
18410
asked Dec 17 '18 at 3:42
SafderSafder
18410
18410
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12
add a comment |
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$X$ is random variable is normal distribution with mean $mu$ and variance $sigma^2$, $Xsim N(mu,sigma^2)$ have probability density function
$$f(x)=dfrac{1}{sqrt{2pisigma^2}}e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2}, -infty<x<infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Ysim N(0,1)$ have probability density function
$$f(y)=dfrac{1}{sqrt{2pi(1)}}e^{-frac{1}{2}left(frac{y-0}{1}right)^2}=dfrac{1}{sqrt{2pi}}e^{-frac{1}{2}y^2}, -infty<y<infty.$$
According to the two equations above,
the transformation is $Y=dfrac{X-mu}{sigma}$.
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
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
});
}
});
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%2f3043503%2fnormal-distribution-to-standard-normal-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$X$ is random variable is normal distribution with mean $mu$ and variance $sigma^2$, $Xsim N(mu,sigma^2)$ have probability density function
$$f(x)=dfrac{1}{sqrt{2pisigma^2}}e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2}, -infty<x<infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Ysim N(0,1)$ have probability density function
$$f(y)=dfrac{1}{sqrt{2pi(1)}}e^{-frac{1}{2}left(frac{y-0}{1}right)^2}=dfrac{1}{sqrt{2pi}}e^{-frac{1}{2}y^2}, -infty<y<infty.$$
According to the two equations above,
the transformation is $Y=dfrac{X-mu}{sigma}$.
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
add a comment |
$begingroup$
$X$ is random variable is normal distribution with mean $mu$ and variance $sigma^2$, $Xsim N(mu,sigma^2)$ have probability density function
$$f(x)=dfrac{1}{sqrt{2pisigma^2}}e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2}, -infty<x<infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Ysim N(0,1)$ have probability density function
$$f(y)=dfrac{1}{sqrt{2pi(1)}}e^{-frac{1}{2}left(frac{y-0}{1}right)^2}=dfrac{1}{sqrt{2pi}}e^{-frac{1}{2}y^2}, -infty<y<infty.$$
According to the two equations above,
the transformation is $Y=dfrac{X-mu}{sigma}$.
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
add a comment |
$begingroup$
$X$ is random variable is normal distribution with mean $mu$ and variance $sigma^2$, $Xsim N(mu,sigma^2)$ have probability density function
$$f(x)=dfrac{1}{sqrt{2pisigma^2}}e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2}, -infty<x<infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Ysim N(0,1)$ have probability density function
$$f(y)=dfrac{1}{sqrt{2pi(1)}}e^{-frac{1}{2}left(frac{y-0}{1}right)^2}=dfrac{1}{sqrt{2pi}}e^{-frac{1}{2}y^2}, -infty<y<infty.$$
According to the two equations above,
the transformation is $Y=dfrac{X-mu}{sigma}$.
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
$endgroup$
$X$ is random variable is normal distribution with mean $mu$ and variance $sigma^2$, $Xsim N(mu,sigma^2)$ have probability density function
$$f(x)=dfrac{1}{sqrt{2pisigma^2}}e^{-frac{1}{2}left(frac{x-mu}{sigma}right)^2}, -infty<x<infty$$
$Y$ is standard normal distribution, have mean $0$ and variance $1$, $Ysim N(0,1)$ have probability density function
$$f(y)=dfrac{1}{sqrt{2pi(1)}}e^{-frac{1}{2}left(frac{y-0}{1}right)^2}=dfrac{1}{sqrt{2pi}}e^{-frac{1}{2}y^2}, -infty<y<infty.$$
According to the two equations above,
the transformation is $Y=dfrac{X-mu}{sigma}$.
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
edited Dec 17 '18 at 7:15
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3798
3798
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
add a comment |
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
I literally knew this but it just didn't click together. Thank you so much.
$endgroup$
– Safder
Dec 17 '18 at 4:14
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
$begingroup$
You're welcome @Safder.
$endgroup$
– Ongky Denny Wijaya
Dec 17 '18 at 4:19
add a comment |
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.
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%2f3043503%2fnormal-distribution-to-standard-normal-distribution%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
$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12