Normal Distribution to Standard Normal Distribution












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: it's z-scores.
    $endgroup$
    – Sean Roberson
    Dec 17 '18 at 4:12
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: it's z-scores.
    $endgroup$
    – Sean Roberson
    Dec 17 '18 at 4:12














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

$X$ 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 $Y=dfrac{X-mu}{sigma}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

$X$ 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 $Y=dfrac{X-mu}{sigma}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

$X$ 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 $Y=dfrac{X-mu}{sigma}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

$X$ 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 $Y=dfrac{X-mu}{sigma}$.






share|cite|improve this answer











$endgroup$



$X$ 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 $Y=dfrac{X-mu}{sigma}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 7:15

























answered Dec 17 '18 at 4:12









Ongky Denny WijayaOngky Denny Wijaya

3818




3818












  • $begingroup$
    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










  • $begingroup$
    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


















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