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
3818
$endgroup$
$begingroup$
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 |
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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
3818
$endgroup$
$begingroup$
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
3818
$endgroup$
$begingroup$
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
3818
$endgroup$
$X$ 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
3818
edited Dec 17 '18 at 7:15
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3818
answered Dec 17 '18 at 4:12
Ongky Denny WijayaOngky Denny Wijaya
3818
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
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 |
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$begingroup$
Hint: it's z-scores.
$endgroup$
– Sean Roberson
Dec 17 '18 at 4:12