How to plot the normal distribution?












0












$begingroup$


According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.



However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?



I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.






probability-theory probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
    $endgroup$
    – J.G.
    Dec 20 '18 at 19:52










  • $begingroup$
    The PDF one, which depends on the number of elements
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:06
















0












$begingroup$


According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.



However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?



I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.






probability-theory probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
    $endgroup$
    – J.G.
    Dec 20 '18 at 19:52










  • $begingroup$
    The PDF one, which depends on the number of elements
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:06














0












0








0





$begingroup$


According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.



However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?



I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.






probability-theory probability-distributions







share|cite|improve this question









$endgroup$




According to 'An Introduction to Probability Theory and Its Applications', Vol. 1 by Feller the number of inversions in a random permutations at large numbers satisfy CLT with dedicated mean and variance.



However, I am practically intrested in how to plot the figure of the normal distribution (what to calculate for it)?



I understand that the figure may depends on the fact of how large the numbers are. Any explanations to clarify the topic are highly welcomed. Thank you in advance.






probability-theory probability-distributions




probability-theory probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 19:43









Mikhail GaichenkovMikhail Gaichenkov

6610




6610












  • $begingroup$
    Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
    $endgroup$
    – J.G.
    Dec 20 '18 at 19:52










  • $begingroup$
    The PDF one, which depends on the number of elements
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:06


















  • $begingroup$
    Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
    $endgroup$
    – J.G.
    Dec 20 '18 at 19:52










  • $begingroup$
    The PDF one, which depends on the number of elements
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:06
















$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52




$begingroup$
Do you want to plot the PDF $frac{1}{sigmasqrt{2pi}}exp-frac{(x-mu)^2}{2sigma^2}$ or the CDF $frac{1}{2}left[1+operatorname{erf}left(frac{x-mu}{sigmasqrt{2}}right)right]$? Either way, the case $mu=0,,sigma=1$ is worth using for definiteness.
$endgroup$
– J.G.
Dec 20 '18 at 19:52












$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06




$begingroup$
The PDF one, which depends on the number of elements
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:06










1 Answer
1






active

oldest

votes


















1












$begingroup$

Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$



In Mathematica, a one-liner:



Plot[PDF[NormalDistribution, x], {x, -4, 4}]


enter image description here



In Python, slightly more verbose:



import numpy as np

import matplotlib.pyplot as plt



x = np.linspace(-4, 4, 101)

y = np.exp(-x*x/2) / np.sqrt(2*np.pi)



plt.plot(x, y)

plt.show()


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 19:56










  • $begingroup$
    Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
    $endgroup$
    – Federico
    Dec 20 '18 at 19:58










  • $begingroup$
    Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:04










  • $begingroup$
    They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
    $endgroup$
    – Federico
    Dec 20 '18 at 20:08












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$



In Mathematica, a one-liner:



Plot[PDF[NormalDistribution, x], {x, -4, 4}]


enter image description here



In Python, slightly more verbose:



import numpy as np

import matplotlib.pyplot as plt



x = np.linspace(-4, 4, 101)

y = np.exp(-x*x/2) / np.sqrt(2*np.pi)



plt.plot(x, y)

plt.show()


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 19:56










  • $begingroup$
    Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
    $endgroup$
    – Federico
    Dec 20 '18 at 19:58










  • $begingroup$
    Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:04










  • $begingroup$
    They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
    $endgroup$
    – Federico
    Dec 20 '18 at 20:08
















1












$begingroup$

Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$



In Mathematica, a one-liner:



Plot[PDF[NormalDistribution, x], {x, -4, 4}]


enter image description here



In Python, slightly more verbose:



import numpy as np

import matplotlib.pyplot as plt



x = np.linspace(-4, 4, 101)

y = np.exp(-x*x/2) / np.sqrt(2*np.pi)



plt.plot(x, y)

plt.show()


enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 19:56










  • $begingroup$
    Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
    $endgroup$
    – Federico
    Dec 20 '18 at 19:58










  • $begingroup$
    Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:04










  • $begingroup$
    They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
    $endgroup$
    – Federico
    Dec 20 '18 at 20:08














1












1








1





$begingroup$

Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$



In Mathematica, a one-liner:



Plot[PDF[NormalDistribution, x], {x, -4, 4}]


enter image description here



In Python, slightly more verbose:



import numpy as np

import matplotlib.pyplot as plt



x = np.linspace(-4, 4, 101)

y = np.exp(-x*x/2) / np.sqrt(2*np.pi)



plt.plot(x, y)

plt.show()


enter image description here






share|cite|improve this answer









$endgroup$



Here is how to plot the density function of $N(0,1)$:
$$
f(x) = frac{e^{-frac{x^2}{2}}}{sqrt{2 pi}} .
$$



In Mathematica, a one-liner:



Plot[PDF[NormalDistribution, x], {x, -4, 4}]


enter image description here



In Python, slightly more verbose:



import numpy as np

import matplotlib.pyplot as plt



x = np.linspace(-4, 4, 101)

y = np.exp(-x*x/2) / np.sqrt(2*np.pi)



plt.plot(x, y)

plt.show()


enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 19:49









FedericoFederico

5,124514




5,124514












  • $begingroup$
    Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 19:56










  • $begingroup$
    Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
    $endgroup$
    – Federico
    Dec 20 '18 at 19:58










  • $begingroup$
    Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:04










  • $begingroup$
    They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
    $endgroup$
    – Federico
    Dec 20 '18 at 20:08


















  • $begingroup$
    Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 19:56










  • $begingroup$
    Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
    $endgroup$
    – Federico
    Dec 20 '18 at 19:58










  • $begingroup$
    Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
    $endgroup$
    – Mikhail Gaichenkov
    Dec 20 '18 at 20:04










  • $begingroup$
    They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
    $endgroup$
    – Federico
    Dec 20 '18 at 20:08
















$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56




$begingroup$
Thank you, Federico! How can I get the N(0,1) if there are mean and variance which depend on the number of elements?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 19:56












$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58




$begingroup$
Translate and rescale: if $Zsim N(mu,sigma)$, then $(Z-mu)/sigmasim N(0,1)$.
$endgroup$
– Federico
Dec 20 '18 at 19:58












$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04




$begingroup$
Could you add the figures here for mean=n(n-1)/4, variance=(2n^3+3n^2-5n)/72 at n=10, 100, 1000?
$endgroup$
– Mikhail Gaichenkov
Dec 20 '18 at 20:04












$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08




$begingroup$
They don't look any different from the one I plotted. They are just translated and stretched. Only the numbers on the axes change
$endgroup$
– Federico
Dec 20 '18 at 20:08


















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