In fact, and the meaning of the question is not understood.

If you read the history of the distribution you will probably conclude that belongs to both.

see

https://higherlogicdownload.s3.amazonaws.com/AMSTAT/1484431b-3202-461e-b7e6-ebce10ca8bcd/UploadedImages/Classroom_Activities/HS_2__Origin_of_the_Normal_Curve.pdf

Does the normal/gaussian distribution belong to physics or mathematics.

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We assume that in general probability and statistics belong to physics rather than mathematics.

The Normal/Gaussian Distribution:

f(x)={Exp(-x-Mue)^2/2 sigm^2)}/sigma.Sqrt(2.Pie) . . . . (1)

can be derived from the universal laws of physics for a given number n of randomly chosen data in less than five minutes.

Accuracy increases as n increases.

In fact, we answered a previous question on how to properly use the Cairo technique B transition matrix chains in integration and differentiation without the need for Newton's calculation of the dx and dy infinitesimal elements.

Also here we can get an accurate answer via the so-called closed B-matrix which is an extension of the original B-matrix for closed systems (no BC boundary conditions and no S source terms) as explained in the analysis next.

Simply the solution is Y(x,sigma)=Bc^N for N large enough.

The matrix Bc is well defined by the same conditions as B except that Bc is closed.

It follows that the one-dimensional Bc nxn is given by,

Row 1: RO 1/2-RO/2 0 0 0. . .  0 0 0 1/2-RO/2

Line 2:1/2-RO/2 RO 1/2-RO/2 0 0 . . .0 0 0 0 0 0 0

. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Line n:1/2-RO/2 0 0 0 . . . . 0 1/2-RO/2 RO

It can be further explained using the following two numerical examples.

Bc 11x11 for RO=0 is expressed as,

Line 1: 0 1/2 0 0 0 0 . . .  0 0 0 1/2

Line 2:1/2 0 1/2- 0 0 0 . . .  0 0 0 0 0

. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Line n:1/2 0 0 . . . .0 0 1/2 0

A suitable computational algorithm can show that the stationary Bc(11x!!)^N for N=10 is given by,

1   0.246093750 9.76562500E-04 0.205078125 9.76562500E-03 0.117187500 4.39453125E-02 4.39453125E-02 0.117187500

2   9.76562500E-04 0.246093750 9.76562500E-04 0.205078125 9.76562500E-03 0.117187500 4.39453125e-02 4.39453125e-02 0.117187500

3   0.205078125 9.76562500E-04 0.246093750 9.76562500E-04 0.205078125 9.76562500E-03 0.117187500 4.39453125e-02 4.39453125e-02 0.1171875009.7653125e-02 0.1171875009.7653125E-02 0.1171875009.7653125E-02 0.1171875009

4   9.76562500E-03 0.205078125 9.76562500E-04 0.246093750 9.76562500E-04 0.205078125 9.76562500E-03 0.117187500 4.3945125e-945 315255e-045 315255e-045

5   0.117187500 9.76562500E-03 0.205078125 9.76562500E-04 0.246093750 9.76562500E-04 0.205078125

6   4.39453125E-02 0.117187500 9.76562500E-03 0.205078125 9.76562500E-04 0.246093750 9.76562500E-04 0.205078125 9.76562500E-03 0.117187500

Similarly, an appropriate computational algorithm can show that the stationary Bc^ for N=10 is given by,

A suitable computational algorithm can show that the stationary Bc(17x17)^N for N=10 is given by,

n, Jan 22 at 2:22 AM

1

.2460938

.000000  .205078      .000000     .1171875       .000000     4.39453E-2   9.765625E-4   9.765625E-3   9.765625E-3   9.765625E-4   4.394531E-2   .000000      .1171875       .00000000      .205078125       .00000000

2    .000000     .246094       .000000  .205078  .000000   .117188    0.00000  4.39453E-2   9.765625E-4   9.765625E-3   9.765625E-3   9.765625E-4   4.394531E-2   .000000      .1171875    .000000    .205078

3  .2050781      .0000000      .2460938      .0000000      .2050781      .0000000     .1171875      .0000000       4.394531E-2   9.765625E-4   9.765625E-3   9.765625E-3   9.765625E-4   4.394531E-2   .0000000      .1171875       .0000000

4

.0000000      .2050781       .0000000      .2460938      .0000000      .2050781       .00000000      .11718750       .00000000       4.39453E-2   9.765625E-4   9.765625E-3   9.765625E-3   9.765625E-4   4.394531E-2   .0000000      .11718750

5

.11718750       .00000000      .20507812       .00000000      .24609375       .00000000      .20507813       .00000000      .11718750       .00000000       4.3945312E-2   9.7656250E-4   9.7656250E-3   9.7656250E-3   9.7656250E-4   4.3945313E-2   .0000000

6     .00000000      0.1171875      .00000000      .20507813     .00000000      .24609375     .00000000      .20507813      .00000000      .11718750      .00000000       4.3945313E-2   9.7656250E-4   9.7656250E-3   9.7656250E-3   9.7656250E-4   4.3945313E-2

7    4.3945313E-2   .00000000      .11718750      .00000000      .20507813       .00000000      .24609375       .00000000      .20507813       .00000000      .11718750       .00000000       4.3945313E-2   9.7656250E-4   9.7656250E-3   9.7656250E-3   9.7656250E-4

8     9.7656250E-4   4.3945313E-2   .00000000     .11718750       .00000000      .20507813       .00000000      .24609375      .00000000      .20507813       .00000000      .11718750       .00000000       4.3945313E-2   9.7656250E-4   9.7656250E-3   9.7656250E-3

9    9.7656250E-3   9.7656250E-4   4.3945313E-2   .00000000     .11718750       .00000000      .205078125       .00000000      .24609375       .00000000      .20507813       .00000000      .11718750       .00000000       4.3945313E-2   9.7656250E-4   9.7656250E-3

Due 11x11 and 17x17 matrix symmetry we present only the upper halves 6 rows and 9 rows respectively.

The precise answer of Eq 1 exist clearly in the LHS diagonal of the above two matrices.

Mean Mue =6 and 9 and the normalisation constant is 0.2461 while sigma=0.401 for both of them.

It worth mention that the summation of LHS diagonal elements in both cases is 1.000000 which is a striking accuracy.

we rigorously show that probability and statistics belong to physics rather than mathematics.

The well-known normal/Gaussian statistical distribution law Y(x)=C1 Exp -C2 (x-Mue)^2 can be derived numerically from the statistical transition matrix of the Cairo technique. We present numerical results for two special cases without loss of generality.

The simplicity and precision of the new, unconventional technique are striking.

III.Applications and numerical results

III.a 11x11 Bc matrix

Equation 1 implies that the matrix Bc is expressed for RO=0 as,

0   1/2  0   0   0   0   0   0   0    0   1/2

1/2  0  1/2 0   0   0   0   0   0    0    0

0   1/2 0 1/2   0   0   0   0   0    0    0

0    0  1/2  0  1/2 0   0   0   0    0    0

0    0   0   1/2 0   0   0   0   0    0    0

0    0   0    0  1/2 0  1/2 0   0    0    0

0    0   0    0   0 1/2  0 1/2  0    0    0

0    0   0    0   0   0 1/2  0 1/2   0    0

0    0   0    0   0   0   0 1/2  0  1/2   0

0    0   0    0   0   0   0   0 1/2   0  1/2

1/2 0   0    0   0   0   0   0   0   1/2   0

and the destination matrix Bc^10 is given by, Matrix Bc(11X11)^N when N = 10,

1  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04

2   9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125

3  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-0

4   9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500

5   0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-0

6    4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02

7   4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03  0.117187500

8   0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125       9.76562500E-03

9   9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-04  0.205078125

10   0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750       9.76562500E-0

11   9.76562500E-04  0.205078125       9.76562500E-03  0.117187500       4.39453125E-02   4.39453125E-02  0.117187500       9.76562500E-03  0.205078125       9.76562500E-04  0.246093750

Notice that the LHS diagonal 11 elements are,

9.76562500E-04   9.76562500E-03  4.39453125E-02  0.117187500  0.205078125  0.246093750    0.205078125  0.117187500   4.39453125E-02     9.76562500E-03   9.76562500E-04 . . . (Distribution 1)

(Distribution 1) is the UNIQUE numerical gaussian distribution law for mean x=6 , sigma= 0.398655057 and maximum of 0.2461

1-the total sum of 11 left diagonals=1.00000000 (area under  the normal/gaussian distribution curve ) which is strikingly accurate.

2-The value of the normalization constant for all B nXn is 0.2461 instead of 1 /SQRT(2 Pie) .

To test the correctness of Distribution 1 compare its numerical results with the gaussian distribution law:To test the correctness of the Distribution 12.41267937E-03 1.27526345E-02 4.65598218E-02 0.117417730 0.204534754 0.246099994 0.204534754 0.117417730 4.65598218E-02 1.27526345E-02 2.41267937E-03 . .. (Distribution 1.)

The above natural distribution derived via B-matrix of the Cairo technique corresponds exactly to normal/gaussian distribution law (Eq 5)

Y(x)=C1 Exp-C2*(x-6)^2 . . . . . (5)

with C1=0.2461 and C2=0.185 which corresponds to 1/sgma^2.

Equation 5 yields,

Y(x)= 2.41267937E-03 1.27526345E-02 4.65598218E-02 0.117417730 0.204534754 0.246099994 0.204534754 0.1174177 4.65598218E-02 1.27526345 02  2.41267937E-03

Which is in excellent agreement with Distribution 1.