Transcript The Standard Normal Distribution

The Standard Normal
Distribution
Fr. Chris
A.P.Statistics
St. Francis High School
Requirements for Any
Probability Distribution
• Always Positive
• Total area under the curve must be 1
(Since 1 represents 100%)
The Standard Normal
Distribution
1
2e
x
2
Any value of x will produce a
POSITIVE result
But is the area below equal 1?
2
x
- ÄÄÄÄ
2Ä
„
ÄÄÄÄÄÄÄÄ
è !!!!ÄÄ
!ÄÄ
!!Ä
2p
0.4
0.3
0.2
0.1
-3
-2
-1
1
2
3
In other words, does



1
2e
x
2
dx  1? YES!
Lets work it out…




1
2e
x
2

dx 

1  2x2
e dx
2


1
 x2
2

e dx
2

But there is a problem…



 x2
2
e dx
Has NO antiderivative!
But Karl Friedrich Gauss figured
a tricky way around this!
Let us introduce an “I” such that:

I=
e
 x2
2
dx

So the area under the
“Standard Normal Curve”
would be
I
2
But let’s concentrate on the I

e
I=

 x2
2
dx
Since x is merely a
variable of integration,
we can also express I as

I=
e

 y2
2
dy
So why not express I2 as

I 
2
e

 x2
2
dx

e
 y2
2
dy

 
or
 e
 
 x2
2
 y2
2
e dxdy
This double integral is actually the
volume under a 3-D “bell”
 
 e
 
-2
0
2
 x2
2
 y2
2
e dxdy 
1
0.75
0.5
0.25
0
-2
0
 
 e
 
2

2 2
 x y
2

dxdy
Rectangles aren’t the only thing
that integrates!
We can now do a clever change of variable
by converting to Polar coordinates!
2
2
 

 
x + y = r2

2 2
 x y
e
2

r
y
dxdy 
š

 e
0 
x
2
 r
2
rdrd
How polar coordinates work,
and how to make the switch...
Any integral can be computed by the limit of Riemann sums over Cartesian rectangles or
Riemann sums over polar rectangles. The area of a Cartesian rectangle of sides
dx and dy is dx*dy but the area of a polar rectangle of sides dr and dt is NOT just dr*dt
Rather dA is a bit more...
Ase c tor  d r
1
2
2
dA  b d  a d
1
2
2
1
2
2
 (a  b)( a  b)d
1
2
Since
1
2
(a  b)  r and (a  b)  dr
dA rdrd
Getting back to our story...
Recall the Pythagorean Theorem?
We can now do a clever change of variable
by converting to Polar coordinates!
2
2
 

 
x + y = r2

2 2
 x y
e
2

r
y
dxdy 
š

 e
0 
x
2
 r
2
rdrd
So now we DO have an antiderivative!
š

 e
0 
2
 r
2
rdrd 

0

š
e
 2
 r
2
d

We can now start evaluating the integral
from negative to positive infinity…

š
e
 2
 r
2


d 

0

š
2
 r
e
2
0
And because of symmetry
2

0
š

2
 r
e
2
d  2

d


0
š
1 1
0 

e2 e 2
d
Almost there!
2

š
1 1
0 

e2 e 2
0
2
d  2

0

0
1
1
 lim b2
1 b e
d
š
š
1 0
š
d  2
d
0
 2
But that wasn’t “I”
I  2
...that was “I” squared!
2
So
I  2
Recall the area under the
Standard Normal Curve
I
2
2

So… 
2
1
So the area under the curve is 1!
Wasn’t Professor Gauss clever?
It is no accident that many
Mathematicians still prefer to call
the “Standard Normal” distribution
The Gaussian Distribution