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 NormalDistribution

1

2πex2

Any value of x will produce a

POSITIVE result

But is the area below equal 1?

-3 -2 -1 1 2 3

0.1

0.2

0.3

0.4

„- x 2ÄÄÄÄÄÄ2

ÄÄÄÄÄÄÄÄÄÄÄÄÄè !!!!!!!2 p

In other words, does

∫−∞

∞1

2πex2 dx=1? YES!

Lets work it out…

∫−∞

∞1

2πex2 dx=∫

−∞

∞12π

e−x2

2 dx

=12π∫

−∞

∞

e−x2

2 dx

But there is a problem…

∫−∞

∞

e−x2

2 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 beI2π

But let’s concentrate on the I

I = e−x2

2 dx∫−∞

∞Since x is merely avariable of integration, we can also express I as

∫−∞

∞

e−y2

2 dyI =

So why not express I2 as

e−y2

2 dye−x2

2 dxI 2 = ∫−∞

∞

∫−∞

∞

or

−∞

∞

∫−∞

∞

∫e−x22 e

−y2

2 dxdy

−∞

∞

∫−∞

∞

∫e−x22 e

−y2

2 dxdy=

−∞

∞

∫−∞

∞

∫e− x2+y2( )2 dxdy

This double integral is actually the volume under a 3-D “bell”

-20

2

-2

0

2

0

0.25

0.5

0.75

1

0

0.25

0.5

0.75

1

Rectangles aren’t the only thing that integrates!

We can now do a clever change of variableby converting to Polar coordinates!

x + y = r

x

2

ry

2 2

−∞

∞

∫−∞

∞

∫e− x2+y2( )2 dxdy=

e− r2( )

2 rdrdθ∫∫π

0 −∞

∞

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...

Asector =12dθ r2

dA=12b

2dθ−12a

2dθ

=12 (a+b)(a−b)dθ

Since 12 (a+b) =r and (a−b)=dr

dA=rdrdθGetting back to our story...

Recall the Pythagorean Theorem?

We can now do a clever change of variableby converting to Polar coordinates!

x + y = r

x

2

ry

2 2

−∞

∞

∫−∞

∞

∫e− x2+y2( )2 dxdy=

e− r2( )

2 rdrdθ∫∫π

0 −∞

∞

So now we DO have an antiderivative!

e− r2( )

2 rdrdθ=∫∫π

0 −∞

∞

∫π

0

−e− r2( )

2

−∞

∞

dθ

∫π

0

−e− r2( )

2

−∞

∞

dθ= ∫π

0

e− r2( )

2

−∞

∞

dθ

And because of symmetry

=2∫π

0

e− r2( )

2

0

∞

dθ=2 1

e02−

1e

∞2dθ∫

π

0

We can now start evaluating the integral from negative to positive infinity…

Almost there!

2 1

e02−

1e

∞2dθ∫

π

0

=2 11

−limb→ ∞

1

eb2

dθ∫π

0

1−0 dθ dθ∫π

0

=2 =2∫π

0

=2π

But that wasn’t “I”

I 2 =2πSo

I = 2π

...that was “I” squared!

Recall the area under the Standard Normal Curve

I2π

So… =2π2π

=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