The Mean and Standard Deviation of the Normal Distribution
Let the random variable X have a normal distribution with probability density function given by
f x  

1
 x  2
2 2
, for - < x < +. We want to show that the mean of the distribution is
2 
 = , and the standard deviation of the distribution is  = .
e
By definition, the mean of the distribution is found as


  EX    xf x dx   x

1
 x  2
2 2
dx .
2 
Before evaluating the integral, we do a linear transformation of the variable to simplify the
x 
calculations. Let z 
, so that x  z   , and dx  dz . With this transformation, we

have



e

z2
2

dz 
2



z2
2

z2
2
 z   e
 ze dz    2 e dz .
2  

The second integral on the right is 1, since it is the integral of the standard normal p.d.f. over the
whole real line. The integrand in the first integral is an odd function1. The integral has
symmetrically located limits; hence the integral is 0. We may also see this by writing the
integral as a sum of integrals over the two halves of the real line, and doing a variable
transformation, as follows:
1


We have
 ze

z2
2
0
dz 


 ze
z2

2

z2
2


dz 
 ze

 we
w

dz   ze

z2
2
0
1
z2
dz . Now let w 
, and dw  zdz . Then
2

dw   we  w dw  0 .
0
0
Therefore, we have  = .
 
To find the standard deviation, we use the fact that  2  E X 2   2 . By definition, the second
moment of the distribution is

  x
E X2 
2
f x dx 


x

1
2
2 

e
 x  2
2 2
dx . We do the same linear transformation of x, and
find
 
EX
2
1

2 
2

2

 z    e
2


z

z2
2

z2
2
1
dz 
2
2


z2
2

 
2

z  2z   e
2
2

z2
2
dz


1

z2
2
 ze dz    2 e dz .
2 
Now the second integral is 0, by the same argument as given above, while the third integral is 1.
Hence, we have

2
e
dz 
2
2
EX  
2

z
2
2
e

z2
2

2
dz   
2
2

z
2
e

z2
2
dz   2 .

We integrate by parts:
Let u  z , so that du  dz . Let dv  ze

z2
2
dz , so that v  e

z2
2
. Then



z2
z2
z2


 

1 2


2
2
2
2
e dz   2 .
 ze    e dz      

  
 2




The last integral is 1, so that
E X 2   2  2 .
Therefore, the variance is
 2  E X 2   2   2   2    2   2 .
2
E X  
2
2
 
 
Therefore, we see that the p.d.f. of the normal distribution may be written as
f x  
1
1
2 

e
 x   2
2 2
, for - < x < +.
A function f(x) is called an odd function if f(-x) = -f(x) for all x.