STT 315, Summer 2006
Lecture 3
Materials Covered: Chapter 4
1. Normal Random Variable.
A continuous random variable X is called to be a normal random variable with mean  and
variance 2, if and only if the probability density function of X is given by

1
f ( x) 
e
 2
( x )2
2 2
   x  ,
,
where -<<, and  > 0. Notation: X~N(, 2).
2. Properties of the Normal Distribution.
(1). If X~N(, 2), then E(X) = , V(X) = 2.
(2). If X1,X2,…,Xn are independent random variables that are normally distributed, then the sum
S = X1+X2+…+Xn will also be normally distributed with E(S) = E(X1) + E(X2) +…+ E(Xn)
and V(S) = V(X1) + V(X2) +…+ V(Xn).
Example: Let X1, X2, X3 be independent random variables that are normally distributed with
means and variances as follows:
Mean
Variance
X1
10
1
X2
20
2
X3
30
3
Find the distribution of the sum S = X1 + X2 + X3.
STT 315, Summer 2006
(3). If X1,X2,…,Xn are independent random variables that are normally distributed, a1, a2,…, an
and b are constants, then the sum Q = a1X1+a2X2+…+anXn + b will also be normally
distributed with
E(Q) = a1E(X1) + a2E(X2) +…+ anE(Xn)
and
V(Q) = a1 V ( X 1 )  a 2 V ( X 2 )  ...  a n V ( X n ) .
2
2
2
Example: Let X1, X2, X3 and X4 be independent random variables that are normally distributed with
means and variances as follows:
Mean
Variance
X1
12
4
X2
-5
2
X3
8
5
X4
10
1
Find the distribution of the sum Q = X1 - 2X2 + 3X3- 4X4+ 5. Find also the standard deviation of Q.
3. The Standard Normal Distribution.

We define the standard normal random variable Z as the normal random variable with mean
=0, and the standard deviation  = 1. Notation Z ~ N(0, 1).
STT 315, Summer 2006

Finding the probability of the N(0,1).
(1). How to find F(z) = P(Zz)?
The normal table only gives the probability of the type P(0Zz), z > 0.
Basic Formula: Let TA denote the table area, then
(1). If z > 0, then F(z) = (TA of 0  Z  z) + 0.5.
(2). If z < 0, then F(z) = 0.5 – (TA of 0  Z  |z| ).
Example: Let Z ~N(0,1), find P(Z  1.1), P(Z0.92 ) and P(Z < -1.23).
(2). How to find P(Z > z)?
P(Z > z) =1- P(Z  z).
STT 315, Summer 2006
Example: Let Z ~N(0,1), find P(Z > 1.1).
(3). How to find P(z1 Z z2)?
P(z1 Z z2) = P(Z  z2) – P(Z  z1) = F(z2) – F(z1)
Example: Let Z ~ N(0, 1), Find P(-1.2  Z 2.32), P(-1.2  Z -1), P(1  Z 2.32)

Finding Values of Z Given a Probability.
Problem: let  be any given number in [0,1], how to find a value z such that P(Zz) = ?
Example: Find z such that
(1). P(0  Z  z) = 0.3997
(2). P(0  Z  z) = 0.3769
(3). P(0  Z  z) = 0.4926
Example: Find z such that P(Zz) = 0.6772, P(Zz) = 0.47.
STT 315, Summer 2006
4. The Transformation of Normal Random Variable.


If X ~N(,), then Z=(X - )/ ~ N(0, 1);
If Z ~ N(0,1), then X=+Z ~ N(,).
Suppose X ~N(,). How to find P(X<a), P(X>b), or P(a<X<b)?
a

P ( X  a )  P Z 

 

b 

P ( X  b )  P Z 

 

b 
a
P ( a  X  b )  P
Z

 
 
Example: Suppose X ~N(383, 144). Find P(394<X<399).
Example: The concentration of impurities in a semiconductor used in the production of
microprocessors for computers is a normally distributed random variable with mean
127 parts per million and standard deviation 22. A semiconductor is acceptable only if
its concentration of impurities is below 150 parts per million. What proportion of the
semiconductors is acceptable for use?
Example: Let X ~N(124, 144). Find the value of x such that P(X>x) = 0.10.