1. Special Continuous Probability Distributions Normal Distributions
Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Special Continuous Probability Distributions
Normal Distributions
Lognormal Distributions
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering

2. Normal Distribution A random variable X is said to have a normal (or
Gaussian) distribution with parameters  and ,
where -  <  <  and  > 0, with probability
density function
for -  < x < 
f(x) x

3. Properties of the Normal Model
the effects of  and 

4. Normal Distribution Mean or expected value of Mean = E(X) = 
Median value of
X0.5 = 
Standard deviation

5. Normal Distribution Standard Normal Distribution If ~ N(, ) and if
then Z ~ N(0, 1).
A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.

6. Normal Distribution
P (X<x’) = P (Z<z’)
f(z)
f(x) x z  X’ Z’

7. Normal Distribution
Standard Normal Distribution Table of Probabilities
Enter table with
and find the
value of 
Excel z  z
f(z)

8. Normal Distribution - Example
The following example illustrates every possible
case of application of the normal distribution.
Let ~ N(100, 10)
Find:
(a) P(X < 105.3)
(b) P(X  91.7)
(c) P(87.1 <  115.7)
(d) the value of x for which P(  x) = 0.05

9. Normal Distribution – Example Solution
a. P( < 105.3) =
= P( < 0.53)
= F(0.53) =
Normal Distribution – Example Solution
f(x)
f(z)
100
105.3 x z
0.53

10. Normal Distribution – Example Solution
b. P(  91.7) =
= P( )
= 1 - P( < -0.83)
= 1- F(-0.83) = =
Normal Distribution – Example Solution
f(x)
f(z) x z
100
91.7
-0.83

11. Normal Distribution – Example Solution
c. P(87.1 <  115.7) = F(115.7) - F(87.1)
= P(-1.29 < Z < 1.57)
= F(1.57) - F(-1.29)
= =
Normal Distribution – Example Solution
f(x)
f(x) x x
87.1
100
115.7
-1.29
1.57

12. Normal Distribution – Example Solution
f(x)
f(z)
0.05
0.05 x
1.64 z
116.4
100

13. Normal Distribution – Example Solution
(d) P(  x) = 0.05
P(  z) = 0.05 implies that z = P(  x) =
therefore
x = 16.4
x = 116.4

14. Normal Distribution – Example Solution
The time it takes a driver to react to the brake lights
on a decelerating vehicle is critical in helping to
avoid rear-end collisions. The article ‘Fast-Rise Brake
Lamp as a Collision-Prevention Device’ suggests
that reaction time for an in-traffic response to a
brake signal from standard brake lights can be
modeled with a normal distribution having mean
value 1.25 sec and standard deviation 0.46 sec.
What is the probability that reaction time is between
1.00 and 1.75 seconds? If we view 2 seconds as a
critically long reaction time, what is the probability
that actual reaction time will exceed this value?

15. Normal Distribution – Example Solution

16. Normal Distribution – Example Solution

17. Lognormal Distribution

18. Lognormal Distribution
Definition - A random variable is said to have the
Lognormal Distribution with parameters  and ,
where  > 0 and  > 0, if the probability density
function of X is:
, for x > 0
, for x  0 x
f(x)

19. Lognormal Distribution - Properties
Rule: If ~ LN(,),
then = ln ( ) ~ N(,)
Probability Distribution Function
where F(z) is the cumulative probability distribution
function of N(0,1)

20. Lognormal Distribution - Properties
Mean or Expected Value
Median
Standard Deviation

21. Lognormal Distribution - Example
A theoretical justification based on a certain material
failure mechanism underlies the assumption that ductile
strength X of a material has a lognormal distribution.
Suppose the parameters are  = 5 and  = 0.1
(a) Compute E( ) and Var( )
(b) Compute P( > 120)
(c) Compute P(110   130)
(d) What is the value of median ductile strength?
(e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120?
(f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

22. Lognormal Distribution –Example Solution

23. Lognormal Distribution –Example Solutionc) d)

24. Lognormal Distribution –Example Solution
Let Y=number of items tested that have strength of at
least 120
y=0,1,2,…,10

25. Lognormal Distribution –Example Solution
f) The value of x, say xms, for which is
determined as follows: