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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
1
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

1
2
1
f ( x) 
e 2
 2
f(x)


2
x


for -  < x < 
x
2
Properties of the Normal Model
the effects of  and 
3
Normal Distribution
• Mean or expected value of
Mean = E(X) = 
• Median value of
X
X
X0.5 = 
• Standard deviation
Var(X )  
4
Normal Distribution
Standard Normal Distribution
• If
X~ N(, )
Z
and if
X 

then Z ~ N(0, 1).
• A normal distribution with  = 0 and  = 1, is called
the standard normal distribution.
5
Normal Distribution
P (X<x’)
=
P (Z<z’)
f(z)
f(x)
X’

x
x'μ
Z
σ
Z’
0
z
6
Normal Distribution
• Standard Normal Distribution Table of Probabilities
http://www.engr.smu.edu/~jerrells/courses/help/normaltable.html
Enter table with
Z
x
f(z)

and find the
value of 
• Excel

0
z
z
7
Normal Distribution - Example
The following example illustrates every possible
case of application of the normal distribution.
Let X ~ N(100, 10)
Find:
(a)
P(X < 105.3)
(b)
P(X  91.7)
(c)
P(87.1 < X  115.7)
(d)
the value of x for which P(X  x) = 0.05
8
Normal Distribution – Example Solution
f(x)
f(z)
100 105.3
a. P( X< 105.3)
=
x
0 0.53
z
 X   105.3  100 
P


10
 

= P(Z< 0.53)
= F(0.53)
= 0.7019
9
Normal Distribution – Example Solution
f(x)
f(z)
91.7 100
b. P( X  91.7)
x
-0.83
0
z
 X   91.7  100 
= P


10
 

= P( Z  -0.83)
= 1 - P( Z < -0.83)
= 1- F(-0.83)
= 1 - 0.2033
= 0.7967
10
Normal Distribution – Example Solution
f(x)
f(x)
x
x
87.1 100 115.7
-1.29 0 1.57
c. P(87.1 < X  115.7) = F(115.7) - F(87.1)
 87.1  100 x  

P

 115.7 
10



= P(-1.29 < Z < 1.57)
= F(1.57) - F(-1.29)
= 0.9418 - 0.0985 = 0.8433
11
Normal Distribution – Example Solution
f(x)
f(z)
0.05
100
116.4
0.05
x
1.64
z
0
1.64 
x  100
10
12
Normal Distribution – Example Solution
(d)
P( X  x) = 0.05
P( Z  z) = 0.05
P( X  x) =
implies that z = 1.64
x  100 
 X   x  100 

 x  100 
P

  P Z 
  1  

10 
10 
 

 10 
therefore
x  100
 1.64
10
x - 100 = 16.4
x = 116.4
13
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?
14
Normal Distribution – Example Solution
P1.00  X  1.75
1.75  1.25 
 1.00  1.25
 P
X

0.46 
 0.46
 P 0.54  X  1.09
 F 1.09  F  0.54
 0.8621  0.2946  0.5675
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Normal Distribution – Example Solution
2  1.25 

P X  2   P  Z 

0.46 

 PZ  1.63
 1  F 1.63
 1  0.9484
 0.0516
16
Lognormal Distribution
17
Lognormal Distribution
Definition - A random variable X is said to have the
Lognormal Distribution with parameters  and ,
where  > 0 and  > 0, if the probability density
function of X is:
f (x) 
f(x)
0
1
x 2
0

e
1
2

ln
x



22
x
,
for x > 0
,
for x  0
18
Lognormal Distribution - Properties
• Rule:
then
If X ~ LN(,),
Y
= ln ( X ) ~ N(,)
• Probability Distribution Function
 ln x   
F ( x)  F 

  
where F(z) is the cumulative probability distribution
function of N(0,1)
19
Lognormal Distribution - Properties
Mean or Expected Value
1 2
 
2
E( X )  e
• Median
x0.5  e

• Standard Deviation
 2μ  σ 2  σ 2 
 e  1
SD(X)  e





1
2
20
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(X) and Var(X)
(b) Compute P(X > 120)
(c) Compute P(110  X 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?
21
Lognormal Distribution –Example Solution
a)
E( X )  e
u
2
Var ( X )  e
b)
2
e
2u  2
5.005
2
(e
e
5.005
 149.16
 1)  223
P( X  120)  1  P( X  120)
ln 120  5.0
 1  P( Z 
)
0.1
 1  F (2.13)
 1  0.0166
 0.9834
22
Lognormal Distribution –Example Solution
ln 110  5.0
ln 130  5.0
c) P (110  X  130)  P (
Z
)
0.1
0.1
 P(2.99  Z  1.32)
 F (1.32)  F (2.99)
 0.0934  0.0014
 0.092
d)
X 0.5  median  e  e  148.41
u
5
23
Lognormal Distribution –Example Solution
e)
P  P( X  120)
 1  P( X  120)
ln 120  5.0
 1  P( Z 
)
0.1
 1  F (2.12)
 1  0.0170
 0.983
Let Y=number of items tested that have strength of at
least 120
y=0,1,2,…,10
24
Lognormal Distribution –Example Solution
Y ~ B(10,0.983)
E (Y )  np  10 * 0.983  9.83
f) The value of x, say xms, for which P( X  xms )  0.05 is
determined as follows:
ln xms  5.0
P( Z 
)  0.05
0.1
P ( Z  1.64)  0.05
ln xms  5.0
 1.64
0 .1
xms  125.964
25