Download Normal (Gaussian) & Lognormal 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
1
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
2
Properties of the Normal Model
the effects of  and 
3
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution
• Mean or expected value of
Mean = E(X) = 
• Median value of
X
X
X0.5 = 
• Standard deviation
Var(X )  
4
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution
P (X<x’)
=
P (Z<z’)
f(z)
f(x)
X’

x
x'μ
Z
σ
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution – Example Solution
f(x)
f(z)
100 105.3
a. P( X< 105.3)
=
x
z
 X   105.3  100 
P


10
 

= P(Z< 0.53)
= F(0.53)
= 0.7019
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
0 0.53
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
15
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
An Application of Probability & Statistics Statistical
Quality Control
Statistical Quality Control is an application of
probabilitistic and statistical techniques to quality
control
17
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Statistical Quality Control - Elements
Analysis
of process
capability
Process
improvement
Statistical
process
control
Acceptance
sampling
18
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Statistical Tolerancing - Convention
Normal Probability
Distribution
0.00135
LTL
-3
0.9973
Nominal

0.00135
UTL
+3
19
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Statistical Tolerancing - Concept
LTL
Nominal
UTL
x
20
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Caution
For a normal distribution, the natural tolerance
limits include 99.73% of the variable, or put
another way, only 0.27% of the process output will
fall outside the natural tolerance limits. Two points
should be remembered:
1. 0.27% outside the natural tolerances sounds
small, but this corresponds to 2700 nonconforming
parts per million.
2. If the distribution of process output is non
normal, then the percentage of output falling
outside   3 may differ considerably from 0.27%.
21
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution - Example
The diameter of a metal shaft used in a disk-drive unit
is normally distributed with mean 0.2508 inches and
standard deviation 0.0005 inches. The specifications
on the shaft have been established as 0.2500  0.0015
inches. We wish to determine what fraction of
the shafts produced conform to specifications.
22
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution - Example Solution
Pmeeting spec   P0.2485  x  0.2515
 Px  0.2515  P0.2485  x 
 0.2515 - 0.2508 
 0.2485 - 0.2508 
 





0.0005
0.0005




 1.40   4.60
 0.91924  0.0000
 0.91924
f(x)
0.2500
0.2485
LSL
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
nominal
0.2508
0.2515
USL
x
23
Normal Distribution - Example Solution
Thus, we would expect the process yield to be
approximately 91.92%; that is, about 91.92% of
the shafts produced conform to specifications. Note
that almost all of the nonconforming shafts are too
large, because the process mean is located very
near to the upper specification limit. Suppose we
can recenter the manufacturing process, perhaps
by adjusting the machine, so that the process mean
is exactly equal to the nominal value of 0.2500.
Then we have
24
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution - Example Solution
P0.2485  x  0.2515  Px  0.2515  P0.2485  x 
 0.2515 - 0.2500 
 0.2485 - 0.2500 
 
  

0.0005
0.0005




 3.00   3.00
 0.99865  0.00135
 0.9973
f(x)
0.2485
LSL
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
nominal
0.2500
x
0.2515
USL
25
26
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
27
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
28
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
29
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution - Example
Using a normal probability distribution as a model
for a quality characteristic with the specification
limits at three standard deviations on either side of
the mean. Now it turns out that in this situation the
probability of producing a product within these
specifications is 0.9973, which corresponds to 2700
parts per million (ppm) defective. This is referred to
as three-sigma quality performance, and it actually
sounds pretty good. However, suppose we have a
product that consists of an assembly of 100
components or parts and all 100 parts must be
non-defective for the product to function
satisfactorily.
30
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Normal Distribution - Example
The probability that any specific unit of product is
non-defective is
0.9973 x 0.9973 x . . . x 0.9973
= (0.9973)100
= 0.7631
That is, about 23.7% of the products produced under
three sigma quality will be defective. This is not an
acceptable situation, because many high technology
products are made up of thousands of components.
An automobile has about 200,000 components and
an airplane has several million!
31
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
Lognormal Distribution
32
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
1
x 2
0

e
1
2

ln
x



22
x
,
for x > 0
,
for x  0
33
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)
34
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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





Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
1
2
35
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?
36
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
37
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
38
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
39
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
Stracener_EMIS 7370/STAT 5340_Sum 08_06.05.08
40