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Math/Stat 370: Engineering Statistics,
Washington State University
Haijun Li
[email protected]
Department of Mathematics
Washington State University
Week 3
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
1 / 15
Normal (or Gaussian) Distribution
Normal density function N(µ, σ 2 ):
−(x−µ)2
1
ϕµ,σ2 (x) = √
e 2σ2 , −∞ < x < ∞.
2πσ
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
2 / 15
Normal (or Gaussian) Distribution
Normal density function N(µ, σ 2 ):
−(x−µ)2
1
ϕµ,σ2 (x) = √
e 2σ2 , −∞ < x < ∞.
2πσ
Normal random variable: X has a normal density function
N(µ, σ 2 ). µ = location parameter, σ = scale parameter.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
2 / 15
Properties of Normal Distributions
1
2
3
4
5
Normal density curve is symmetric about µ.
The curve reaches the maximum at µ.
E(X ) = µ.
V (X ) = σ 2 .
P(µ − σ < X < µ + σ) ≈ 0.68.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
3 / 15
Normal Approximation
Many histograms of large samples have similar shapes as
that of normal distributions.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
4 / 15
Standardizing Normal Random Variables
Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
5 / 15
Standardizing Normal Random Variables
Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1.
If X has N(µ, σ 2 ), then
Z =
X −µ
, (called the standardization!)
σ
has the standard normal N(0, 1).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
5 / 15
Standardizing Normal Random Variables
Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1.
If X has N(µ, σ 2 ), then
Z =
X −µ
, (called the standardization!)
σ
has the standard normal N(0, 1).
CDF of N(0, 1): Φ(z) = P(Z ≤ z).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
5 / 15
Standardizing Normal Random Variables
Standard Normal Z : N(0, 1), µ = 0, σ 2 = 1.
If X has N(µ, σ 2 ), then
Z =
X −µ
, (called the standardization!)
σ
has the standard normal N(0, 1).
CDF of N(0, 1): Φ(z) = P(Z ≤ z).
Appendix A Table I lists all the values of Φ(z) from z = −3.9
to z = 3.9.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
5 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) =
0.8849 − 0.0107 = 0.8742.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) =
0.8849 − 0.0107 = 0.8742.
4
P(|X | > 3.1).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) =
0.8849 − 0.0107 = 0.8742.
4
P(|X | > 3.1).
P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) =
0.8849 − 0.0107 = 0.8742.
4
P(|X | > 3.1).
P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194.
5
Find z such that P(−z < Z < z) = 0.95.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Standard Normal Probabilities
Example: Find the following.
1
P(Z ≤ −2.3).
P(Z ≤ −2.3) = Φ(−2.3) = 0.0107.
2
P(Z > 1.2)
P(Z > 1.2) = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151.
3
P(−2.3 < Z < 1.2).
P(−2.3 < Z < 1.2) = Φ(1.2) − Φ(−2.3) =
0.8849 − 0.0107 = 0.8742.
4
P(|X | > 3.1).
P(|X | > 3.1) = 2Φ(−3.1) = 2 × 0.00097 = 0.00194.
5
Find z such that P(−z < Z < z) = 0.95.
First, we must have that P(Z < z) = 0.975. Second, we find
from the normal table that z = 1.96.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
6 / 15
Normal Probabilities
If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization
leads to


Z
z }| {
a − µ
X −µ
b − µ

P(a < X < b) = P 
<
<
 σ
σ
σ 
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
7 / 15
Normal Probabilities
If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization
leads to


Z
z }| {
a − µ
X −µ
b − µ

P(a < X < b) = P 
<
<
 σ
σ
σ 
= P
Haijun Li
a−µ
b−µ
<Z <
σ
σ
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
7 / 15
Normal Probabilities
If X has N(µ, σ 2 ) and Z has N(0, 1), then the standardization
leads to


Z
z }| {
a − µ
X −µ
b − µ

P(a < X < b) = P 
<
<
 σ
σ
σ 
a−µ
b−µ
= P
<Z <
σ
σ
b−µ
a−µ
= Φ
−Φ
.
σ
σ
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
7 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
What is the probability that the diameter of a dot exceeds
0.0026 inch?
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
2
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
What is the probability that the diameter of a dot between
0.0014 and 0.0026 inch?
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
2
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
What is the probability that the diameter of a dot between
0.0014 and 0.0026 inch?
P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) =
0.933 − 0.067 = 0.866.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
2
3
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
What is the probability that the diameter of a dot between
0.0014 and 0.0026 inch?
P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) =
0.933 − 0.067 = 0.866.
What is the diameter that is exceeded 90% of time?
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
2
3
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
What is the probability that the diameter of a dot between
0.0014 and 0.0026 inch?
P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) =
0.933 − 0.067 = 0.866.
What is the diameter that is exceeded 90% of time?
Solution: Find
P(X > t) = 0.9. That is,
t such that t−0.002
t−0.002
1 − Φ 0.0004 = 0.9 or Φ 0.0004 = 0.1.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Example: The diameter X of the dot produced produced by a
printer is normally distributed with a mean diameter of µ = 0.002
inch and a standard deviation of σ = 0.0004 inch.
1
2
3
What is the probability that the diameter of a dot exceeds
0.0026 inch?
P(X > 0.0026) = 1 − Φ (0.0026 − 0.002)/0.0004 =
1 − Φ(1.5) = 0.067.
What is the probability that the diameter of a dot between
0.0014 and 0.0026 inch?
P(0.0014 ≤ X ≤ 0.0026) = Φ(1.5) − Φ(−1.5) =
0.933 − 0.067 = 0.866.
What is the diameter that is exceeded 90% of time?
Solution: Find
P(X > t) = 0.9. That is,
t such that t−0.002
t−0.002
1 − Φ 0.0004 = 0.9 or Φ 0.0004 = 0.1.
From the normal table, t−0.002
= −1.28 and t = 0.001482
0.0004
inch.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
8 / 15
Log-normal Distribution
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
distribution with mean θ and variance ω 2 , N(θ, ω 2 ).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
9 / 15
Log-normal Distribution
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
2
2
distribution with mean θ and variance
ω ,N(θ, ω ).
P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ
.
ω
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
9 / 15
Log-normal Distribution
0/02 5:20 PM Page 136 RK UL 6 RK UL 6:Desktop Folder:TEMP WORK:MONTGOMERY:REVISES UPLO D C
136
X ≥ 0 has a log-normal distribution if ln(X ) has a normal
2
2
distribution with mean θ and variance
ω ,N(θ, ω ).
P(X ≤ x) = P(ln(X ) ≤ ln(x)) = Φ ln(x)−θ
.
ω
Compound returns of stock prices and component lifetimes
CHAPTER 4 CONTINUOUS RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
can be modeled using log-normal distributions.
1
ω 2 = 0.25
ω2 = 1
ω 2 = 2.25
0.9
0.8
0.7
0.6
f (x)
0.5
0.4
0.3
0.2
0.1
0
–0.1
Haijun Li
0
1
2
3
x
4
5
Math/Stat 370: Engineering Statistics, Washington State University
6
Week 3
9 / 15
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
10 / 15
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
10 / 15
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
10 / 15
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
2
What lifetime is exceeded by 99% of lasers?
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
10 / 15
Example
The lifetime X of a semiconductor laser has a lognormal
distribution with θ = 10 and ω = 1.5 hours.
1
What is the probability that lifetime exceeds 10,000 hours?
P(X > 10, 000) = 1 − P(X ≤ 10, 000) =
= 1 − Φ(−0.53) = 0.702.
1 − Φ ln 10,000−10
1.5
2
What lifetime is exceeded by 99% of lasers?
Solution: Find t such that P(X > t) = 0.99. That is,
ln t − 10
P(X ≤ t) = 0.01, and Φ
= 0.01.
1.5
Solve it, we have
hours.
Haijun Li
ln t−10
1.5
= −2.33 and t = e6.505 = 668.48
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
10 / 15
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
11 / 15
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
If k is a positive integer, then the gamma function
Γ(k ) = (k − 1)!.
Widely used in reliability engineering and
telecommunication.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
11 / 15
Gamma Distribution
A random variable X is said to have a gamma distribution if
its pdf
Z ∞
λk x k −1 e−λx
f (x) =
, x > 0, where Γ(k ) =
x k −1 e−x dx,
Γ(k )
0
If k is a positive integer, then the gamma function
Γ(k ) = (k − 1)!.
Widely used in reliability engineering and
telecommunication.
E(X ) = λk , V (X ) = λk2 .
k = shape parameter, λ = rate parameter.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
11 / 15
k = shape parameter, λ = rate parameter, θ = λ−1 = scale
parameter.
A special case: When k = n (integer), the distribution is
called an Erlang distribution (widely used in
telecommunication).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
12 / 15
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
13 / 15
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
13 / 15
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
13 / 15
k = 1: Exponential distribution. The pdf is given by
f (x) = λe−λx , x ≥ 0.
The CDF of the exponential is given by F (x) = 1 − e−λx .
E(X ) = λ1 , V (X ) = λ12 . Note that λ−1 represents the scale.
f (x)
Failure rate h(x) = 1−F
= λ, a constant.
(x)
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
13 / 15
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
14 / 15
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
14 / 15
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
14 / 15
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
The failure rate
f (x)
β x β−1
h(x) =
.
=
1 − F (x)
δ δ
h(x) is increasing if β > 1; decreasing if β < 1; and a
constant if β = 1.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
14 / 15
Weibull Distribution
A random variable X with pdf
β x β−1 −( x )β
e δ , x > 0,
f (x) =
δ δ
is said to have a Weibull distribution with parameters β
(shape), and δ (scale).
If β = 1, the Weibull becomes an exponential.
x β
The CDF F (x) = 1 − e−( δ ) .
The failure rate
f (x)
β x β−1
h(x) =
.
=
1 − F (x)
δ δ
h(x) is increasing if β > 1; decreasing if β < 1; and a
constant if β = 1.
E(X ) = δΓ(1 + β1 ), V (X ) = δ 2 Γ(1 + β2 ) − δ 2 [Γ(1 + β1 )]2 .
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
14 / 15
Weibull PDF
k = β = shape parameter, λ = δ = scale parameter.
Haijun Li
Math/Stat 370: Engineering Statistics, Washington State University
Week 3
15 / 15