Normal Distribution
Normal Distribution
Proposition
If X has a normal distribution with mean µ and stadard deviation
σ, then
X −µ
Z=
σ
has a standard normal distribution. Thus
a−µ
b−µ
≤Z ≤
)
σ
σ
b−µ
a−µ
= Φ(
) − Φ(
)
σ
σ
P(a ≤ X ≤ b) = P(
P(X ≤ a) = Φ(
a−µ
)
σ
P(X ≥ b) = 1 − Φ(
b−µ
)
σ
Normal Distribution
Normal Distribution
Example (Problem 38):
There are two machines available for cutting corks intended for use
in wine bottles. The first produces corks with diameters that are
normally distributed with mean 3cm and standard deviation 0.1cm.
The second produces corks with diameters that have a normal
distribution with mean 3.04cm and standard deviation 0.02cm.
Acceptable corks have diameters between 2.9cm and 3.1cm.
Which machine is more likely to produce an acceptable cork?
Normal Distribution
Example (Problem 38):
There are two machines available for cutting corks intended for use
in wine bottles. The first produces corks with diameters that are
normally distributed with mean 3cm and standard deviation 0.1cm.
The second produces corks with diameters that have a normal
distribution with mean 3.04cm and standard deviation 0.02cm.
Acceptable corks have diameters between 2.9cm and 3.1cm.
Which machine is more likely to produce an acceptable cork?
2.9 − 3
3.1 − 3
≤Z ≤
)
0.1
0.1
= P(−1 ≤ Z ≤ 1) = 0.8413 − 0.1587 = 0.6826
2.9 − 3.04
3.1 − 3.04
P(2.9 ≤ X2 ≤ 3.1) = P(
≤Z ≤
)
0.02
0.02
= P(−7 ≤ Z ≤ 3) = 0.9987 − 0 = 0.9987
P(2.9 ≤ X1 ≤ 3.1) = P(
Normal Distribution
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
µ − 1.5σ − µ
µ + 1.5σ − µ
≤Z ≤
)
σ
σ
= P(−1.5 ≤ Z ≤ 1.5)
P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P(
= 0.9332 − 0.0668 = 0.8664
Normal Distribution
Example (Problem 44):
If bolt thread length is normally distributed, what is the probability
that the thread length of a randomly selected bolt is (a)within 1.5
SDs of its mean value? (b)between 1 and 2 SDs from its mean
value?
µ − 1.5σ − µ
µ + 1.5σ − µ
≤Z ≤
)
σ
σ
= P(−1.5 ≤ Z ≤ 1.5)
P(µ − 1.5σ ≤ X1 ≤ µ + 1.5σ) = P(
= 0.9332 − 0.0668 = 0.8664
µ+σ−µ
µ + 2σ − µ
≤Z ≤
)
σ
σ
= 2P(1 ≤ Z ≤ 2)
2 · P(µ + σ ≤ X1 ≤ µ + 2σ) = 2P(
= 2(0.9772 − 0.8413) = 0.0.2718
Normal Distribution
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Example (Problem 39)
The width of a line etched on an integrated circuit chip is normally
distributed with mean 3.000 µm and standard deviation 0.140.
What width value separates the widest 10% of all such lines from
the other 90%?
Normal Distribution
Proposition
{(100p)th percentile for N(µ, σ 2 )} =
µ + {(100p)th percentile for N(0, 1)} · σ
Example (Problem 39)
The width of a line etched on an integrated circuit chip is normally
distributed with mean 3.000 µm and standard deviation 0.140.
What width value separates the widest 10% of all such lines from
the other 90%?
ηN(3,0.1402 ) (90) = 3.0+0.140·ηN(0,1) (90) = 3.0+0.140·1.28 = 3.1792
Normal Distribution
Normal Distribution
Proposition
Let X be a binomial rv based on n trials with success probability p.
Then if the binomial probability histogram is not too skewed, X has
√
approximately a normal distribution with µ = np and σ = npq,
where q = 1 − p. In particular, for x = a posible value of X ,
area under the normal curve
P(X ≤ x) = B(x; n, p) ≈
to the left of x+0.5
x+0.5 − np
= Φ( √
)
npq
In practice, the approximation is adequate provided that both
np ≥ 10 and nq ≥ 10, since there is then enough symmetry in the
underlying binomial distribution.
Normal Distribution
Normal Distribution
A graphical explanation for
P(X ≤ x) = B(x; n, p) ≈
= Φ(
area under the normal curve
to the left of x+0.5
x+0.5 − np
)
√
npq
Normal Distribution
A graphical explanation for
P(X ≤ x) = B(x; n, p) ≈
= Φ(
area under the normal curve
to the left of x+0.5
x+0.5 − np
)
√
npq
Normal Distribution
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus
np = 20 > 10 and nq = 180 > 10
Normal Distribution
Example (Problem 54)
Suppose that 10% of all steel shafts produced by a certain process
are nonconforming but can be reworked (rather than having to be
scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can
be reworked. What is the (approximate) probability that X is
between 15 and 25 (inclusive)?
In this problem n = 200, p = 0.1 and q = 1 − p = 0.9. Thus
np = 20 > 10 and nq = 180 > 10
P(15 ≤ X ≤ 25) = Bin(25; 200, 0.1) − Bin(14; 200, 0.1)
25 + 0.5 − 20
15 + 0.5 − 20
≈ Φ( √
) − Φ( √
)
200 · 0.1 · 0.9
200 · 0.1 · 0.9
= Φ(0.3056) − Φ(−0.2500)
= 0.6217 − 0.4013
= 0.2204
Exponential Distribution
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Remark:
1. Usually we use X ∼ EXP(λ) to denote that the random variable
X has an exponential distribution with parameter λ.
Exponential Distribution
Definition
X is said to have an exponential distribution with parameter
λ(λ > 0) if the pdf of X is
(
λe −λx x ≥ 0
f (x; λ) =
0
otherwise
Remark:
1. Usually we use X ∼ EXP(λ) to denote that the random variable
X has an exponential distribution with parameter λ.
2. In some sources, the pdf of exponential distribution is given by
(
1 − θx
x ≥0
e
f (x; θ) = θ
0
otherwise
The difference is that λ → 1θ .
Exponential Distribution
Exponential Distribution
Exponential Distribution
Exponential Distribution
Proposition
If X ∼ EXP(λ), then
E (X ) =
1
λ
and
V (X ) =
1
λ2
And the cdf for X is
(
1 − e −λx
F (x; λ) =
0
x ≥0
x <0
Exponential Distribution
Exponential Distribution
Proof:
Z
E (X ) =
0
=
=
=
=
=
∞
xλe −λx dx
Z
1 ∞
(λx)e −λx d(λx)
λ 0
Z
1 ∞ −y
ye dy
y = λx
λ 0
Z ∞
1
−y ∞
[−ye |0 +
e −y dy ] integration by parts:u = y , v = −e −
λ
0
1
[0 + (−e −y |∞
0 )]
λ
1
λ
Exponential Distribution
Exponential Distribution
Proof (continued):
Z ∞
E (X 2 ) =
x 2 λe −λx dx
0
Z ∞
1
= 2
(λx)2 e −λx d(λx)
λ 0
Z ∞
1
= 2
y 2 e −y dy
y = λx
λ 0
Z ∞
1
= 2 [−y 2 e −y |∞
2ye −y dy ]
integration by parts
0 +
λ
0
Z ∞
1
= 2 [0 + 2(−ye −y |∞
+
e −y dy )] integration by parts
0
λ
0
1
−y ∞
= 2 2[0 + (−ye |0 )]
λ
2
= 2
λ
Exponential Distribution
Exponential Distribution
Proof (continued):
1
1
2
V (X ) = E (X 2 ) − [E (X )]2 = 2 − ( )2 = 2
λ
λ
λ
Z x
F (x) =
λe −λy dy
0
Z x
=
e −λy d(λy )
Z0 x
=
e −z dz
0
= −e −z |x0
= 1 − e −x
z = λy
Exponential Distribution
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
b. What is the probability that path length exceeds 3.0?
Exponential Distribution
Example (Problem 108)
The article “Determination of the MTF of Positive Photoresists
Using the Monte Carlo method” (Photographic Sci. and
Engr., 1983: 254-260) proposes the exponential distribution
with parameter λ = 0.93 as a model for the distribution of a
photon’s free path length (µm) under certain circumstances.
Suppose this is the correct model.
a. What is the expected path length, and what is the standard
deviation of path length?
b. What is the probability that path length exceeds 3.0?
c. What value is exceeded by only 10% of all path lengths?
Exponential Distribution
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
e.g.
the number of customers visiting Costco in each hour =⇒
Poisson distribution;
Exponential Distribution
Proposition
Suppose that the number of events occurring in any time interval
of length t has a Poisson distribution with parameter αt (where α,
the rate of the event process, is the expected number of events
occurring in 1 unit of time) and that numbers of occurrences in
nonoverlappong intervals are independent of one another. Then
the distribution of elapsed time between the occurrence of two
successive events is exponential with parameter λ = α.
e.g.
the number of customers visiting Costco in each hour =⇒
Poisson distribution;
the time between every two successive customers visiting Costco
=⇒ Exponential distribution.
Exponential Distribution
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
The probability that more than 3 days elapse between calls is
P(X > 3) = 1 − P(X ≤ 3) = 1 − F (3; 0.5) = e −0.5·3 = 0.223.
Exponential Distribution
Example (Example 4.22)
Suppose that calls are received at a 24-hour hotline according to a
Poisson process with rate α = 0.5 call per day.
Then the number of days X between successive calls has an
exponential distribution with parameter value 0.5.
The probability that more than 3 days elapse between calls is
P(X > 3) = 1 − P(X ≤ 3) = 1 − F (3; 0.5) = e −0.5·3 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.
Exponential Distribution
Exponential Distribution
“Memoryless” Property
Let X = the time certain component lasts (in hours)
and we assume the component lifetime is exponentially distributed
with parameter λ. Then what is the probability that the
component can last at least an additional t hours after working for
t0 hours, i.e. what is P(X ≥ t + t0 | X ≥ t0 )?
Exponential Distribution
“Memoryless” Property
Let X = the time certain component lasts (in hours)
and we assume the component lifetime is exponentially distributed
with parameter λ. Then what is the probability that the
component can last at least an additional t hours after working for
t0 hours, i.e. what is P(X ≥ t + t0 | X ≥ t0 )?
P({X ≥ t + t0 } ∩ {X ≥ t0 })
P(X ≥ t0 )
P(X ≥ t + t0 )
=
P(X ≥ t0 )
1 − F (t + t0 ; λ)
=
F (t0 ; λ)
P(X ≥ t + t0 | X ≥ t0 ) =
= e −λt
Exponential Distribution
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Therefore, we have
P(X ≥ t) = P(X ≥ t + t0 | X ≥ t0 )
for any positive t and t0 .
Exponential Distribution
“Memoryless” Property
However, we have
P(X ≥ t) = 1 − F (t; λ) = e −λt
Therefore, we have
P(X ≥ t) = P(X ≥ t + t0 | X ≥ t0 )
for any positive t and t0 .
In words, the distribution of additional lifetime is exactly the same
as the original distribution of lifetime, so at each point in time the
component shows no effect of wear. In other words, the
distribution of remaining lifetime is independent of current age.
Gamma Distribution
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
√
3. Γ( 12 ) = π.
Gamma Distribution
Definition
For α > 0, the gamma function Γ(α) is defined by
Z ∞
Γ(α) =
x α−1 e −x dx
0
Properties for gamma function:
1. For any α > 1, Γ(α) = (α − 1) · Γ(α − 1) [via integration by
parts];
2. For any positive integer, n, Γ(n) = (n − 1)!;
√
3. Γ( 12 ) = π.
e.g. Γ(4) = (4 − 1)! = 6 and Γ( 25 ) =
3
2
· Γ( 32 ) = 32 [ 12 · Γ( 21 )] =
3√
4 π
Gamma Distribution
Gamma Distribution
Definition
A continuous random variable X is said to have a gamma
distribution if the pdf of X is
(
1
x α−1 e −x/β x ≥ 0
α
f (x; α, β) = β Γ(α)
0
otherwise
where the parameters α and β satisfy α > 0, β > 0. The standard
gamma distribution has β = 1, so the pdf of a standard gamma
rv is
(
1
x α−1 e −x x ≥ 0
f (x; α) = Γ(α)
0
otherwise
Gamma Distribution
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
3. When X is a standard gamma rv (β = 1), the cdf of X ,
Z
F (x; α) =
0
x
y α−1 e −y
dy
Γ(α)
is called the incomplete gamma function.
Gamma Distribution
Remark:
1. We use X ∼ GAM(α, β) to denote that the rv X has a gamma
distribution with parameter α and β.
2. If we let α = 1 and β = 1/λ, then we get the exponential
distribution:
( 1
1
x 1−1 e −x/ λ = λe −λx x ≥ 0
1
1
Γ(1)
f (x; 1, ) = λ
λ
0
otherwise
3. When X is a standard gamma rv (β = 1), the cdf of X ,
Z
F (x; α) =
0
x
y α−1 e −y
dy
Γ(α)
is called the incomplete gamma function.
There are extensive tables of F (x; α) available (Appendix Table
A.4).
Gamma Distribution
Gamma Distribution
Gamma Distribution
Gamma Distribution
Proposition
If X ∼ GAM(α, β), then
E (X ) = αβ
and
V (X ) = αβ 2
Furthermore, for any x > 0, the cdf of X is given by
x
P(X ≤ x) = F (x; α, β) = F
;α
β
where F (•; α) is the incomplete gamma function.
Gamma Distribution
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
c. the expected survival time.
Gamma Distribution
Example:
The survival time (in days) of a white rat that was subjected to a
certain level of X-ray radiation is a random variable
X ∼ GAM(5, 4). Then what is
a. the probability that the survival time is at most 16 days;
b. the probability that the survival time is between 16 days and
20 days (not inclusive);
c. the expected survival time.
Chi-Squared Distribution
Chi-Squared Distribution
Definition
Let ν be a positive integer. Then a random variable X is said to
have a chi-squared distribution with parameter ν if the pdf of X
is the gamma density with α = ν/2 and β = 2. The pdf of a
chi-squared rv is thus
(
1
x (ν/2)−1 e −x/2 x ≥ 0
ν/2
f (x; ν) = 2 Γ(ν/2)
0
x <0
The parameter ν is called the number of degrees of freedom
(df) of X . The symbol χ2 is often used in place of “chi-squared”.
Chi-Squared Distribution
Chi-Squared Distribution
Remark:
1. Usually, we use X ∼ χ2 (ν) to denote that X is a chi-squared rv
with parameter ν;
Chi-Squared Distribution
Remark:
1. Usually, we use X ∼ χ2 (ν) to denote that X is a chi-squared rv
with parameter ν;
2. If X1 , X2 , . . . , Xn is n independent standard normal rv’s, then
X12 + X22 + · · · + Xn2 has the same distribution as χ2 (n).
Chi-Squared Distribution
Chi-Squared Distribution
Weibull Distribution
Weibull Distribution
Definition
A random variable X is said to have a Weibull distribution with
parameters α and β (α > 0, β > 0) if the pdf of X is
(
α α−1 −(x/β)α
e
x ≥0
βα x
f (x; α, β) =
0
x <0
Weibull Distribution
Definition
A random variable X is said to have a Weibull distribution with
parameters α and β (α > 0, β > 0) if the pdf of X is
(
α α−1 −(x/β)α
e
x ≥0
βα x
f (x; α, β) =
0
x <0
Remark:
1. The family of Weibull distributions was introduced by the
Swedish physicist Waloddi Weibull in 1939.
Weibull Distribution
Definition
A random variable X is said to have a Weibull distribution with
parameters α and β (α > 0, β > 0) if the pdf of X is
(
α α−1 −(x/β)α
e
x ≥0
βα x
f (x; α, β) =
0
x <0
Remark:
1. The family of Weibull distributions was introduced by the
Swedish physicist Waloddi Weibull in 1939.
2. We use X ∼ WEB(α, β) to denote that the rv X has a Weibull
distribution with parameters α and β.
Weibull Distribution
Weibull Distribution
Remark:
Weibull Distribution
Remark:
3. When α = 1, the pdf becomes
(
1 −x/β
e
f (x; β) = β
0
x ≥0
x <0
which is the pdf for an exponential distribution with parameter
λ = β1 . Thus we see that the exponential distribution is a special
case of both the gamma and Weibull distributions.
Weibull Distribution
Remark:
3. When α = 1, the pdf becomes
(
1 −x/β
e
f (x; β) = β
0
x ≥0
x <0
which is the pdf for an exponential distribution with parameter
λ = β1 . Thus we see that the exponential distribution is a special
case of both the gamma and Weibull distributions.
4. There are gamma distributions that are not Weibull distributios
and vice versa, so one family is not a subset of the other.
Weibull Distribution
Weibull Distribution
Weibull Distribution
Weibull Distribution
Weibull Distribution
Weibull Distribution
Proposition
Let X be a random variable such that X ∼ WEI(α, β). Then
( 2 )
2
1
1
and V (X ) = β 2 Γ 1 +
− Γ 1+
E (X ) = βΓ 1 +
α
α
α
The cdf of X is
(
α
1 − e −(x/β)
F (x; α, β) =
0
x ≥0
x <0
Weibull Distribution
Weibull Distribution
Example:
The shear strength (in pounds) of a spot weld is a Weibull
distributed random variable, X ∼ WEB(400, 2/3).
a. Find P(X > 410).
Weibull Distribution
Example:
The shear strength (in pounds) of a spot weld is a Weibull
distributed random variable, X ∼ WEB(400, 2/3).
a. Find P(X > 410).
b. Find P(X > 410 | X > 390).
Weibull Distribution
Example:
The shear strength (in pounds) of a spot weld is a Weibull
distributed random variable, X ∼ WEB(400, 2/3).
a. Find P(X > 410).
b. Find P(X > 410 | X > 390).
c. Find E (X ) and V (X ).
Weibull Distribution
Example:
The shear strength (in pounds) of a spot weld is a Weibull
distributed random variable, X ∼ WEB(400, 2/3).
a. Find P(X > 410).
b. Find P(X > 410 | X > 390).
c. Find E (X ) and V (X ).
d. Find the 95th percentile.
Weibull Distribution
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibull
distribution.
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibull
distribution.
Example (Problem 74):
Let X = the time (in 10−1 weeks) from shipment of a
defective product until the customer returns the
product. Suppose that the minimum return time is γ = 3.5 and
that the excess X − 3.5 over the minimum has a Weibull
distribution with parameters α = 2 and β = 1.5.
a. What is the cdf of X ?
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibull
distribution.
Example (Problem 74):
Let X = the time (in 10−1 weeks) from shipment of a
defective product until the customer returns the
product. Suppose that the minimum return time is γ = 3.5 and
that the excess X − 3.5 over the minimum has a Weibull
distribution with parameters α = 2 and β = 1.5.
a. What is the cdf of X ?
b. What are the expected return time and variance of return
time?
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibull
distribution.
Example (Problem 74):
Let X = the time (in 10−1 weeks) from shipment of a
defective product until the customer returns the
product. Suppose that the minimum return time is γ = 3.5 and
that the excess X − 3.5 over the minimum has a Weibull
distribution with parameters α = 2 and β = 1.5.
a. What is the cdf of X ?
b. What are the expected return time and variance of return
time?
c. Compute P(X > 5).
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibull
distribution.
Example (Problem 74):
Let X = the time (in 10−1 weeks) from shipment of a
defective product until the customer returns the
product. Suppose that the minimum return time is γ = 3.5 and
that the excess X − 3.5 over the minimum has a Weibull
distribution with parameters α = 2 and β = 1.5.
a. What is the cdf of X ?
b. What are the expected return time and variance of return
time?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
Lognormal Distribution
Lognormal Distribution
Definition
A nonnegative rv X is said to have a lognormal distribution if the
rv Y = ln(X ) has a normal distribution. The resulting pdf of a
lognormal rv when ln(X ) is normally distributed with parameters µ
and σ is
(
2
2
√ 1
e −[ln(x)−µ] /(2σ ) x ≤ 0
2πσx
f (x; µ, σ) =
0
x <0
Lognormal Distribution
Definition
A nonnegative rv X is said to have a lognormal distribution if the
rv Y = ln(X ) has a normal distribution. The resulting pdf of a
lognormal rv when ln(X ) is normally distributed with parameters µ
and σ is
(
2
2
√ 1
e −[ln(x)−µ] /(2σ ) x ≤ 0
2πσx
f (x; µ, σ) =
0
x <0
Remark:
1. We use X ∼ LOGN(µ, σ 2 ) to denote that rv X have a
lognormal distribution with parameters µ and σ.
Lognormal Distribution
Definition
A nonnegative rv X is said to have a lognormal distribution if the
rv Y = ln(X ) has a normal distribution. The resulting pdf of a
lognormal rv when ln(X ) is normally distributed with parameters µ
and σ is
(
2
2
√ 1
e −[ln(x)−µ] /(2σ ) x ≤ 0
2πσx
f (x; µ, σ) =
0
x <0
Remark:
1. We use X ∼ LOGN(µ, σ 2 ) to denote that rv X have a
lognormal distribution with parameters µ and σ.
2. Notice here that the parameter µ is not the mean and σ 2 is not
the variance, i.e.
µ 6= E (X )
and
σ 2 6= V (X )
Lognormal Distribution
Lognormal Distribution
Lognormal Distribution
Lognormal Distribution
Proposition
If X ∼ LOGN(µ, σ 2 ), then
E (X ) = e µ+σ
2 /2
2
2
and V (X ) = e 2µ+σ · (e σ − 1)
The cdf of X is
F (x; µ, σ) = P(X ≤ x) = P[ln(X ) ≤ ln(x)]
ln(x) − µ
ln(x) − µ
=Φ
=P Z ≤
σ
σ
where Φ(z) is the cdf of the standard normal rv Z .
x ≤0
Lognormal Distribution
Lognormal Distribution
Example (Problem 115)
Let Ii be the input current to a transistor and I0 be the output
current. Then the current gain is proportional to ln(I0 /Ii ).
Suppose the constant of proportionality is 1 (which amounts to
choosing a particular unit of measurement), so that current gain =
X = ln(I0 /Ii ). Assume X is normally distributed with µ = 1 and
σ = 0.05.
Lognormal Distribution
Example (Problem 115)
Let Ii be the input current to a transistor and I0 be the output
current. Then the current gain is proportional to ln(I0 /Ii ).
Suppose the constant of proportionality is 1 (which amounts to
choosing a particular unit of measurement), so that current gain =
X = ln(I0 /Ii ). Assume X is normally distributed with µ = 1 and
σ = 0.05.
a. What is the probability that the output current is more than
twice the input current?
Lognormal Distribution
Example (Problem 115)
Let Ii be the input current to a transistor and I0 be the output
current. Then the current gain is proportional to ln(I0 /Ii ).
Suppose the constant of proportionality is 1 (which amounts to
choosing a particular unit of measurement), so that current gain =
X = ln(I0 /Ii ). Assume X is normally distributed with µ = 1 and
σ = 0.05.
a. What is the probability that the output current is more than
twice the input current?
b. What are the expected value and variance of the ratio of
output to input current?
Lognormal Distribution
Example (Problem 115)
Let Ii be the input current to a transistor and I0 be the output
current. Then the current gain is proportional to ln(I0 /Ii ).
Suppose the constant of proportionality is 1 (which amounts to
choosing a particular unit of measurement), so that current gain =
X = ln(I0 /Ii ). Assume X is normally distributed with µ = 1 and
σ = 0.05.
a. What is the probability that the output current is more than
twice the input current?
b. What are the expected value and variance of the ratio of
output to input current?
c. What value r is such that only 5% chance we will have the
ratio of output to input current exceed r ?
Beta Distribution
Beta Distribution
Definition
A random variable X is said to have a beta distribution with
parameters α, β(both positive), A, and B if the pdf of X is

α−1 β−1
 1
Γ(α+β)
x−A
B−x
·
·
A≤x ≤B
·
B−A
B−A
f (x; α, β, A, B) = B−A Γ(α)·Γ(β)
0
otherwise
The case A = 0, B = 1 gives the standard beta distribution.
Beta Distribution
Definition
A random variable X is said to have a beta distribution with
parameters α, β(both positive), A, and B if the pdf of X is

α−1 β−1
 1
Γ(α+β)
x−A
B−x
·
·
A≤x ≤B
·
B−A
B−A
f (x; α, β, A, B) = B−A Γ(α)·Γ(β)
0
otherwise
The case A = 0, B = 1 gives the standard beta distribution.
Remark: We use X ∼ BETA(α, β, A, B) to denote that rv X has a
beta distribution with parameters α, β, A, and B.
Beta Distribution
Beta Distribution
Proposition
If X ∼ BETA(α, β, A, B), then
E (X ) = A + (B − A) ·
α
(B − A)2 αβ
and V (X ) =
α+β
(α + β)2 (α + β + 1)
Beta Distribution
Beta Distribution
Beta Distribution
Beta Distribution
Example (Problem 127)
An individual’s credit score is a number calculated based on that
person’s credit history which helps a lender determine how much
he/she should be loaned or what credit limit should be established
for a credit card. An article in the Los Angeles Times gave data
which suggested that a beta distribution with parameters
A = 150, B = 850, α = 8, β = 2 would provide a reasonable
approximation to the distribution of American credit scores.
[Note: credit scores are integer-valued].
Beta Distribution
Example (Problem 127)
An individual’s credit score is a number calculated based on that
person’s credit history which helps a lender determine how much
he/she should be loaned or what credit limit should be established
for a credit card. An article in the Los Angeles Times gave data
which suggested that a beta distribution with parameters
A = 150, B = 850, α = 8, β = 2 would provide a reasonable
approximation to the distribution of American credit scores.
[Note: credit scores are integer-valued].
a. Let X represent a randomly selected American credit score.
What are the mean value and standard deviation of this
random variable? What is the probability that X is within 1
standard deviation of its mean value?
Beta Distribution
Example (Problem 127)
An individual’s credit score is a number calculated based on that
person’s credit history which helps a lender determine how much
he/she should be loaned or what credit limit should be established
for a credit card. An article in the Los Angeles Times gave data
which suggested that a beta distribution with parameters
A = 150, B = 850, α = 8, β = 2 would provide a reasonable
approximation to the distribution of American credit scores.
[Note: credit scores are integer-valued].
a. Let X represent a randomly selected American credit score.
What are the mean value and standard deviation of this
random variable? What is the probability that X is within 1
standard deviation of its mean value?
b. What is the approximate probability that a randomly selected
score will exceed 750 (which lenders consider a very good
score)?