beta n weibul distri
TRANSCRIPT
![Page 1: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/1.jpg)
Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is
f (x ;α, β) =
{αβα xα−1e−(x/β)α
x ≥ 0
0 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.
![Page 2: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/2.jpg)
Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is
f (x ;α, β) =
{αβα xα−1e−(x/β)α
x ≥ 0
0 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.
![Page 3: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/3.jpg)
Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is
f (x ;α, β) =
{αβα xα−1e−(x/β)α
x ≥ 0
0 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.
2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.
![Page 4: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/4.jpg)
Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters α and β (α > 0, β > 0) if the pdf of X is
f (x ;α, β) =
{αβα xα−1e−(x/β)α
x ≥ 0
0 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X ∼WEB(α, β) to denote that the rv X has a Weibulldistribution with parameters α and β.
![Page 5: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/5.jpg)
Weibull Distribution
Remark:3. When α = 1, the pdf becomes
f (x ;β) =
{1β e−x/β x ≥ 0
0 x < 0
which is the pdf for an exponential distribution with parameterλ = 1
β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.
![Page 6: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/6.jpg)
Weibull Distribution
Remark:
3. When α = 1, the pdf becomes
f (x ;β) =
{1β e−x/β x ≥ 0
0 x < 0
which is the pdf for an exponential distribution with parameterλ = 1
β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.
![Page 7: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/7.jpg)
Weibull Distribution
Remark:3. When α = 1, the pdf becomes
f (x ;β) =
{1β e−x/β x ≥ 0
0 x < 0
which is the pdf for an exponential distribution with parameterλ = 1
β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.
4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.
![Page 8: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/8.jpg)
Weibull Distribution
Remark:3. When α = 1, the pdf becomes
f (x ;β) =
{1β e−x/β x ≥ 0
0 x < 0
which is the pdf for an exponential distribution with parameterλ = 1
β . Thus we see that the exponential distribution is a specialcase of both the gamma and Weibull distributions.4. There are gamma distributions that are not Weibull distributiosand vice versa, so one family is not a subset of the other.
![Page 9: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/9.jpg)
Weibull Distribution
![Page 10: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/10.jpg)
Weibull Distribution
![Page 11: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/11.jpg)
Weibull Distribution
![Page 12: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/12.jpg)
Weibull Distribution
![Page 13: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/13.jpg)
Weibull Distribution
Proposition
Let X be a random variable such that X ∼WEI(α, β). Then
E (X ) = βΓ
(1 +
1
α
)and V (X ) = β2
{Γ
(1 +
2
α
)−[
Γ
(1 +
1
α
)]2}
The cdf of X is
F (x ;α, β) =
{1− e−(x/β)α
x ≥ 0
0 x < 0
![Page 14: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/14.jpg)
Weibull Distribution
Proposition
Let X be a random variable such that X ∼WEI(α, β). Then
E (X ) = βΓ
(1 +
1
α
)and V (X ) = β2
{Γ
(1 +
2
α
)−[
Γ
(1 +
1
α
)]2}
The cdf of X is
F (x ;α, β) =
{1− e−(x/β)α
x ≥ 0
0 x < 0
![Page 15: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/15.jpg)
Weibull Distribution
Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed 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.
![Page 16: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/16.jpg)
Weibull Distribution
Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed 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.
![Page 17: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/17.jpg)
Weibull Distribution
Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed 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.
![Page 18: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/18.jpg)
Weibull Distribution
Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed 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.
![Page 19: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/19.jpg)
Weibull Distribution
Example:The shear strength (in pounds) of a spot weld is a Weibulldistributed 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.
![Page 20: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/20.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 21: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/21.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.
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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 22: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/22.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 23: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/23.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 24: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/24.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 25: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/25.jpg)
Weibull Distribution
In practical situations, γ = min(X ) > 0 and X − γ has a Weibulldistribution.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 andthat the excess X − 3.5 over the minimum has a Weibulldistribution 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 returntime?
c. Compute P(X > 5).
d. Compute P(5 ≤ X ≤ 8).
![Page 26: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/26.jpg)
Lognormal Distribution
DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is
f (x ;µ, σ) =
{1√
2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0
0 x < 0
Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.
µ 6= E (X ) and σ2 6= V (X )
![Page 27: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/27.jpg)
Lognormal Distribution
DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is
f (x ;µ, σ) =
{1√
2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0
0 x < 0
Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.
µ 6= E (X ) and σ2 6= V (X )
![Page 28: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/28.jpg)
Lognormal Distribution
DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is
f (x ;µ, σ) =
{1√
2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0
0 x < 0
Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.
2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.
µ 6= E (X ) and σ2 6= V (X )
![Page 29: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/29.jpg)
Lognormal Distribution
DefinitionA nonnegative rv X is said to have a lognormal distribution if therv Y = ln(X ) has a normal distribution. The resulting pdf of alognormal rv when ln(X ) is normally distributed with parameters µand σ is
f (x ;µ, σ) =
{1√
2πσxe−[ln(x)−µ]2/(2σ2) x ≤ 0
0 x < 0
Remark:1. We use X ∼ LOGN(µ, σ2) to denote that rv X have alognormal distribution with parameters µ and σ.2. Notice here that the parameter µ is not the mean and σ2 is notthe variance, i.e.
µ 6= E (X ) and σ2 6= V (X )
![Page 30: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/30.jpg)
Lognormal Distribution
![Page 31: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/31.jpg)
Lognormal Distribution
![Page 32: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/32.jpg)
Lognormal Distribution
Proposition
If X ∼ LOGN(µ, σ2), then
E (X ) = eµ+σ2/2 and V (X ) = e2µ+σ2 · (eσ2 − 1)
The cdf of X is
F (x ;µ, σ) = P(X ≤ x) = P[ln(X ) ≤ ln(x)]
= P
(Z ≤ ln(x)− µ
σ
)= Φ
(ln(x)− µ
σ
)x ≤ 0
where Φ(z) is the cdf of the standard normal rv Z .
![Page 33: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/33.jpg)
Lognormal Distribution
Proposition
If X ∼ LOGN(µ, σ2), then
E (X ) = eµ+σ2/2 and V (X ) = e2µ+σ2 · (eσ2 − 1)
The cdf of X is
F (x ;µ, σ) = P(X ≤ x) = P[ln(X ) ≤ ln(x)]
= P
(Z ≤ ln(x)− µ
σ
)= Φ
(ln(x)− µ
σ
)x ≤ 0
where Φ(z) is the cdf of the standard normal rv Z .
![Page 34: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/34.jpg)
Lognormal Distribution
Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.
a. What is the probability that the output current is more thantwice the input current?
b. What are the expected value and variance of the ratio ofoutput to input current?
c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?
![Page 35: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/35.jpg)
Lognormal Distribution
Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.
a. What is the probability that the output current is more thantwice the input current?
b. What are the expected value and variance of the ratio ofoutput to input current?
c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?
![Page 36: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/36.jpg)
Lognormal Distribution
Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.
a. What is the probability that the output current is more thantwice the input current?
b. What are the expected value and variance of the ratio ofoutput to input current?
c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?
![Page 37: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/37.jpg)
Lognormal Distribution
Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.
a. What is the probability that the output current is more thantwice the input current?
b. What are the expected value and variance of the ratio ofoutput to input current?
c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?
![Page 38: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/38.jpg)
Lognormal Distribution
Example (Problem 115)Let Ii be the input current to a transistor and I0 be the outputcurrent. Then the current gain is proportional to ln(I0/Ii ).Suppose the constant of proportionality is 1 (which amounts tochoosing a particular unit of measurement), so that current gain =X = ln(I0/Ii ). Assume X is normally distributed with µ = 1 andσ = 0.05.
a. What is the probability that the output current is more thantwice the input current?
b. What are the expected value and variance of the ratio ofoutput to input current?
c. What value r is such that only 5% chance we will have theratio of output to input current exceed r?
![Page 39: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/39.jpg)
Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is
f (x ;α, β,A,B) =
1B−A ·
Γ(α+β)Γ(α)·Γ(β) ·
(x−AB−A
)α−1·(
B−xB−A
)β−1A ≤ x ≤ B
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 abeta distribution with parameters α, β, A, and B.
![Page 40: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/40.jpg)
Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is
f (x ;α, β,A,B) =
1B−A ·
Γ(α+β)Γ(α)·Γ(β) ·
(x−AB−A
)α−1·(
B−xB−A
)β−1A ≤ x ≤ B
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 abeta distribution with parameters α, β, A, and B.
![Page 41: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/41.jpg)
Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters α, β(both positive), A, and B if the pdf of X is
f (x ;α, β,A,B) =
1B−A ·
Γ(α+β)Γ(α)·Γ(β) ·
(x−AB−A
)α−1·(
B−xB−A
)β−1A ≤ x ≤ B
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 abeta distribution with parameters α, β, A, and B.
![Page 42: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/42.jpg)
Beta Distribution
Proposition
If X ∼ BETA(α, β,A,B), then
E (X ) = A + (B − A) · α
α + βand V (X ) =
(B − A)2αβ
(α + β)2(α + β + 1)
![Page 43: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/43.jpg)
Beta Distribution
Proposition
If X ∼ BETA(α, β,A,B), then
E (X ) = A + (B − A) · α
α + βand V (X ) =
(B − A)2αβ
(α + β)2(α + β + 1)
![Page 44: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/44.jpg)
Beta Distribution
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Beta Distribution
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Beta Distribution
Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation 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 thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?
b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?
![Page 47: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/47.jpg)
Beta Distribution
Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation 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 thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?
b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?
![Page 48: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/48.jpg)
Beta Distribution
Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation 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 thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?
b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?
![Page 49: Beta n Weibul Distri](https://reader033.vdocuments.mx/reader033/viewer/2022050707/552ed2b84a795906588b4a7f/html5/thumbnails/49.jpg)
Beta Distribution
Example (Problem 127)An individual’s credit score is a number calculated based on thatperson’s credit history which helps a lender determine how muchhe/she should be loaned or what credit limit should be establishedfor a credit card. An article in the Los Angeles Times gave datawhich suggested that a beta distribution with parametersA = 150,B = 850, α = 8, β = 2 would provide a reasonableapproximation 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 thisrandom variable? What is the probability that X is within 1standard deviation of its mean value?
b. What is the approximate probability that a randomly selectedscore will exceed 750 (which lenders consider a very goodscore)?