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Normal Distribution
Proposition
If X has a normal distribution with mean and stadard deviation, then
Z =X
has a standard normal distribution. Thus
P(a X b) = P(a Z b
)
= (b
) (a
)
P(X a) = (a
) P(X b) = 1 (b
)

Normal Distribution
Proposition
If X has a normal distribution with mean and stadard deviation, then
Z =X
has a standard normal distribution. Thus
P(a X b) = P(a Z b
)
= (b
) (a
)
P(X a) = (a
) P(X b) = 1 (b
)

Normal Distribution
Example (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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?
P(2.9 X1 3.1) = P(2.9 3
0.1 Z 3.1 3
0.1)
= P(1 Z 1) = 0.8413 0.1587 = 0.6826
P(2.9 X2 3.1) = P(2.9 3.04
0.02 Z 3.1 3.04
0.02)
= P(7 Z 3) = 0.9987 0 = 0.9987

Normal DistributionExample (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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?
P(2.9 X1 3.1) = P(2.9 3
0.1 Z 3.1 3
0.1)
= P(1 Z 1) = 0.8413 0.1587 = 0.6826
P(2.9 X2 3.1) = P(2.9 3.04
0.02 Z 3.1 3.04
0.02)
= P(7 Z 3) = 0.9987 0 = 0.9987

Normal DistributionExample (Problem 38):There are two machines available for cutting corks intended for usein wine bottles. The first produces corks with diameters that arenormally distributed with mean 3cm and standard deviation 0.1cm.The second produces corks with diameters that have a normaldistribution 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?
P(2.9 X1 3.1) = P(2.9 3
0.1 Z 3.1 3
0.1)
= P(1 Z 1) = 0.8413 0.1587 = 0.6826
P(2.9 X2 3.1) = P(2.9 3.04
0.02 Z 3.1 3.04
0.02)
= P(7 Z 3) = 0.9987 0 = 0.9987

Normal Distribution
Example (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?
P( 1.5 X1 + 1.5) = P( 1.5
Z + 1.5
)
= P(1.5 Z 1.5)= 0.9332 0.0668 = 0.8664
2 P(+ X1 + 2) = 2P(+
Z + 2
)
= 2P(1 Z 2)= 2(0.9772 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?
P( 1.5 X1 + 1.5) = P( 1.5
Z + 1.5
)
= P(1.5 Z 1.5)= 0.9332 0.0668 = 0.8664
2 P(+ X1 + 2) = 2P(+
Z + 2
)
= 2P(1 Z 2)= 2(0.9772 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?
P( 1.5 X1 + 1.5) = P( 1.5
Z + 1.5
)
= P(1.5 Z 1.5)= 0.9332 0.0668 = 0.8664
2 P(+ X1 + 2) = 2P(+
Z + 2
)
= 2P(1 Z 2)= 2(0.9772 0.8413) = 0.0.2718

Normal DistributionExample (Problem 44):If bolt thread length is normally distributed, what is the probabilitythat the thread length of a randomly selected bolt is (a)within 1.5SDs of its mean value? (b)between 1 and 2 SDs from its meanvalue?
P( 1.5 X1 + 1.5) = P( 1.5
Z + 1.5
)
= P(1.5 Z 1.5)= 0.9332 0.0668 = 0.8664
2 P(+ X1 + 2) = 2P(+
Z + 2
)
= 2P(1 Z 2)= 2(0.9772 0.8413) = 0.0.2718

Normal Distribution
Proposition
{(100p)th percentile for N(, 2)} = + {(100p)th percentile for N(0, 1)}
Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 m and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe other 90%?
N(3,0.1402)(90) = 3.0+0.140N(0,1)(90) = 3.0+0.1401.28 = 3.1792

Normal Distribution
Proposition
{(100p)th percentile for N(, 2)} = + {(100p)th percentile for N(0, 1)}
Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 m and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe other 90%?
N(3,0.1402)(90) = 3.0+0.140N(0,1)(90) = 3.0+0.1401.28 = 3.1792

Normal Distribution
Proposition
{(100p)th percentile for N(, 2)} = + {(100p)th percentile for N(0, 1)}
Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 m and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe other 90%?
N(3,0.1402)(90) = 3.0+0.140N(0,1)(90) = 3.0+0.1401.28 = 3.1792

Normal Distribution
Proposition
{(100p)th percentile for N(, 2)} = + {(100p)th percentile for N(0, 1)}
Example (Problem 39)The width of a line etched on an integrated circuit chip is normallydistributed with mean 3.000 m and standard deviation 0.140.What width value separates the widest 10% of all such lines fromthe other 90%?
N(3,0.1402)(90) = 3.0+0.140N(0,1)(90) = 3.0+0.1401.28 = 3.1792

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 hasapproximately a normal distribution with = np and =
npq,
where q = 1 p. In particular, for x = a posible value of X ,
P(X x) = B(x ; n, p) (
area under the normal curve
to the left of x+0.5
)= (
x+0.5 np
npq)
In practice, the approximation is adequate provided that bothnp 10 and nq 10, since there is then enough symmetry in theunderlying binomial 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 hasapproximately a normal distribution with = np and =
npq,
where q = 1 p. In particular, for x = a posible value of X ,
P(X x) = B(x ; n, p) (
area under the normal curve
to the left of x+0.5
)= (
x+0.5 np
npq)
In practice, the approximation is adequate provided that bothnp 10 and nq 10, since there is then enough symmetry in theunderlying binomial distribution.

Normal Distribution
A graphical explanation for
P(X x) = B(x ; n, p) (
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) (
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) (
area under the normal curve
to the left of x+0.5
)= (
x+0.5 np
npq)

Normal Distribution
Example (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10
P(15 X 25) = Bin(25; 200, 0.1) Bin(14; 200, 0.1)
( 25 + 0.5 20200 0.1 0.9
) ( 15 + 0.5 20200 0.1 0.9
)
= (0.3056) (0.2500)= 0.6217 0.4013= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?
In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10
P(15 X 25) = Bin(25; 200, 0.1) Bin(14; 200, 0.1)
( 25 + 0.5 20200 0.1 0.9
) ( 15 + 0.5 20200 0.1 0.9
)
= (0.3056) (0.2500)= 0.6217 0.4013= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10
P(15 X 25) = Bin(25; 200, 0.1) Bin(14; 200, 0.1)
( 25 + 0.5 20200 0.1 0.9
) ( 15 + 0.5 20200 0.1 0.9
)
= (0.3056) (0.2500)= 0.6217 0.4013= 0.2204

Normal DistributionExample (Problem 54)Suppose that 10% of all steel shafts produced by a certain processare nonconforming but can be reworked (rather than having to bescrapped). Consider a random sample of 200 shafts, and let Xdenote the number among these that are nonconforming and canbe reworked. What is the (approximate) probability that X isbetween 15 and 25 (inclusive)?In this problem n = 200, p = 0.1 and q = 1 p = 0.9. Thusnp = 20 > 10 and nq = 180 > 10
P(15 X 25) = Bin(25; 200, 0.1) Bin(14; 200, 0.1)
( 25 + 0.5 20200 0.1 0.9
) ( 15 + 0.5 20200 0.1 0.9
)
= (0.3056) (0.2500)= 0.6217 0.4013= 0.2204

Exponential Distribution
DefinitionX is said to have an exponential distribution with parameter( > 0) if the pdf of X is
f (x ;) =
{ex x 00 otherwise
Remark:1. Usually we use X EXP() to denote that the random variableX has an exponential distribution with parameter .2. In some sources, the pdf of exponential distribution is given by
f (x ; ) =
{1 e x x 0
0 otherwise
The difference is that 1 .

Exponential Distribution
DefinitionX is said to have an exponential distribution with parameter( > 0) if the pdf of X is
f (x ;) =
{ex x 00 otherwise
Remark:1. Usually we use X EXP() to denote that the random variableX has an exponential distribution with parameter .2. In some sources, the pdf of exponential distribution is given by
f (x ; ) =
{1 e x x 0
0 otherwise
The difference is that 1 .

Exponential Distribution
DefinitionX is said to have an exponential distribution with parameter( > 0) if the pdf of X is
f (x ;) =
{ex x 00 otherwise
Remark:1. Usually we use X EXP() to denote that the random variableX has an exponential distribution with parameter .
2. In some sources, the pdf of exponential distribution is given by
f (x ; ) =
{1 e x x 0
0 otherwise
The difference is that 1 .

Exponential Distribution
DefinitionX is said to have an exponential distribution with parameter( > 0) if the pdf of X is
f (x ;) =
{ex x 00 otherwise
Remark:1. Usually we use X EXP() to denote that the random variableX has an exponential distribution with parameter .2. In some sources, the pdf of exponential distribution is given by
f (x ; ) =
{1 e x x 0
0 otherwise
The difference is that 1 .

Exponential Distribution

Exponential Distribution

Exponential Distribution
Proposition
If X EXP(), then
E (X ) =1
and V (X ) =
1
2
And the cdf for X is
F (x ;) =
{1 ex x 00 x < 0

Exponential Distribution
Proposition
If X EXP(), then
E (X ) =1
and V (X ) =
1
2
And the cdf for X is
F (x ;) =
{1 ex x 00 x < 0

Exponential Distribution
Proof:
E (X ) =
0
xexdx
=1
0
(x)exd(x)
=1
0
yey dy y = x
=1
[yey 0 +
0
ey dy ] integration by parts:u = y , v = ey
=1
[0 + (ey 0 )]
=1

Exponential Distribution
Proof:
E (X ) =
0
xexdx
=1
0
(x)exd(x)
=1
0
yey dy y = x
=1
[yey 0 +
0
ey dy ] integration by parts:u = y , v = ey
=1
[0 + (ey 0 )]
=1

Exponential Distribution
Proof (continued):
E (X 2) =
0
x2exdx
=1
2
0
(x)2exd(x)
=1
2
0
y 2ey dy y = x
=1
2[y 2ey 0 +
0
2yey dy ] integration by parts
=1
2[0 + 2(yey 0 +
0
ey dy)] integration by parts
=1
22[0 + (yey 0 )]
=2
2

Exponential Distribution
Proof (continued):
E (X 2) =
0
x2exdx
=1
2
0
(x)2exd(x)
=1
2
0
y 2ey dy y = x
=1
2[y 2ey 0 +
0
2yey dy ] integration by parts
=1
2[0 + 2(yey 0 +
0
ey dy)] integration by parts
=1
22[0 + (yey 0 )]
=2
2

Exponential Distribution
Proof (continued):
V (X ) = E (X 2) [E (X )]2 = 22 ( 1
)2 =
1
2
F (x) =
x0ey dy
=
x0
ey d(y)
=
x0
ezdz z = y
= ez x0= 1 ex

Exponential Distribution
Proof (continued):
V (X ) = E (X 2) [E (X )]2 = 22 ( 1
)2 =
1
2
F (x) =
x0ey dy
=
x0
ey d(y)
=
x0
ezdz z = y
= ez x0= 1 ex

Exponential Distribution
Example (Problem 108)The article Determination of the MTF of Positive PhotoresistsUsing the Monte Carlo method (Photographic Sci. andEngr., 1983: 254260) proposes the exponential distributionwith parameter = 0.93 as a model for the distribution of aphotons free path length (m) under certain circumstances.Suppose this is the correct model.
a. What is the expected path length, and what is the standarddeviation 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
Example (Problem 108)The article Determination of the MTF of Positive PhotoresistsUsing the Monte Carlo method (Photographic Sci. andEngr., 1983: 254260) proposes the exponential distributionwith parameter = 0.93 as a model for the distribution of aphotons free path length (m) under certain circumstances.Suppose this is the correct model.
a. What is the expected path length, and what is the standarddeviation 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
Example (Problem 108)The article Determination of the MTF of Positive PhotoresistsUsing the Monte Carlo method (Photographic Sci. andEngr., 1983: 254260) proposes the exponential distributionwith parameter = 0.93 as a model for the distribution of aphotons free path length (m) under certain circumstances.Suppose this is the correct model.
a. What is the expected path length, and what is the standarddeviation 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
Example (Problem 108)The article Determination of the MTF of Positive PhotoresistsUsing the Monte Carlo method (Photographic Sci. andEngr., 1983: 254260) proposes the exponential distributionwith parameter = 0.93 as a model for the distribution of aphotons free path length (m) under certain circumstances.Suppose this is the correct model.
a. What is the expected path length, and what is the standarddeviation 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
Example (Problem 108)The article Determination of the MTF of Positive PhotoresistsUsing the Monte Carlo method (Photographic Sci. andEngr., 1983: 254260) proposes the exponential distributionwith parameter = 0.93 as a model for the distribution of aphotons free path length (m) under certain circumstances.Suppose this is the correct model.
a. What is the expected path length, and what is the standarddeviation 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
Proposition
Suppose that the number of events occurring in any time intervalof length t has a Poisson distribution with parameter t (where ,the rate of the event process, is the expected number of eventsoccurring in 1 unit of time) and that numbers of occurrences innonoverlappong intervals are independent of one another. Thenthe distribution of elapsed time between the occurrence of twosuccessive events is exponential with parameter = .
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
Proposition
Suppose that the number of events occurring in any time intervalof length t has a Poisson distribution with parameter t (where ,the rate of the event process, is the expected number of eventsoccurring in 1 unit of time) and that numbers of occurrences innonoverlappong intervals are independent of one another. Thenthe distribution of elapsed time between the occurrence of twosuccessive events is exponential with parameter = .
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
Proposition
Suppose that the number of events occurring in any time intervalof length t has a Poisson distribution with parameter t (where ,the rate of the event process, is the expected number of eventsoccurring in 1 unit of time) and that numbers of occurrences innonoverlappong intervals are independent of one another. Thenthe distribution of elapsed time between the occurrence of twosuccessive events is exponential with parameter = .
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
Proposition
Suppose that the number of events occurring in any time intervalof length t has a Poisson distribution with parameter t (where ,the rate of the event process, is the expected number of eventsoccurring in 1 unit of time) and that numbers of occurrences innonoverlappong intervals are independent of one another. Thenthe distribution of elapsed time between the occurrence of twosuccessive events is exponential with parameter = .
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
Example (Example 4.22)Suppose that calls are received at a 24hour hotline according to aPoisson process with rate = 0.5 call per day.Then the number of days X between successive calls has anexponential 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) = e0.53 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution
Example (Example 4.22)Suppose that calls are received at a 24hour hotline according to aPoisson process with rate = 0.5 call per day.
Then the number of days X between successive calls has anexponential 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) = e0.53 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution
Example (Example 4.22)Suppose that calls are received at a 24hour hotline according to aPoisson process with rate = 0.5 call per day.Then the number of days X between successive calls has anexponential 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) = e0.53 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution
Example (Example 4.22)Suppose that calls are received at a 24hour hotline according to aPoisson process with rate = 0.5 call per day.Then the number of days X between successive calls has anexponential 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) = e0.53 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution
Example (Example 4.22)Suppose that calls are received at a 24hour hotline according to aPoisson process with rate = 0.5 call per day.Then the number of days X between successive calls has anexponential 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) = e0.53 = 0.223.
The expected time between successive calls is 1/0.5 = 2 days.

Exponential Distribution
Memoryless PropertyLet X = the time certain component lasts (in hours)and we assume the component lifetime is exponentially distributedwith parameter . Then what is the probability that thecomponent can last at least an additional t hours after working fort0 hours, i.e. what is P(X t + t0  X t0)?
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;)
F (t0;)
= et

Exponential Distribution
Memoryless PropertyLet X = the time certain component lasts (in hours)and we assume the component lifetime is exponentially distributedwith parameter . Then what is the probability that thecomponent can last at least an additional t hours after working fort0 hours, i.e. what is P(X t + t0  X t0)?
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;)
F (t0;)
= et

Exponential Distribution
Memoryless PropertyLet X = the time certain component lasts (in hours)and we assume the component lifetime is exponentially distributedwith parameter . Then what is the probability that thecomponent can last at least an additional t hours after working fort0 hours, i.e. what is P(X t + t0  X t0)?
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;)
F (t0;)
= et

Exponential Distribution
Memoryless PropertyHowever, we have
P(X t) = 1 F (t;) = et
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 sameas the original distribution of lifetime, so at each point in time thecomponent shows no effect of wear. In other words, thedistribution of remaining lifetime is independent of current age.

Exponential Distribution
Memoryless PropertyHowever, we have
P(X t) = 1 F (t;) = et
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 sameas the original distribution of lifetime, so at each point in time thecomponent shows no effect of wear. In other words, thedistribution of remaining lifetime is independent of current age.

Exponential Distribution
Memoryless PropertyHowever, we have
P(X t) = 1 F (t;) = et
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 sameas the original distribution of lifetime, so at each point in time thecomponent shows no effect of wear. In other words, thedistribution of remaining lifetime is independent of current age.

Exponential Distribution
Memoryless PropertyHowever, we have
P(X t) = 1 F (t;) = et
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 sameas the original distribution of lifetime, so at each point in time thecomponent shows no effect of wear. In other words, thedistribution of remaining lifetime is independent of current age.

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];2. For any positive integer, n, (n) = (n 1)!;3. ( 12 ) =
.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];2. For any positive integer, n, (n) = (n 1)!;3. ( 12 ) =
.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];
2. For any positive integer, n, (n) = (n 1)!;3. ( 12 ) =
.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];2. For any positive integer, n, (n) = (n 1)!;
3. ( 12 ) =.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];2. For any positive integer, n, (n) = (n 1)!;3. ( 12 ) =
.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionFor > 0, the gamma function () is defined by
() =
0
x1exdx
Properties for gamma function:1. For any > 1, () = ( 1) ( 1) [via integration byparts];2. For any positive integer, n, (n) = (n 1)!;3. ( 12 ) =
.
e.g. (4) = (4 1)! = 6 and ( 52 ) =32 (
32 ) =
32 [
12 (
12 )] =
34

Gamma Distribution
DefinitionA continuous random variable X is said to have a gammadistribution if the pdf of X is
f (x ;, ) =
{1
() x1ex/ x 0
0 otherwise
where the parameters and satisfy > 0, > 0. The standardgamma distribution has = 1, so the pdf of a standard gammarv is
f (x ;) =
{1
() x1ex x 0
0 otherwise

Gamma Distribution
DefinitionA continuous random variable X is said to have a gammadistribution if the pdf of X is
f (x ;, ) =
{1
() x1ex/ x 0
0 otherwise
where the parameters and satisfy > 0, > 0. The standardgamma distribution has = 1, so the pdf of a standard gammarv is
f (x ;) =
{1
() x1ex x 0
0 otherwise

Gamma Distribution
Remark:1. We use X GAM(, ) to denote that the rv X has a gammadistribution with parameter and .2. If we let = 1 and = 1/, then we get the exponentialdistribution:
f (x ; 1,1
) =
{1
1
(1)x11ex/
1 = ex x 0
0 otherwise
3. When X is a standard gamma rv ( = 1), the cdf of X ,
F (x ;) =
x0
y1ey
()dy
is called the incomplete gamma function.There are extensive tables of F (x ;) available (Appendix TableA.4).

Gamma Distribution
Remark:1. We use X GAM(, ) to denote that the rv X has a gammadistribution with parameter and .
2. If we let = 1 and = 1/, then we get the exponentialdistribution:
f (x ; 1,1
) =
{1
1
(1)x11ex/
1 = ex x 0
0 otherwise
3. When X is a standard gamma rv ( = 1), the cdf of X ,
F (x ;) =
x0
y1ey
()dy
is called the incomplete gamma function.There are extensive tables of F (x ;) available (Appendix TableA.4).

Gamma Distribution
Remark:1. We use X GAM(, ) to denote that the rv X has a gammadistribution with parameter and .2. If we let = 1 and = 1/, then we get the exponentialdistribution:
f (x ; 1,1
) =
{1
1
(1)x11ex/
1 = ex x 0
0 otherwise
3. When X is a standard gamma rv ( = 1), the cdf of X ,
F (x ;) =
x0
y1ey
()dy
is called the incomplete gamma function.There are extensive tables of F (x ;) available (Appendix TableA.4).

Gamma Distribution
Remark:1. We use X GAM(, ) to denote that the rv X has a gammadistribution with parameter and .2. If we let = 1 and = 1/, then we get the exponentialdistribution:
f (x ; 1,1
) =
{1
1
(1)x11ex/
1 = ex x 0
0 otherwise
3. When X is a standard gamma rv ( = 1), the cdf of X ,
F (x ;) =
x0
y1ey
()dy
is called the incomplete gamma function.
There are extensive tables of F (x ;) available (Appendix TableA.4).

Gamma Distribution
Remark:1. We use X GAM(, ) to denote that the rv X has a gammadistribution with parameter and .2. If we let = 1 and = 1/, then we get the exponentialdistribution:
f (x ; 1,1
) =
{1
1
(1)x11ex/
1 = ex x 0
0 otherwise
3. When X is a standard gamma rv ( = 1), the cdf of X ,
F (x ;) =
x0
y1ey
()dy
is called the incomplete gamma function.There are extensive tables of F (x ;) available (Appendix TableA.4).

Gamma Distribution

Gamma Distribution

Gamma Distribution
Proposition
If X GAM(, ), then
E (X ) = and V (X ) = 2
Furthermore, for any x > 0, the cdf of X is given by
P(X x) = F (x ;, ) = F(
x
;
)where F (;) is the incomplete gamma function.

Gamma Distribution
Proposition
If X GAM(, ), then
E (X ) = and V (X ) = 2
Furthermore, for any x > 0, the cdf of X is given by
P(X x) = F (x ;, ) = F(
x
;
)where F (;) is the incomplete gamma function.

Gamma Distribution
Example:The survival time (in days) of a white rat that was subjected to acertain level of Xray radiation is a random variableX 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 and20 days (not inclusive);
c. the expected survival time.

Gamma Distribution
Example:The survival time (in days) of a white rat that was subjected to acertain level of Xray radiation is a random variableX 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 and20 days (not inclusive);
c. the expected survival time.

Gamma Distribution
Example:The survival time (in days) of a white rat that was subjected to acertain level of Xray radiation is a random variableX 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 and20 days (not inclusive);
c. the expected survival time.

Gamma Distribution
Example:The survival time (in days) of a white rat that was subjected to acertain level of Xray radiation is a random variableX 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 and20 days (not inclusive);
c. the expected survival time.

Gamma Distribution
Example:The survival time (in days) of a white rat that was subjected to acertain level of Xray radiation is a random variableX 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 and20 days (not inclusive);
c. the expected survival time.

ChiSquared Distribution
DefinitionLet be a positive integer. Then a random variable X is said tohave a chisquared distribution with parameter if the pdf of Xis the gamma density with = /2 and = 2. The pdf of achisquared rv is thus
f (x ; ) =
{1
2/2(/2)x (/2)1ex/2 x 0
0 x < 0
The parameter is called the number of degrees of freedom(df) of X . The symbol 2 is often used in place of chisquared.

ChiSquared Distribution
DefinitionLet be a positive integer. Then a random variable X is said tohave a chisquared distribution with parameter if the pdf of Xis the gamma density with = /2 and = 2. The pdf of achisquared rv is thus
f (x ; ) =
{1
2/2(/2)x (/2)1ex/2 x 0
0 x < 0
The parameter is called the number of degrees of freedom(df) of X . The symbol 2 is often used in place of chisquared.

ChiSquared Distribution
Remark:1. Usually, we use X 2() to denote that X is a chisquared rvwith parameter ;2. If X1,X2, . . . ,Xn is n independent standard normal rvs, thenX 21 + X
22 + + X 2n has the same distribution as 2(n).

ChiSquared Distribution
Remark:1. Usually, we use X 2() to denote that X is a chisquared rvwith parameter ;
2. If X1,X2, . . . ,Xn is n independent standard normal rvs, thenX 21 + X
22 + + X 2n has the same distribution as 2(n).

ChiSquared Distribution
Remark:1. Usually, we use X 2() to denote that X is a chisquared rvwith parameter ;2. If X1,X2, . . . ,Xn is n independent standard normal rvs, thenX 21 + X
22 + + X 2n has the same distribution as 2(n).

ChiSquared Distribution

ChiSquared Distribution

Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters and ( > 0, > 0) if the pdf of X is
f (x ;, ) =
{ x
1e(x/)
x 00 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X WEB(, ) to denote that the rv X has a Weibulldistribution with parameters and .

Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters and ( > 0, > 0) if the pdf of X is
f (x ;, ) =
{ x
1e(x/)
x 00 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X WEB(, ) to denote that the rv X has a Weibulldistribution with parameters and .

Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters and ( > 0, > 0) if the pdf of X is
f (x ;, ) =
{ x
1e(x/)
x 00 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.
2. We use X WEB(, ) to denote that the rv X has a Weibulldistribution with parameters and .

Weibull Distribution
DefinitionA random variable X is said to have a Weibull distribution withparameters and ( > 0, > 0) if the pdf of X is
f (x ;, ) =
{ x
1e(x/)
x 00 x < 0
Remark:1. The family of Weibull distributions was introduced by theSwedish physicist Waloddi Weibull in 1939.2. We use X WEB(, ) to denote that the rv X has a Weibulldistribution with parameters and .

Weibull Distribution
Remark:3. When = 1, the pdf becomes
f (x ;) =
{1 ex/ x 0
0 x < 0
which is the pdf for an exponential distribution with parameter = 1 . 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.

Weibull Distribution
Remark:
3. When = 1, the pdf becomes
f (x ;) =
{1 ex/ x 0
0 x < 0
which is the pdf for an exponential distribution with parameter = 1 . 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.

Weibull Distribution
Remark:3. When = 1, the pdf becomes
f (x ;) =
{1 ex/ x 0
0 x < 0
which is the pdf for an exponential distribution with parameter = 1 . 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.

Weibull Distribution
Remark:3. When = 1, the pdf becomes
f (x ;) =
{1 ex/ x 0
0 x < 0
which is the pdf for an exponential distribution with parameter = 1 . 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.

Weibull Distribution

Weibull Distribution

Weibull Distribution

Weibull Distribution

Weibull Distribution
Proposition
Let X be a random variable such that X WEI(, ). Then
E (X ) =
(1 +
1
)and V (X ) = 2
{
(1 +
2
)[
(1 +
1
)]2}
The cdf of X is
F (x ;, ) =
{1 e(x/) x 00 x < 0

Weibull Distribution
Proposition
Let X be a random variable such that X WEI(, ). Then
E (X ) =
(1 +
1
)and V (X ) = 2
{
(1 +
2
)[
(1 +
1
)]2}
The cdf of X is
F (x ;, ) =
{1 e(x/) x 00 x < 0

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.

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.

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.

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.

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.

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.
Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

Weibull Distribution
In practical situations, = min(X ) > 0 and X has a Weibulldistribution.Example (Problem 74):Let X = the time (in 101 weeks) from shipment of adefective product until the customer returns the
product. Suppose that the minimum return time is = 3.5 andthat the excess X 3.5 over the minimum has a Weibulldistribution with parameters = 2 and = 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).

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 and is
f (x ;, ) =
{1
2xe[ln(x)]
2/(22) x 00 x < 0
Remark:1. We use X LOGN(, 2) to denote that rv X have alognormal distribution with parameters and .2. Notice here that the parameter is not the mean and 2 is notthe variance, i.e.
6= E (X ) and 2 6= V (X )

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 and is
f (x ;, ) =
{1
2xe[ln(x)]
2/(22) x 00 x < 0
Remark:1. We use X LOGN(, 2) to denote that rv X have alognormal distribution with parameters and .2. Notice here that the parameter is not the mean and 2 is notthe variance, i.e.
6= E (X ) and 2 6= V (X )

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 and is
f (x ;, ) =
{1
2xe[ln(x)]
2/(22) x 00 x < 0
Remark:1. We use X LOGN(, 2) to denote that rv X have alognormal distribution with parameters and .
2. Notice here that the parameter is not the mean and 2 is notthe variance, i.e.
6= E (X ) and 2 6= V (X )

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 and is
f (x ;, ) =
{1
2xe[ln(x)]
2/(22) x 00 x < 0
Remark:1. We use X LOGN(, 2) to denote that rv X have alognormal distribution with parameters and .2. Notice here that the parameter is not the mean and 2 is notthe variance, i.e.
6= E (X ) and 2 6= V (X )

Lognormal Distribution

Lognormal Distribution

Lognormal Distribution
Proposition
If X LOGN(, 2), then
E (X ) = e+2/2 and V (X ) = e2+
2 (e2 1)
The cdf of X is
F (x ;, ) = P(X x) = P[ln(X ) ln(x)]
= P
(Z ln(x)
)=
(ln(x)
)x 0
where (z) is the cdf of the standard normal rv Z .

Lognormal Distribution
Proposition
If X LOGN(, 2), then
E (X ) = e+2/2 and V (X ) = e2+
2 (e2 1)
The cdf of X is
F (x ;, ) = P(X x) = P[ln(X ) ln(x)]
= P
(Z ln(x)
)=
(ln(x)
)x 0
where (z) is the cdf of the standard normal rv Z .

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 = 1 and = 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?

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 = 1 and = 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?

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 = 1 and = 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?

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 = 1 and = 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?

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 = 1 and = 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?

Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters , (both positive), A, and B if the pdf of X is
f (x ;, ,A,B) =
1BA (+)()() (
xABA
)1(
BxBA
)1A x B
0 otherwise
The case A = 0,B = 1 gives the standard beta distribution.
Remark: We use X BETA(, ,A,B) to denote that rv X has abeta distribution with parameters , , A, and B.

Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters , (both positive), A, and B if the pdf of X is
f (x ;, ,A,B) =
1BA (+)()() (
xABA
)1(
BxBA
)1A x B
0 otherwise
The case A = 0,B = 1 gives the standard beta distribution.
Remark: We use X BETA(, ,A,B) to denote that rv X has abeta distribution with parameters , , A, and B.

Beta Distribution
DefinitionA random variable X is said to have a beta distribution withparameters , (both positive), A, and B if the pdf of X is
f (x ;, ,A,B) =
1BA (+)()() (
xABA
)1(
BxBA
)1A x B
0 otherwise
The case A = 0,B = 1 gives the standard beta distribution.
Remark: We use X BETA(, ,A,B) to denote that rv X has abeta distribution with parameters , , A, and B.

Beta Distribution
Proposition
If X BETA(, ,A,B), then
E (X ) = A + (B A) +
and V (X ) =(B A)2
( + )2( + + 1)

Beta Distribution
Proposition
If X BETA(, ,A,B), then
E (X ) = A + (B A) +
and V (X ) =(B A)2
( + )2( + + 1)

Beta Distribution

Beta Distribution

Beta Distribution
Example (Problem 127)An individuals credit score is a number calculated based on thatpersons 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, = 8, = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integervalued].
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)?

Beta Distribution
Example (Problem 127)An individuals credit score is a number calculated based on thatpersons 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, = 8, = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integervalued].
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)?

Beta Distribution
Example (Problem 127)An individuals credit score is a number calculated based on thatpersons 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, = 8, = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integervalued].
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)?

Beta Distribution
Example (Problem 127)An individuals credit score is a number calculated based on thatpersons 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, = 8, = 2 would provide a reasonableapproximation to the distribution of American credit scores.[Note: credit scores are integervalued].
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)?