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Chapter 4. Continuous Probability DistributionsPowerPoint Presentation

Chapter 4. Continuous Probability Distributions

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Chapter 4. Continuous Probability Distributions. 4.1 The Uniform Distribution 4.2 The Exponential Distribution 4.3 The Gamma Distribution 4.4 The Weibull Distribution 4.5 The Beta Distribution. 4.1 The Uniform Distribution 4.1.1 Definition of the Uniform Distribution(1/2).

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### Chapter 4. Continuous Probability Distributions

4.1 The Uniform Distribution

4.2 The Exponential Distribution

4.3 The Gamma Distribution

4.4 The Weibull Distribution

4.5 The Beta Distribution

4.1 The Uniform Distribution4.1.1 Definition of the Uniform Distribution(1/2)

- It has a flat pdf over a region.
- if X takes on values between a and b,
and

- Mean and Variance

- if X takes on values between a and b,

Figure 4.1 Probability density function of aU(a, b) distribution

4.2 The Exponential Distribution4.2.1 Definition of the Exponential Distribution

- Pdf:
- Cdf:
- Mean and variance:

Figure 4.3Probability density function of an exponential distribution with parameter l= 1

- The exponential distribution often arises, in practice, as being of the amount of time until some specific event occurs.
For example,

the amount of time until an earthquake occurs,

the amount of time until a new war breaks out, or

the amount of time until a telephone call you receive turns out to be a wrong number, etc.

4.2.2 The memoryless property of the Exponential Distribution

- For any non-negative x and y
- The exponential distribution is the only continuous distribution that has the memoryless property

- Memoryless property of exponential random variable: Distribution
This is equivalent to

When X is an exponential random variable,

The memoryless condition is satisfied since

- Example: Suppose that a number of miles that a car run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5,000 mile trip, what is the probability that he will be able to complete his trip without to replace the battery?
(Sol)

Let X be a random variable including the remaining lifetime (in thousand miles) of the battery. Then,

What if X is not exponential random variable?

That is, an additional information of t should be known.

- Proposition: If are independent exponential random variables having respective parameters , then is the exponential random variable with
parameter

(proof)

- Example: A series system is one that all of its components to function in order for the system itself to be functional. For an n component series system in which the component lifetimes are independent exponential random variables with respective parameters .
What is the probability the system serves for a time t?

(Sol)

Let X be a random variable indicating the system lifetime.

Then, X is an exponential random variable with parameter

Hence,

4.2.3 The Poisson process to function in order for the system itself to be functional. For an n component series system in which the component lifetimes are independent exponential random variables with respective parameters .

- A stochastic process is a sequence of random events
- A Poisson process with parameter is a stochastic process
where the time (or space) intervals between event-occurrences follow the Exponential distribution with parameter .

- If X is the number of events occurring within a fixed time (or space) interval of length t, then

Figure 4.7 to function in order for the system itself to be functional. For an n component series system in which the component lifetimes are independent exponential random variables with respective parameters .A Poisson process. The number of events occurring in a time interval of length t has a Poisson distribution with mean lt

- The Poisson Process to function in order for the system itself to be functional. For an n component series system in which the component lifetimes are independent exponential random variables with respective parameters .
Suppose that events are occurring at random time points and

let N(t) denote the number of points that occur in time interval [0,t].

Then, A Poisson process having rate is defined if

(a) N(0)=0,

(b) the number of events that occur in disjoint time intervals are independent,

(c) the distribution of N(t) depends only on the length of interval,

(d)

and

(e)

- Let us break the interval [0,t] into n non-overlapping subintervals each of length t/n. Now, there will be k events in [0,t] if either
(1) N(t) equals to k and there is at most one event in each subinterval, or

(2) N(t) equals to k and at least one of subintervals contain 2 or more events. Then,

Since the number of events in the different subintervals are independent, it follows that

Hence, as n approach infinity, subintervals each of length t/n. Now, there will be k events in [0,t] if either

That is, the number of events in any interval of length t has a Poisson distribution with mean

For a Poisson process, let denote the time of the first event and for n>1, denote the elapsed time between the (n-1)st and nth event. Then, the sequence

is called the sequence of inter-arrival times.

The distribution of subintervals each of length t/n. Now, there will be k events in [0,t] if either

The event takes place if and only if no events of

the Poisson process occur in [0,t] and thus,

Likewise,

Repeating the same argument yields are independent exponential random variables with mean

- Example 32 subintervals each of length t/n. Now, there will be k events in [0,t] if either (Steel Girder Fractures, p.209)
- 42 fractures on average on a 10m long girder
between-fracture length:

10/43=0.23m on average

- If the between-fracture length (X) follows an exponential distribution, how would you define the gap?
- How would you define the number of fractures (Y) per 1m steel girder?
- How are the Exponential distribution and the Poisson distribution related?

- 42 fractures on average on a 10m long girder

Figure 4.9 subintervals each of length t/n. Now, there will be k events in [0,t] if eitherPoisson process modeling fracturelocations on a steel girder

- P(the length of a gap is less than 10cm)=? subintervals each of length t/n. Now, there will be k events in [0,t] if either
- P(a 25-cm segment of a girder contains at least two fractures)=?

Figure 4.10 subintervals each of length t/n. Now, there will be k events in [0,t] if eitherThe number of fractures in a 25-cm segment of the steel girder has a Poisson distribution with mean1.075

4.3 The Gamma Distribution subintervals each of length t/n. Now, there will be k events in [0,t] if either4.3.1 Definition of the Gamma distribution

- Useful for reliability theory and life-testing and has several important sister distributions
- The Gamma function:
- The Gamma pdf with parameters k>0 and >0:
- Mean and variance:

Curves of Gamma pdf subintervals each of length t/n. Now, there will be k events in [0,t] if either

- Calculation of Gamma function subintervals each of length t/n. Now, there will be k events in [0,t] if either
When k is an integer,

When k=1, the gamma distribution is reduced to the exponential

with mean

- Properties of Gamma random variables subintervals each of length t/n. Now, there will be k events in [0,t] if either
If are independent gamma random variables with respective parameters then

is a gamma random variable with parameters

- The gamma random variable with parameters is equivalent to the exponential random variable with parameter
- If are independent exponential random variables, each having rate , then
is a gamma random variable with parameters

4.3.2 Examples of the Gamma distribution subintervals each of length t/n. Now, there will be k events in [0,t] if either

- Example 32 (Steel Girder Fractures) :
- Y= the number of fractures within 1m of the girder

Figure 4.15 subintervals each of length t/n. Now, there will be k events in [0,t] if eitherDistance to fifth fracture has agamma distribution with parameters k = 5 and l= 4.3

4.4 Weibull distribution subintervals each of length t/n. Now, there will be k events in [0,t] if either4.4.1 Definition of the Weibull distribution

- Useful for modeling failure and waiting times
(see Examples 33 & 34)

- The pdf:
- Mean and variance:

Curves of the Weibull distribution subintervals each of length t/n. Now, there will be k events in [0,t] if either

The c.d.f. of Weibull distribution: subintervals each of length t/n. Now, there will be k events in [0,t] if either

The pth quantile of Weibull distribution:

Find x such that

4.5 The Beta distribution subintervals each of length t/n. Now, there will be k events in [0,t] if either

- Useful for modeling proportions and personal probability
(See Examples 35 & 36.)

- Pdf:
- Mean and variance:

Figure 4.21 subintervals each of length t/n. Now, there will be k events in [0,t] if eitherProbability density functions of the beta distribution

Figure 4.22 subintervals each of length t/n. Now, there will be k events in [0,t] if eitherProbability density functions of the beta distribution

- Exponential and Weibull random variables have as their set of possible values.
- In engineering applications of probability theory, it is occasionally helpful to have available family of distributions whose set of possible values is finite interval.
- One of such family is the beta family of distributions.

- Example: the beta distribution and rainstorms. their set of possible values.
Data gathered by the U.S. Weather Service in Alberquerque, concern the fraction of the total rainfall falling during the first 5 minutes of storms occurring during both summer and nonsummer seasons. The data for 14 nonsummer storms can be described reasonably well by a standard beta distribution with a=2.0 and b=8.8.

- Let X be the fraction of the storm’s rainfall falling during the first 5 minutes. Then, the probability that more than 20% of the storm’s rainfall during the first 5 minutes is determined by

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