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Chapter 4: Continuous Random Variables and Probability Distributions Continuous Random Variables Probability Distributions and Probability Density Functions Cumulative Distribution Functions Mean and Variance of a Continuous Random Variable Continuous Uniform Distribution Normal Distribution Normal Approximation to the Binomial and Poisson Distributions 4-8 Exponential Distribution 4-9 Erlang and Gamma Distributions 4-10 Weibull Distribution 4-11 Lognormal Distribution 4-12 Beta Distribution Chapter Learning Objectives 4-1 4-2 4-3 4-4 4-5 4-6 4-7 After careful study of this chapter you should be able to: 1.Determine probabilities from probability density functions 2.Determine probabilities from cumulative distribution functions and cumulative distribution functions from probability density functions, and the reverse 3.Calculate means and variances for continuous random variables 4.Understand the assumptions for some common continuous probability distributions 5.Select an appropriate continuous probability distribution to calculate probabilities in specific applications 6.Calculate probabilities, determine means and variances for some common continuous probability distributions 7.Standardize normal random variables 8.Use the table for the cumulative distribution function of a standard normal distribution to calculate probabilities 9.Approximate probabilities for some binomial and Poisson distributions 1 2 Probability Distributions and Probability Density Functions Continuous Random Variables 3 4 Probability Density Function Defined Probability Density Functions and Histograms 5 6 Another Probability Density Function Example A Probability Density Function Example Example 4-1 Example 4-2 Let the continuous random variable X denote the current measured in a thin copper wire in milliamperes. Assume that the range of X is [0, 20 mA], and assume that the probability density function of X is f(x) = 0.05 for 0 x 20. What is the probability that a current measurement is less than 10 milliamperes? The probability density function is shown in the figure below (it is assumed that f(x) = 0 wherever it is not specifically defined). The probability requested is indicated by the shaded area in the figure, which can be computed using the equation from slide #5.. 7 8 A Cumulative Distribution Function Example The Cumulative Distribution Function in the Continuous Case Example 4-4 9 Cumulative Distribution Function Examples • Here is the cdf for Example 4-2: 10 Mean and Variance of a Continuous Random Variable • Here is the cdf for Example 4-1: 11 12 The Mean of a Function of a Continuous Random Variable Example of Mean and Variance of a Continuous Random Variable Example 4-6 13 The Continuous Uniform Distribution 14 The Mean and Variance of the Continuous Uniform Distribution • Use these formula for the situation in example 4-1 (and 4-6) to verify the results from slide #13 that when a=0 and b=20: = 10.0 15 and 2 = 33.33 16 The cdf of the General Continuous Uniform Distribution The Normal Distribution 17 18 Well-Known Normal Distribution Probabilities Example of Normal Distribution pdfs • For any normal random variable: 19 20 How to Use a Table of the cdf of the Standard Normal Distribution The Standard Normal Distribution Example 4-11 Figure 4-13 Standard normal probability density function. 21 Cumulative Standard Normal Distribution 1. Example 4-12: Standard Normal Exercises P(Z > 1.26) = 0.1038 2. P(Z < -0.86) = 0.195 3. P(Z > -1.37) = 0.915 4. P(-1.25 < 0.37) = 22 0.5387 z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 0.841345 0.864334 0.884930 0.903199 0.919243 0.933193 0.945201 0.955435 0.964070 0.971283 0.843752 0.866500 0.886860 0.904902 0.920730 0.934478 0.946301 0.956367 0.964852 0.971933 0.846136 0.868643 0.888767 0.906582 0.922196 0.935744 0.947384 0.957284 0.965621 0.972571 0.848495 0.870762 0.890651 0.908241 0.923641 0.936992 0.948449 0.958185 0.966375 0.973197 0.850830 0.872857 0.892512 0.909877 0.925066 0.938220 0.949497 0.959071 0.967116 0.973810 0.853141 0.874928 0.894350 0.911492 0.926471 0.939429 0.950529 0.959941 0.967843 0.974412 0.855428 0.876976 0.896165 0.913085 0.927855 0.940620 0.951543 0.960796 0.968557 0.975002 0.857690 0.878999 0.897958 0.914657 0.929219 0.941792 0.952540 0.961636 0.969258 0.975581 0.859929 0.881000 0.899727 0.916207 0.930563 0.942947 0.953521 0.962462 0.969946 0.976148 0.862143 0.882977 0.901475 0.917736 0.931888 0.944083 0.954486 0.963273 0.970621 0.976705 5. P(Z -4.6) 0 6. Find z for P(Z z) = 0.05, z = -1.65 7. Find z for (-z < Z < z) = 0.99, z = 2.58 23 Figure 4-14 Graphical displays for standard normal distributions. 24 An Example on Standardizing a Normal Random Variable Using The Standard Normal cdf to Determine Probabilities for Other Normal Distributions Example 4-13 25 Another Example (4-14) on Standardizing A Normal Random Variable 26 The Normal Approximation to the Binomial Distribution 27 28 An Example (4-17 & 4-18) of a Normal Approximation to the Binomial Guidelines for Using the Normal to Approximate the Binomial or Hypergeometric 150+.05-160 -0.75) 0.75) 0.773 29 The Normal Approximation to the Poisson Distribution 30 An Example of a Normal Approximation to the Poisson Example 4-20 31 32 An Exponential Distribution Example (Example 4-21) The Exponential Distribution The exponential distribution cdf is: F(x) = 1-e-Ox 33 34 An Exponential Distribution Example (Example 4-21 continued) An Exponential Distribution Example (Example 4-21 continued) 35 36 An Unusual Property of the Exponential Distribution Exponential Application in Reliability Lack of Memory Property • The reliability of electronic components is often modeled by the exponential distribution. A chip might have mean time to failure of 40,000 operating hours • The memoryless property implies that the component does not wear out – the probability of failure in the next hour is constant, regardless of the component age e.g. In Example 4-21, suppose that there are no log-ons from 12:00 to 12:15; the probability that there are no log-ons from 12:15 to 12:21 is still 0.082. Because we have already been waiting for 15 minutes, we feel that we are “due.” That is, the probability of a log-on in the next 6 minutes should be greater than 0.082. However, for an exponential distribution this is not true. • The reliability of mechanical components do have a memory – the probability of failure in the next hour increases as the component ages. The Weibull distribution is used to model this situation 37 38 Example pdfs for the Erlang Distribution Correction! The Erlang Distribution 2 Misc. Erlang pdfs for different values of r (the "order") {O=0.5} The random variable X that equals the interval length until r counts occur in a Poisson process with mean > 0 has and Erlang random variable with parameters and r. The probability density function of X is 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 for x > 0 and r =1, 2, 3, …. 0.1 0.0 0.0 39 1.0 2.0 r=1 3.0 r=2 4.0 r=4 5.0 r=8 6.0 7.0 8.0 40 Example pdfs for the Gamma Distribution The Gamma Distribution Figure 4-25 Gamma probability density functions for selected values of r and O. 41 42 Example pdfs for the Weibull Distribution The Weibull Distribution Figure 4-26 Weibull probability density functions for selected values of D and E. 43 44 Example pdfs for the Lognormal Distribution The Lognormal Distribution 45 Beta Distribution Beta Shapes are Flexible • A continuous distribution that is flexible, but bounded over the [0, 1] interval is useful for probability models. Examples are: • Distribution shape guidelines: – Proportion of solar radiation absorbed by a material – Proportion of the max time to complete a task * D E D 1 E 1 for 0 d x d 1 x 1 x * D * E is a beta random variable with parameters D ! 0 and E ! 0. If X has a beta distribution with parameters and , the mean and variance of X are: P EX V2 V X Figure 4-28 Beta probability density functions for selected values of the parameters and – If = , symmetrical about x = 0.5 – If = = 1, uniform – If = < 1, symmetric & U- shaped – If = > 1, symmetric & mound-shaped – If , skewed The random variable X with probability density function f x = 46 D D E DE D E D E 1 2 47 48 Extended Range for the Beta Distribution The beta random variable X is defined for the [0, 1] interval. That interval can be changed to [a, b]. Then the random variable W is defined as a linear function of X: W = a + (b –a)X With mean and variance: E(W) = a + (b –a) E(X) V(W) = (b - a)2V(X) 49