Chapter 4: Continuous Random Variables and Probability s 4-1 Continuous Random Variables 4-2 Probability s and Probability Density Functions 4-3 Cumulative Functions 4-4 Mean and Variance of a Continuous Random Variable 4-5 Continuous Uniform 4-6 Normal 4-7 Normal Approximation to the Binomial and Poisson s 4-8 Exponential 4-9 Erlang and Gamma s 4-10 Weibull 4-11 Lognormal 4-12 Beta 1 Chapter Learning Objectives 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 2 Continuous Random Variables Probability s and Probability Density Functions 3 4
Probability Density Function Defined Probability Density Functions and Histograms 5 6 A Probability Density Function Example Example 4-1 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? Another Probability Density Function Example Example 4-2 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
The Cumulative Function in the Continuous Case A Cumulative Function Example Example 4-4 9 10 Cumulative Function Examples Here is the cdf for Example 4-2: Here is the cdf for Example 4-1: Mean and Variance of a Continuous Random Variable 11 12
Example of Mean and Variance of a Continuous Random Variable Example 4-6 The Mean of a Function of a Continuous Random Variable 13 14 The Continuous Uniform The Mean and Variance of the Continuous Uniform 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 and 2 = 33.33 15 16
The cdf of the General Continuous Uniform The Normal 17 18 Example of Normal pdfs Well-Known Normal Probabilities For any normal random variable: 19 20
The Standard Normal How to Use a Table of the cdf of the Standard Normal Example 4-11 21 Figure 4-13 Standard normal probability density function. 22 Cumulative Standard Normal z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.0 0.841345 0.843752 0.846136 0.848495 0.850830 0.853141 0.855428 0.857690 0.859929 0.862143 1.1 0.864334 0.866500 0.868643 0.870762 0.872857 0.874928 0.876976 0.878999 0.881000 0.882977 1.2 0.884930 0.886860 0.888767 0.890651 0.892512 0.894350 0.896165 0.897958 0.899727 0.901475 1.3 0.903199 0.904902 0.906582 0.908241 0.909877 0.911492 0.913085 0.914657 0.916207 0.917736 1.4 0.919243 0.920730 0.922196 0.923641 0.925066 0.926471 0.927855 0.929219 0.930563 0.931888 1.5 0.933193 0.934478 0.935744 0.936992 0.938220 0.939429 0.940620 0.941792 0.942947 0.944083 1.6 0.945201 0.946301 0.947384 0.948449 0.949497 0.950529 0.951543 0.952540 0.953521 0.954486 1.7 0.955435 0.956367 0.957284 0.958185 0.959071 0.959941 0.960796 0.961636 0.962462 0.963273 1.8 0.964070 0.964852 0.965621 0.966375 0.967116 0.967843 0.968557 0.969258 0.969946 0.970621 1.9 0.971283 0.971933 0.972571 0.973197 0.973810 0.974412 0.975002 0.975581 0.976148 0.976705 Example 4-12: Standard Normal Exercises 1. 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) = 0.5387 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 Figure 4-14 Graphical displays for standard normal distributions. 23 24
Using The Standard Normal cdf to Determine Probabilities for Other Normal s An Example on Standardizing a Normal Random Variable Example 4-13 25 26 Another Example (4-14) on Standardizing A Normal Random Variable The Normal Approximation to the Binomial 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 30 The Normal Approximation to the Poisson An Example of a Normal Approximation to the Poisson Example 4-20 31 32
The Exponential An Exponential Example (Example 4-21) The exponential distribution cdf is: F(x) = 1-e -x 33 34 An Exponential Example (Example 4-21 continued) An Exponential Example (Example 4-21 continued) 35 36
An Unusual Property of the Exponential Lack of Memory Property Exponential Application in Reliability 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 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. 37 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 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 38 The Erlang 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 Example pdfs for the Erlang 1.0 0.9 0.8 Misc. Erlang pdfs for different values of r (the "order") {=0.5} Correction! 2 0.7 0.6 0.5 0.4 0.3 for x > 0 and r =1, 2, 3,. 0.2 0.1 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 39 r=1 r=2 r=4 r=8 40
The Gamma Example pdfs for the Gamma Figure 4-25 Gamma probability density functions for selected values of r and. 41 42 The Weibull Example pdfs for the Weibull Figure 4-26 Weibull probability density functions for selected values of and. 43 44
The Lognormal Example pdfs for the Lognormal 45 46 Beta A continuous distribution that is flexible, but bounded over the [0, 1] interval is useful for probability models. Examples are: Proportion of solar radiation absorbed by a material Proportion of the max time to complete a task The random variable X with probability density function 1 1 = 1 for 0 1 f x x x x is a beta random variable with parameters 0 and 0. If X has a beta distribution with parameters and, the mean and variance of X are: E X 2 V X 2 1 47 Beta Shapes are Flexible shape guidelines: If =, symmetrical about x = 0.5 If = = 1, uniform If = < 1, symmetric & U- shaped If = > 1, symmetric & mound-shaped If, skewed Figure 4-28 Beta probability density functions for selected values of the parameters and 48
Extended Range for the Beta 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) 2 V(X) 49