# Joint Probability Distributions

date post

25-Feb-2016Category

## Documents

view

67download

2

Embed Size (px)

description

### Transcript of Joint Probability Distributions

Slide 1

Chapter 4 Title and Outline15Joint Probability Distributions5-1 Two or More Random Variables 5-1.1 Joint Probability Distributions 5-1.2 Marginal Probability Distributions 5-1.3 Conditional Probability Distributions 5-1.4 Independence 5-1.5 More Than Two Random Variables

5-2 Covariance and Correlation5-3 Common Joint Distributions5-3.1 Multinomial Probability Distribution5-3.2 Bivariate Normal Distribution5-4 Linear Functions of Random Variables5-5 General Functions of Random Variables

CHAPTER OUTLINE1Learning Objective for Chapter 5After careful study of this chapter, you should be able to do the following:Use joint probability mass functions and joint probability density functions to calculate probabilities.Calculate marginal and conditional probability distributions from joint probability distributions.Interpret and calculate covariances and correlations between random variables.Use the multinomial distribution to determine probabilities.Understand properties of a bivariate normal distribution and be able to draw contour plots for the probability density function.Calculate means and variances for linear combinations of random variables, and calculate probabilities for linear combinations of normally distributed random variables.Determine the distribution of a general function of a random variable.

2Chapter 5 Learning Objectives John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.2Concept of Joint ProbabilitiesSome random variables are not independent of each other, i.e., they tend to be related.Urban atmospheric ozone and airborne particulate matter tend to vary together.Urban vehicle speeds and fuel consumption rates tend to vary inversely.The length (X) of a injection-molded part might not be independent of the width (Y). Individual parts will vary due to random variation in materials and pressure.A joint probability distribution will describe the behavior of several random variables, say, X and Y. The graph of the distribution is 3-dimensional: x, y, and f(x,y).Chapter 5 Introduction3 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.3Example 5-1: Signal BarsYou use your cell phone to check your airline reservation. The airline system requires that you speak the name of your departure city to the voice recognition system.Let Y denote the number of times that you have to state your departure city.Let X denote the number of bars of signal strength on you cell phone.Sec 5-1.1 Joint Probability Distributions4Figure 5-1 Joint probability distribution of X and Y. The table cells are the probabilities. Observe that more bars relate to less repeating.

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.4Joint Probability Mass Function DefinedSec 5-1.1 Joint Probability Distributions5

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.5Joint Probability Density Function DefinedSec 5-1.1 Joint Probability Distributions6

Figure 5-2 Joint probability density function for the random variables X and Y. Probability that (X, Y) is in the region R is determined by the volume of fXY(x,y) over the region R.The joint probability density function for the continuous random variables X and Y, denotes as fXY(x,y), satisfies the following properties: John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.6Joint Probability Mass Function GraphSec 5-1.1 Joint Probability Distributions7

Figure 5-3 Joint probability density function for the continuous random variables X and Y of different dimensions of an injection-molded part. Note the asymmetric, narrow ridge shape of the PDF indicating that small values in the X dimension are more likely to occur when small values in the Y dimension occur. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.7Example 5-2: Server Access Time-1Let the random variable X denote the time (msecs) until a computer server connects to your machine. Let Y denote the time until the server authorizes you as a valid user. X and Y measure the wait from a common starting point (x < y). The range of x and y are shown here.Sec 5-1.1 Joint Probability Distributions8

Figure 5-4 The joint probability density function of X and Y is nonzero over the shaded region where x < y. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.8Example 5-2: Server Access Time-2The joint probability density function is:

We verify that it integrates to 1 as follows:Sec 5-1.1 Joint Probability Distributions9

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.9Example 5-2: Server Access Time-2Now calculate a probability:Sec 5-1.1 Joint Probability Distributions10

Figure 5-5 Region of integration for the probability that X < 1000 and Y < 2000 is darkly shaded.

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.10Marginal Probability Distributions (discrete)For a discrete joint PDF, there are marginal distributions for each random variable, formed by summing the joint PMF over the other variable.

Sec 5-1.2 Marginal Probability Distributions11

Figure 5-6 From the prior example, the joint PMF is shown in green while the two marginal PMFs are shown in blue. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.11Marginal Probability Distributions (continuous)Rather than summing a discrete joint PMF, we integrate a continuous joint PDF.The marginal PDFs are used to make probability statements about one variable.If the joint probability density function of random variables X and Y is fXY(x,y), the marginal probability density functions of X and Y are:Sec 5-1.2 Marginal Probability Distributions12

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.12Example 5-4: Server Access Time-1For the random variables times in Example 5-2, find the probability that Y exceeds 2000.Integrate the joint PDF directly using the picture to determine the limits.Sec 5-1.2 Marginal Probability Distributions13

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.13Example 5-4: Server Access Time-2Alternatively, find the marginal PDF and then integrate that to find the desired probability.Sec 5-1.2 Marginal Probability Distributions14

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.14Mean & Variance of a Marginal DistributionMeans E(X) and E(Y) are calculated from the discrete and continuous marginal distributions.

Sec 5-1.2 Marginal Probability Distributions15

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.15Mean & Variance for Example 5-1Sec 5-1.2 Marginal Probability Distributions16

E(X) = 2.35V(X) = 6.15 2.352 = 6.15 5.52 = 0.6275 E(Y) = 2.49V(Y) = 7.61 2.492 = 7.61 16.20 = 1.4099 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.16Conditional Probability DistributionsFrom Example 5-1P(Y=1|X=3) = 0.25/0.55 = 0.455P(Y=2|X=3) = 0.20/0.55 = 0.364P(Y=3|X=3) = 0.05/0.55 = 0.091P(Y=4|X=3) = 0.05/0.55 = 0.091 Sum = 1.001

Sec 5-1.3 Conditional Probability Distributions17

Note that there are 12 probabilities conditional on X, and 12 more probabilities conditional upon Y. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.17Conditional Probability Density Function DefinedSec 5-1.3 Conditional Probability Distributions18

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.18Example 5-6: Conditional Probability-1From Example 5-2, determine the conditional PDF for Y given X=x.Sec 5-1.3 Conditional Probability Distributions19

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.19Example 5-6: Conditional Probability-2Now find the probability that Y exceeds 2000 given that X=1500:Sec 5-1.3 Conditional Probability Distributions20

Figure 5-8 again The conditional PDF is nonzero on the solid line in the shaded region. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.20Example 5-7: Conditional Discrete PMFsConditional discrete PMFs can be shown as tables.Sec 5-1.3 Conditional Probability Distributions21

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.21Mean & Variance of Conditional Random VariablesThe conditional mean of Y given X = x, denoted as E(Y|x) or Y|x is:

The conditional variance of Y given X = x, denoted as V(Y|x) or 2Y|x is:Sec 5-1.3 Conditional Probability Distributions22

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.22Example 5-8: Conditional Mean & VarianceFrom Example 5-2 & 6, what is the conditional mean for Y given that x = 1500? Integrate by parts.Sec 5-1.3 Conditional Probability Distributions23

If the connect time is 1500 ms, then the expected time to be authorized is 2000 ms. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.23Example 5-9For the discrete random variables in Exercise 5-1, what is the conditional mean of Y given X=1?Sec 5-1.3 Conditional Probability Distributions24

The mean number of attempts given one bar is 3.55 with variance of 0.7475. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.24Joint Random Variable IndependenceRandom variable independence means that knowledge of the values of X does not change any of the probabilities associated with the values of Y.X and Y vary independently.Dependence implies that the values of X are influenced by the values of Y.Do you think that a persons height and weight are independent?Sec 5-1.4 Independence25 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.25Exercise 5-10: Independent Random VariablesIn a plastic molding operation, each part is classified as to whether it conforms to color and length specifications.Sec 5-1.4 Independence26

Figure 5-10(a) shows marginal & joint probabilities, fXY(x, y) = fX(x) * fY(y)Figure 5-10(b) show the conditional probabilities, fY|x(y) = fY(y) John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.26Properties of IndependenceFor random variables X and Y, if any one of the following properties is true, the others are also true. Then X and Y are independent.Sec 5-1.4 Independence27

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.27Rectangular Range for (X, Y)A rectangular range for X and Y is a necessary, but not sufficient, condition for the independence of the variables.If the range of X and Y is not rectangular, then the range of one variable is limited by the value of the other variable.If the range of X and Y is rectangular, then one of the properties of (5-7) must be demonstrated to prove independence.Sec 5-1.4 Independence28 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.28Example 5-11: Independent Random VariablesSuppose the Example 5-2 is modified such that the joint PDF is:

Are X and Y independent? Is the product of the marginal PDFs equal the joint PDF? Yes by inspection.

Find this probability:Sec 5-1.4 Independence29

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.29Example 5-12: Machined DimensionsLet the random variables X and Y denote the lengths of 2 dimensions of a machined part. Assume that X and Y are independent and normally distributed. Find the desired probability.Sec 5-1.4 Independence30

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.30Example 5-13: More Than Two Random VariablesMany dimensions of a machined part are routinely measured during production. Let the random variables X1, X2, X3 and X4 denote the lengths of four dimensions of a part.What we have learned about joint, marginal and conditional PDFs in two variables extends to many (p) random variables.Sec 5-1.5 More Than Two Random Variables31 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.31Joint Probability Density Function RedefinedSec 5-1.5 More Than Two Random Variables32

The joint probability density function for the continuous random variables X1, X2, X3, Xp, denoted as satisfies the following properties:

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.32Example 5-14: Component LifetimesIn an electronic assembly, let X1, X2, X3, X4 denote the lifetimes of 4 components in hours. The joint PDF is:

What is the probability that the device operates more than 1000 hours?The joint PDF is a product of exponential PDFs.P(X1 > 1000, X2 > 1000, X3 > 1000, X4 > 1000) = e-1-2-1.5-3 = e-7.5 = 0.00055Sec 5-1.5 More Than Two Random Variables33

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.33Example 5-15: Probability as a Ratio of VolumesSuppose the joint PDF of the continuous random variables X and Y is constant over the region x2 + y2 =4. The graph is a round cake with radius of 2 and height of 1/4.A cake round of radius 1 is cut from the center of the cake, the region x2 + y2 =1.What is the probability that a randomly selected bit of cake came from the center round?Volume of the cake is 1. The volume of the round is *1/4 = . The desired probability is .Sec 5-1.5 More Than Two Random Variables34 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.34Marginal Probability Density FunctionSec 5-1.5 More Than Two Random Variables35

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.35Mean & Variance of a Joint PDFThe mean and variance of Xi can be determined from either the marginal PDF, or the joint PDF as follows:Sec 5-1.5 More Than Two Random Variables36

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.36Example 5-16There are 10 points in this discrete joint PDF.Note that x1+x2+x3 = 3List the marginal PDF of X2Sec 5-1.5 More Than Two Random Variables37

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.37Reduced DimensionalitySec 5-1.5 More Than Two Random Variables38

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.38Conditional Probability DistributionsConditional probability distributions can be developed for multiple random variables by extension of the ideas used for two random variables.Suppose p = 5 and we wish to find the distribution conditional on X4 and X5.Sec 5-1.5 More Than Two Random Variables39

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.39Independence with Multiple VariablesThe concept of independence can be extended to multiple variables.Sec 5-1.5 More Than Two Random Variables40

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.40Example 5-17In Chapter 3, we showed that a negative binomial random variable with parameters p and r can be represented as a sum of r geometric random variables X1, X2,, Xr , each with parameter p.Because the binomial trials are independent, X1, X2,, Xr are independent random variables. Sec 5-1.5 More Than Two Random Variables41 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.41Example 5-18: Layer ThicknessSuppose X1,X2,X3 represent the thickness in m of a substrate, an active layer and a coating layer of a chemical product. Assume that these variables are independent and normally distributed with parameters and specified limits as tabled.Sec 5-1.5 More Than Two Random Variables42

What proportion of the product meets all specifications? Answer: 0.7783, 3 layer product.Which one of the three thicknesses has the least probability of meeting specs?Answer: Layer 3 has least prob. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.42CovarianceCovariance is a measure of the relationship between two random variables.First, we need to describe the expected value of a function of two random variables. Let h(X, Y) denote the function of interest.Sec 5-2 Covariance & Correlation43

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.43Example 5-19: E(Function of 2 Random Variables)Sec 5-2 Covariance & Correlation44

Figure 5-12 Discrete joint distribution of X and Y.Task: Calculate E[(X-X)(Y-Y)] = covariance

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.44Covariance DefinedSec 5-2 Covariance & Correlation45

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.45Covariance and Scatter PatternsSec 5-2 Covariance & Correlation46

Figure 5-13 Joint probability distributions and the sign of cov(X, Y). Note that covariance is a measure of linear relationship. Variables with non-zero covariance are correlated. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.46Example 5-20: Intuitive CovarianceSec 5-2 Covariance & Correlation47The probability distribution of Example 5-1 is shown.

By inspection, note that the larger probabilities occur as X and Y move in opposite directions. This indicates a negative covariance.

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.47Correlation ( = rho)Sec 5-2 Covariance & Correlation48

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.48Example 5-21: Covariance & CorrelationDetermine the covariance and correlation.Sec 5-2 Covariance & Correlation49

Figure 5-14 Discrete joint distribution, f(x, y). John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.49Example 5-21Sec 5-2 Covariance & Correlation50

Calculate the correlation.

Figure 5-15 Discrete joint distribution. Steepness of line connecting points is immaterial. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.50Independence Implies = 0If X and Y are independent random variables,XY = XY = 0 (5-17)

XY = 0 is necessary, but not a sufficient condition for independence. Figure 5-13d (x, y plots as a circle) provides an example.Figure 5-13b (x, y plots as a square) indicates independence, but a non-rectangular pattern would indicate dependence.Sec 5-2 Covariance & Correlation51 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.51Example 5-23: Independence Implies Zero CovarianceSec 5-2 Covariance & Correlation52

Figure 5-15 A planar joint distribution.

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.52Common Joint DistributionsThere are two common joint distributionsMultinomial probability distribution (discrete), an extension of the binomial distributionBivariate normal probability distribution (continuous), a two-variable extension of the normal distribution. Although they exist, we do not deal with more than two random variables.There are many lesser known and custom joint probability distributions as you have already seen.

Sec 5-3 Common Joint Distributions53 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.53Multinomial Probability DistributionSuppose a random experiment consists of a series of n trials. Assume that:The outcome of each trial can be classifies into one of k classes.The probability of a trial resulting in one of the k outcomes is constant, denoted as p1, p2, , pk.The trials are independent.The random variables X1, X2,, Xk denote the number of outcomes in each class and have a multinomial distribution and probability mass function:Sec 5-3.1 Multinomial Probability Distribution54

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.54Example 5-24: Digital ChannelOf the 20 bits received over a digital channel, 14 are of excellent quality, 3 are good, 2 are fair, 1 is poor. The sequence received was EEEEEEEEEEEEEEGGGFFP.The probability of that sequence is 0.6140.330.0820.021 = 2.708*10-9However, the number of different ways of receiving those bits is a lot!

The combined result is a multinomial distribution.Sec 5-3.1 Multinomial Probability Distribution55

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.55Example 5-25: Digital ChannelRefer again to the prior Example 5-24.What is the probability that 12 bits are E, 6 bits are G, 2 are F, and 0 are P?Sec 5-3.1 Multinomial Probability Distribution56

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.56Multinomial Means and VariancesThe marginal distributions of the multinomial are binomial.

If X1, X2,, Xk have a multinomial distribution, the marginal probability distributions of Xi is binomial with:E(Xi) = npi and V(Xi) = npi(1-pi) (5-19)Sec 5-3.1 Multinomial Probability Distribution57 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.57Example 5-26: Marginal Probability DistributionsRefer again to the prior Example 5-25.The classes are now {G}, {F}, and {E, P}.Now the multinomial changes to:Sec 5-3.1 Multinomial Probability Distribution58

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.58Example 5-27: Bivariate Normal DistributionEarlier, we discussed the two dimensions of an injection-molded part as two random variables (X and Y). Let each dimension be modeled as a normal random variable. Since the dimensions are from the same part, they are typically not independent and hence correlated.Now we have five parameters to describe the bivariate normal distribution:X, X, Y, Y, XYSec 5-3.2 Bivariate Normal Distribution59 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.59Bivariate Normal Distribution DefinedSec 5-3.2 Bivariate Normal Distribution60

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.60Role of CorrelationSec 5-3.2 Bivariate Normal Distribution61

Figure 5-17 These illustrations show the shapes and contour lines of two bivariate normal distributions. The left distribution has independent X, Y random variables ( = 0). The right distribution has dependent X, Y random variables with positive correlation ( > 0, actually 0.9). The center of the contour ellipses is the point (X, Y). John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.61Example 5-28: Standard Bivariate Normal DistributionSec 5-3.2 Bivariate Normal Distribution62

Figure 5-18 This is a standard bivariate normal because its means are zero, its standard deviations are one, and its correlation is zero since X and Y are independent. The density function is:

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.62Marginal Distributions of the Bivariate NormalIf X and Y have a bivariate normal distribution with joint probability density function fXY(x,y;X,Y,X,Y,), the marginal probability distributions of X and Y are normal with means X and Y and X and Y, respectively. (5-21)

Sec 5-3.2 Bivariate Normal Distribution63

Figure 5-19 The marginal probability density functions of a bivariate normal distribution are simply projections of the joint onto each of the axis planes. Note that the correlation () has no effect on the marginal distributions. John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.63Conditional Distributions of the Joint NormalIf X and Y have a bivariate normal distribution with joint probability density fXY(x,y;X,Y,X,Y,), the conditional probability distribution of Y given X = x is normal with mean and variance as follows:Sec 5-3.2 Bivariate Normal Distribution64

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.64Correlation of Bivariate Normal Random VariablesIf X and Y have a bivariate normal distribution with joint probability density function fXY(x,y;X,Y,X,Y,), the correlation between X and Y is .(5-22)Sec 5-3.2 Bivariate Normal Distribution65 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.65Bivariate Normal Correlation & IndependenceIn general, zero correlation does not imply independence. But in the special case that X and Y have a bivariate normal distribution, if = 0, then X and Y are independent. (5-23)Sec 5-3.2 Bivariate Normal Distribution66 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.66Example 5-29: Injection-Molded PartThe injection- molded part dimensions has parameters as tabled and is graphed as shown.The probability of X and Y being within limits is the volume within the PDF between the limit values.This volume is determined by numerical integration beyond the scope of this text.Sec 5-3.2 Bivariate Normal Distribution67

Figure 5-3 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.67Linear Functions of Random VariablesA function of random variables is itself a random variable.A function of random variables can be formed by either linear or nonlinear relationships. We limit our discussion here to linear functions.Given random variables X1, X2,,Xp and constants c1, c2, , cp Y= c1X1 + c2X2 + + cpXp (5-24) is a linear combination of X1, X2,,Xp.Sec 5-4 Linear Functions of Random Variables68 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.68Mean & Variance of a Linear FunctionLet Y= c1X1 + c2X2 + + cpXp and use Equation 5-10:Sec 5-4 Linear Functions of Random Variables69

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.69Example 5-30: Negative Binomial DistributionLet Xi be a geometric random variable with parameter p with = 1/p and 2 = (1-p)/p2Let Y = X1 + X2 ++Xr, a linear combination of r independent geometric random variables.Then Y is a negative binomial random variable with = r/p and 2 = r(1-p)/p2 by Equations 5-25 and 5-27.Thus, a negative binomial random variable is a sum of r identically distributed and independent geometric random variables.Sec 5-4 Linear Functions of Random Variables70 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.70Example 5-31: Error PropagationA semiconductor product consists of three layers. The variances of the thickness of each layer is 25, 40 and 30 nm. What is the variance of the finished product?Answer:Sec 5-4 Linear Functions of Random Variables71

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.71Mean & Variance of an AverageSec 5-4 Linear Functions of Random Variables72

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.72Reproductive Property of the Normal DistributionSec 5-4 Linear Functions of Random Variables73

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.73Example 5-32: Linear Function of Independent NormalsLet the random variables X1 and X2 denote the independent length and width of a rectangular manufactured part. Their parameters are shown in the table.What is the probability that the perimeter exceeds 14.5 cm?Sec 5-4 Linear Functions of Random Variables74

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.74Example 5-33: Beverage VolumeSoft drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid ounces, and the standard deviation is 0.1 fl oz. Assume that the fill volumes are independent, normal random variables. What is the probability that the average volume of 10 cans is less than 12 fl oz?Sec 5-4 Linear Functions of Random Variables75

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.75General Functions of Random VariablesA linear combination is not the only way that we can form a new random variable (Y) from one or more of random variables (Xi) that we already know. We will focus on single variables, i.e., Y = h(X), the transform function.The transform function must be monotone:Each value of x produces exactly one value of y.Each value of y translates to only one value of x.This methodology produces the probability mass or density function of Y from the function of X.

Sec 5-5 General Functions of Random Variables76 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.76General Function of a Discrete Random VariableSuppose that X is a discrete random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability mass function of the random variable Y isfY(y) = fX[u(y)](5-30)

Sec 5-5 General Functions of Random Variables77 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.77Example 5-34: Function of a Discrete Random VariableLet X be a geometric random variable with PMF: fX(x) = p(1-p)x-1 for x = 1, 2, Find the probability distribution of Y = X2.Solution: Since X > 0, the transformation is one-to-one.The inverse transform function is X = sqrt(Y).fY(y) = p(1-p)sqrt(y)-1 for y = 1, 4, 9, 16,Sec 5-5 General Functions of Random Variables78 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.78General Function of a Continuous Random VariableSuppose that X is a continuous random variable with probability distribution fX(x). Let Y = h(X) define a one-to-one transformation between the values of X and Y so that the equation y = h(x) can be solved uniquely for x in terms of y. Let this solution be x = u(y), the inverse transform function. Then the probability density function of the random variable Y isfY(y) = fX[u(y)]|J|(5-31) where J = u(y) is called the Jacobian of the transformation and the absolute value is used.

Sec 5-5 General Functions of Random Variables79 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.79Example 5-35: Function of a Continuous Random VariableLet X be a continuous random variable with probability distribution:

Find the probability distribution of Y = h(X) = 2X + 4Sec 5-5 General Functions of Random Variables80

John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.80Important Terms & Concepts for Chapter 5Bivariate distributionBivariate normal distributionConditional meanConditional probability density functionConditional probability mass functionConditional varianceContour plotsCorrelationCovarianceError propagation

General functions of random variablesIndependenceJoint probability density functionJoint probability mass functionLinear functions of random variablesMarginal probability distributionMultinomial distributionReproductive property of the normal distribution

Chapter 5 Summary81 John Wiley & Sons, Inc. Applied Statistics and Probability for Engineers, by Montgomery and Runger.81Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy =12312310.010.020.2510.010.020.2520.020.030.2020.020.030.2030.020.100.0530.020.100.0540.150.100.0540.150.100.05

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =10.010.020.2510.010.020.250.2820.020.030.2020.020.030.200.2530.020.100.0530.020.100.050.1740.150.100.0540.150.100.050.30f(x) =0.200.250.551.00

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =10.010.020.2510.010.020.250.280.280.2820.020.030.2020.020.030.200.250.501.0030.020.100.0530.020.100.050.170.511.5340.150.100.0540.150.100.050.301.204.80f(x) =0.200.250.551.002.497.61x*f(x) =0.200.501.652.35x2*f(x) =0.201.004.956.15

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =10.010.020.2510.010.020.250.280.280.2820.020.030.2020.020.030.200.250.501.0030.020.100.0530.020.100.050.170.511.5340.150.100.0540.150.100.050.301.204.80f(x) =0.200.250.551.002.497.61x*f(x) =0.200.501.652.356.2001x2*f(x) =0.201.004.956.151.40995.52250.6275

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y =123123f(y) =y*f(y) =y2*f(y) =123f(y) =12310.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00020.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00030.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00040.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.000f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55x*f(x) =0.200.501.652.356.200110.0500.0800.455x2*f(x) =0.201.004.956.151.409920.1000.1200.3645.522530.1000.4000.0910.627540.7500.4000.091Sum of f(y|x) =1.0001.0001.000-0.4550.3640.091

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.3640.74750.091

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.3640.74750.091

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.3640.74750.091NormalRandom VariablesXYMean10.53.2Variance0.00250.0036

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesXYMean10.53.2Variance0.00250.0036P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(X2 = 3) =PXXX(0,3,0)Note the index pattern

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.4Note the strong positive correlation.00.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using Equations 5-14 & 5-1590.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02Using Excel0.03582= (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02Using Excel0.03582= (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2With parametersX > 0- < X < -1 < < 1Y > 1- < Y < Bivariate NormalRandom VariablesXYMean37.7Std Dev0.040.08Correlation0.8Upper Limit3.057.80Lower Limit2.957.60

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02Using Excel0.03582= (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2With parametersX > 0- < X < -1 < < 19.7467943448Y > 1- < Y < Bivariate NormalRandom VariablesParameters ofRandom VariablesXYX1X2Mean37.7Mean25Std Dev0.040.08Std Dev0.10.2Correlation0.8Upper Limit3.057.80Lower Limit2.957.60

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02Using Excel0.03582= (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2With parametersX > 0- < X < -1 < < 19.7467943448Y > 1- < Y < Bivariate NormalRandom VariablesParameters ofRandom VariablesXYX1X2Mean37.7Mean25Std Dev0.040.08Std Dev0.10.2Correlation0.8Upper Limit3.057.800.4472135955Lower Limit2.957.601.11803398870.1317762386Using Excel0.1318= 1 - NORMDIST(14.5, 14, SQRT(0.2), TRUE)

Sheet2

Sheet3

Sheet1x =y = number of times city name is statedx = number of bars of signal strengthy = number of times city name is statedx = number of bars of signal strengthf(x|y) for y =Sum of f(x|y) =y = number of times city name is statedx = number of bars of signal strengthy =123123f(y) =y*f(y) =y2*f(y) =123f(y) =123123f(y) =10.010.020.2510.010.020.250.280.280.2810.010.020.250.280.0360.0710.8931.00010.010.020.250.2820.020.030.2020.020.030.200.250.501.0020.020.030.200.250.0800.1200.8001.00020.020.030.200.2530.020.100.0530.020.100.050.170.511.5330.020.100.050.170.1180.5880.2941.00030.020.100.050.1740.150.100.0540.150.100.050.301.204.8040.150.100.050.300.5000.3330.1671.00040.150.100.050.30f(x) =0.200.250.551.002.497.61f(x) =0.200.250.55f(x) =0.200.250.55y*f(y|x=1)y2*f(y|x=1)x*f(x) =0.200.501.652.356.200110.0500.0800.45510.0500.0800.4550.050.05x2*f(x) =0.201.004.956.151.409920.1000.1200.36420.1000.1200.3640.200.405.522530.1000.4000.09130.1000.4000.0910.300.900.627540.7500.4000.09140.7500.4000.0913.0012.00Sum of f(y|x) =1.0001.0001.000Sum of f(y|x) =1.0001.0001.0003.5513.35-0.45512.60250.00055308440.3640.74750.091NormalRandom VariablesNormalRandom VariablesXYX1X2X3Mean10.53.2Mean ()10,0001,00080Variance0.00250.0036Std dev ()250204P(X2 = 0) =PXXX(0,0,3) +PXXX(1,0,2) +PXXX(2,0,1) +PXXX(3,0,0)Lower limit9,20095075P(X2 = 1) =PXXX(0,1,2) +PXXX(1,1,1) +PXXX(2,1,0)Upper limit10,8001,05085P(X2 = 2) =PXXX(0,2,1) +PXXX(1,2,0)P(in limits)0.998630.987580.78870P(X2 = 3) =PXXX(0,3,0)Note the index patternP(all in limits) =0.7778348902xyf(x, y)x-Xy-Y ProdJoint110.1-1.4-1.00.14120.2-1.40.00.00310.20.6-1.0-0.12320.20.60.00.00330.30.61.00.18Marginal10.3covariance =0.2030.710.320.430.3MeanX =2.4Y = 2.0Ex 5-21xyf(x, y)x-Xy-Y Prod000.2-1.8-1.20.42Joint110.1-0.8-0.20.01120.1-0.80.8-0.07210.10.2-0.2-0.00220.10.20.80.02330.41.21.80.88Marginal00.2covariance =1.26010.2correlation =0.92620.230.400.210.220.230.4MeanX =1.8Y = 1.8StDevX =1.166190379Y =1.166190379Ex 5-22xyf(x, y)x*y*fJoint170.21.4290.610.83110.26.6Marginals10.218.8= E(XY)20.60.80= cov(X,Y)30.21.00= XY70.2Using (5-14) & (5-15)90.6110.2MeanX =2Y = 9.0StDevX =0.632455532Y =1.2649110641Ex5-24xP(x)E0.60G0.30F0.08P0.02Using Excel0.03582= (FACT(20)/(FACT(12)*FACT(6)*FACT(2))) * 0.6^12*0.3^6*0.08^2With parameters0.0010.0316227766X > 0- < X < -1 < < 19.7467943448Y > 1- < Y < Bivariate NormalRandom VariablesParameters ofRandom VariablesXYX1X2Mean37.7Mean25Std Dev0.040.08Std Dev0.10.2Correlation0.8Upper Limit3.057.800.4472135955Lower Limit2.957.601.11803398870.1317762386Using Excel0.1318= 1 - NORMDIST(14.5, 14, SQRT(0.2), TRUE)Using Excel0.000783= NORMDIST(12, 12.1, SQRT(0.001), TRUE)

Sheet2

Sheet3