x 2 a x f 1x2 x in .NET Assign QR Code JIS X 0510 in .NET x 2 a x f 1x2 x x 2 a x f 1x2 xQR Code barcode library in .netuse .net vs 2010 qrcode integrated tocreate qr code for .net2 2f 1x2 2 decode qr barcode with .netUsing Barcode scanner for .net framework Control to read, scan read, scan image in .net framework applications.2 a x f 1x2 x 2 recognize bar code for .netUsing Barcode recognizer for VS .NET Control to read, scan read, scan image in VS .NET applications.2 a xf 1x2 .net Framework bar code encoder in .netgenerate, create barcode none for .net projectsa f 1x2 Control qr code image in .net c#using barcode drawer for visual .net control to generate, create denso qr bar code image in visual .net applications.2 a x f 1x2 x Control qr code iso/iec18004 data with .netto display qr codes and qr-codes data, size, image with .net barcode sdkEither formula for V1x2 can be used. Figure 3-6 illustrates that two probability distributions can differ even though they have identical means and variances. EXAMPLE 3-9 In Example 3-4, there is a chance that a bit transmitted through a digital transmission channel is received in error. Let X equal the number of bits in error in the next four bits transmitted. The possible values for X are 50, 1, 2, 3, 46 . Based on a model for the errors that is presented in the following section, probabilities for these values will be determined. Suppose that the probabilities are P1X P1X 02 12 0.6561 0.2916 P1X P1X 22 32 0.0486 0.0036 P1X 42 0.0001Control qr bidimensional barcode image in visual basicuse .net framework qr bidimensional barcode integrated toreceive denso qr bar code on visual basic4 (a)Barcode creation for .netusing visual .net tointegrate bar code in asp.net web,windows application4 (b)Visual .net upc-a printing for .netusing barcode integration for vs .net control to generate, create upc-a image in vs .net applications.Figure 3-6 The probability distributions illustrated in Parts (a) and (b) differ even though they have equal means and equal variances.Use barcode 39 in .netusing barcode integrated for visual studio .net control to generate, create ansi/aim code 39 image in visual studio .net applications.CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Visual .net Crystal code 128 code set b generator on .netusing visual .net crystal tointegrate code-128c for asp.net web,windows applicationNow E1X2 0f 102 1f 112 2f 122 010.65612 110.29162 0.4 3f 132 4f 142 210.04862 310.00362 410.00012 Royal Mail Barcode generation on .netusing barcode creation for visual studio .net crystal control to generate, create royal mail barcode image in visual studio .net crystal applications.Although X never assumes the value 0.4, the weighted average of the possible values is 0.4. To calculate V1X 2, a table is convenient.Control upc-a data on visual c#.netto create upc code and upc code data, size, image with c#.net barcode sdkx 0 1 2 3 4Control data matrix barcodes size in .netto get data matrix barcodes and data matrix 2d barcode data, size, image with .net barcode sdk0.4 0.4 0.6 1.6 2.6 3.6Control qr codes image in visual c#.netgenerate, create qr code iso/iec18004 none with visual c# projects0.42 2 0.16 0.36 2.56 6.76 12.96Pdf417 barcode library on visual basicuse aspx.net crystal pdf417 development toprint barcode pdf417 with visual basicf 1x2 0.6561 0.2916 0.0486 0.0036 0.0001 Control code 128b data with visual c#.net code128b data in visual c#.netf 1x21x Office Excel 1d barcode implement for office exceluse office excel linear barcode printer tobuild 1d with office excel0.42 2Code-128 implement on visual c#.netusing barcode drawer for windows forms crystal control to generate, create code-128c image in windows forms crystal applications.0.104976 0.104976 0.124416 0.024336 0.001296Data Matrix Barcodes encoding for visual basicusing .net winforms crystal toinclude data matrix ecc200 for asp.net web,windows applicationV1X2 a f 1xi 21xi 0.42 2The alternative formula for variance could also be used to obtain the same result. EXAMPLE 3-10 Two new product designs are to be compared on the basis of revenue potential. Marketing feels that the revenue from design A can be predicted quite accurately to be \$3 million. The revenue potential of design B is more difficult to assess. Marketing concludes that there is a probability of 0.3 that the revenue from design B will be \$7 million, but there is a 0.7 probability that the revenue will be only \$2 million. Which design do you prefer Let X denote the revenue from design A. Because there is no uncertainty in the revenue from design A, we can model the distribution of the random variable X as \$3 million with probability 1. Therefore, E1X 2 \$3 million. Let Y denote the revenue from design B. The expected value of Y in millions of dollars is E1Y 2 \$710.32 \$210.72 \$3.5Because E(Y) exceeds E(X), we might prefer design B. However, the variability of the result from design B is larger. That is,5.25 millions of dollars squared 3.52 2 10.323.52 2 10.72Because the units of the variables in this example are millions of dollars, and because the variance of a random variable squares the deviations from the mean, the units of 2 are millions of dollars squared. These units make interpretation difficult. Because the units of standard deviation are the same as the units of the random variable, the standard deviation is easier to interpret. In this example, we can summarize our results as the average deviation of Y from its mean is \$2.29 million. 3-4 MEAN AND VARIANCE OF A DISCRETE RANDOM VARIABLE EXAMPLE 3-11The number of messages sent per hour over a computer network has the following distribution:x f 1x2 number of messages 10 0.08 11 0.15 12 0.30 13 0.20 14 0.20 15 0.07 Determine the mean and standard deviation of the number of messages sent per hour. V1X 2 E1X 2 10 10.0822V1X 2 p p 1.36 11 10.15215 10.07212.5 12.52 1.85The variance of a random variable X can be considered to be the expected value of a specific function of X, namely, h1X 2 1X 2 2 . In general, the expected value of any function h1X 2 of a discrete random variable is defined in a similar manner.Expected Value of a Function of a Discrete Random Variable If X is a discrete random variable with probability mass function f 1x2, E3h1X 2 4 a xh1x 2 f 1x 2 (3-4)EXAMPLE 3-12In Example 3-9, X is the number of bits in error in the next four bits transmitted. What is the expected value of the square of the number of bits in error Now, h1X 2 X 2 . Therefore, E3h1X2 4 02 0.6561 12 32 0.2916 0.0036 22 42 0.0486 0.0001 0.52In the previous example, the expected value of X 2 does not equal E1X 2 squared. However, in aE1X 2 b. This the special case that h1X 2 aX b for any constants a and b, E 3h1X 2 4 can be shown from the properties of sums in the definition in Equation 3-4.EXERCISES FOR SECTION 3-43-37. If the range of X is the set {0, 1, 2, 3, 4} and P(X x) 0.2 determine the mean and variance of the random variable. 3-38. Determine the mean and variance of the random variable in Exercise 3-13. 3-39. Determine the mean and variance of the random variable in Exercise 3-15. 3-40. Determine the mean and variance of the random variable in Exercise 3-17. 3-41. Determine the mean and variance of the random variable in Exercise 3-19. 3-42. Determine the mean and variance of the random variable in Exercise 3-20. 3-43. Determine the mean and variance of the random variable in Exercise 3-22. 3-44. Determine the mean and variance of the random variable in Exercise 3-23. 3-45. The range of the random variable X is 3 0, 1, 2, 3, x4 , where x is unknown. If each value is equally likely and the mean of X is 6, determine x.