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because X1 and X2 are independent. Let Y1 h1(X1, X2) X1 X2. The inverse solutions of y1 x1 x2 and y2 x1 x2 y2 11 y1 2 , and it follows that x1 y1 x2 y1 Therefore y2 11 y1 2 2 y2 11 y1 2 2 11 11 y1 1 y1 2 y1 2 y2 c y2 c 11 11 1 y1 2 2 d, x1 y2 x2 y2 c c 11 11

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and from Equation S5-4 the joint probability distribution of Y1 and Y2 is fY1Y2 1 y1, y2 2 fX1X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 0 J 0 y2 4e 23 y1 y2 11 y12 y2 11 y124 ` ` 11 y1 2 2 4e

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3. Click OK, Next to complete the installation. Install Other Software 1. Add any drivers that your system requires. 2. Install Visual Studio 6, if required, and apply the latest Visual Studio 6 service pack. 3. Install Microsoft Office, if required, and apply the latest Office service pack. Install SQL Server 2000 Developer Edition In this section, you will create a service account, perform the SQL Server installation, add a SQL login for yourself, and verify that the SQL Debugger is operational. Create the Service Account You will create an account for the SQL services. The creation of this account will be done differently, depending on whether you are in a domain environment or not. NONDOMAIN E NVIRONMENT 1. Right-click My Computer, and click Manage. 2. Expand Computer Management, System Tools, Local Users and Computers.

y2 11

y1 2 2

8

for y1 0, y2 0. We need to nd the distribution of Y1 ability distribution of Y1, or fY1 1 y1 2 fY1Y2 1y1, y2 2 dy2 4e

X1 X2 . This is the marginal prob-

Dim dr as DataRow = dt.NewRow( ) dt.Rows.Add(dr) dt.AcceptChanges( ) dr("CustomerID")="ABCDE" dt.Rows.AcceptChanges( ) dr("CustomerID")="VWXYZ" dt.RejectChanges (back to "ABCDE") dr.Delete( ) dt.RejectChanges

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y1 2 2

An important application of Equation S5-4 is to obtain the distribution of the sum of two independent random variables X1 and X2. Let Y1 X1 X2 and let Y2 X2. The inverse solutions are x1 y1 y2 and x2 y2. Therefore, x1 y1 x2 y1 1 0 x1 y2 x2 y2 1 1

RowState = Detached RowState = Added RowState = Unchanged RowState = Modified RowState = Unchanged RowState = Modified RowState = Unchanged RowState = Deleted RowState = Unchanged

and J 1. From Equation S5-4, the joint probability density function of Y1 and Y2 is fY1Y2 1 y1, y2 2 f X1 1 y1 y2 2 f X2 1 y2 2

Therefore, the marginal probability density function of Y1 is fY1 1 y1 2 fX1 1 y1 y2 2 fX2 1 y2 2 dy2

82 84 86 88 89 90 92 94 95 96 Enter Cell Data Select Cells Faster Data Entry with AutoFill Change the Font and Size Change Number Formats Apply Conditional Formatting Add Columns and Rows Freeze a Column or Row Name a Range Delete Data or Cells

A DataRow can also contain different versions of the data, which can be filtered and viewed using the RowVersion property. This can be handy when it s desirable to look at the deleted or changed rows of a DataTable. Table 8.6 shows the list of available RowVersions. This will be covered in more detail in the following sections of this chapter.

The notation is simpler if the variable of integration y2 is replaced with x and y1 is replaced with y. Then the following result is obtained.

P(E) = 30(0.01) = 0.30

It is frequently necessary to assign probabilities to events that are composed of several outcomes from the sample space. This is straightforward for a discrete sample space. EXAMPLE 2-9 Assume that 30% of the laser diodes in a batch of 100 meet the minimum power requirements of a speci c customer. If a laser diode is selected randomly, that is, each laser diode is equally likely to be selected, our intuitive feeling is that the probability of meeting the customer s requirements is 0.30. Let E denote the subset of 30 diodes that meet the customer s requirements. Because E contains 30 outcomes and each outcome has probability 0.01, we conclude that the probability of E is 0.3. The conclusion matches our intuition. Figure 2-10 illustrates this example. For a discrete sample space, the probability of an event can be de ned by the reasoning used in the example above.

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