document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1188511>Question 1188511</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The bivariate distribution of X and Y is described below:\r<BR>\n" );
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document.write( "X	<BR>\n" );
document.write( "Y	1	2<BR>\n" );
document.write( "1	0.22	0.47<BR>\n" );
document.write( "2	0.14	0.17\r<BR>\n" );
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document.write( "A. Find the marginal probability distribution of X.\r<BR>\n" );
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document.write( "1: <BR>\n" );
document.write( "2: \r<BR>\n" );
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document.write( "B. Find the marginal probability distribution of Y.\r<BR>\n" );
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document.write( "1: <BR>\n" );
document.write( "2: \r<BR>\n" );
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document.write( "C. Compute the mean and variance of X.\r<BR>\n" );
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document.write( "Mean = <BR>\n" );
document.write( "Variance = \r<BR>\n" );
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document.write( "C. Compute the mean and variance of Y.\r<BR>\n" );
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document.write( "Mean = <BR>\n" );
document.write( "Variance = <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849303 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849303\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's the solution:\r<BR>\n" );
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document.write( "**A. Marginal Probability Distribution of X:**\r<BR>\n" );
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document.write( "The marginal probability distribution of X is found by summing the joint probabilities for each value of X across all values of Y.\r<BR>\n" );
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document.write( "*   P(X=1) = P(X=1, Y=1) + P(X=1, Y=2) = 0.22 + 0.47 = 0.69<BR>\n" );
document.write( "*   P(X=2) = P(X=2, Y=1) + P(X=2, Y=2) = 0.14 + 0.17 = 0.31\r<BR>\n" );
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document.write( "**B. Marginal Probability Distribution of Y:**\r<BR>\n" );
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document.write( "The marginal probability distribution of Y is found by summing the joint probabilities for each value of Y across all values of X.\r<BR>\n" );
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document.write( "*   P(Y=1) = P(X=1, Y=1) + P(X=2, Y=1) = 0.22 + 0.14 = 0.36<BR>\n" );
document.write( "*   P(Y=2) = P(X=1, Y=2) + P(X=2, Y=2) = 0.47 + 0.17 = 0.64\r<BR>\n" );
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document.write( "**C. Mean and Variance of X:**\r<BR>\n" );
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document.write( "*   **Mean (E[X]):** E[X] = Σ [x * P(X=x)] = (1 * 0.69) + (2 * 0.31) = 0.69 + 0.62 = 1.31<BR>\n" );
document.write( "*   **Variance (Var[X]):** Var[X] = E[X²] - (E[X])²\r<BR>\n" );
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document.write( "    *   E[X²] = Σ [x² * P(X=x)] = (1² * 0.69) + (2² * 0.31) = 0.69 + 1.24 = 1.93<BR>\n" );
document.write( "    *   Var[X] = 1.93 - (1.31)² = 1.93 - 1.7161 = 0.2139\r<BR>\n" );
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document.write( "**D. Mean and Variance of Y:**\r<BR>\n" );
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document.write( "*   **Mean (E[Y]):** E[Y] = Σ [y * P(Y=y)] = (1 * 0.36) + (2 * 0.64) = 0.36 + 1.28 = 1.64<BR>\n" );
document.write( "*   **Variance (Var[Y]):** Var[Y] = E[Y²] - (E[Y])²\r<BR>\n" );
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document.write( "    *   E[Y²] = Σ [y² * P(Y=y)] = (1² * 0.36) + (2² * 0.64) = 0.36 + 2.56 = 2.92<BR>\n" );
document.write( "    *   Var[Y] = 2.92 - (1.64)² = 2.92 - 2.6896 = 0.2304\r<BR>\n" );
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document.write( "**Answers:**\r<BR>\n" );
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document.write( "*   **A. Marginal Probability Distribution of X:**<BR>\n" );
document.write( "    *   1: 0.69<BR>\n" );
document.write( "    *   2: 0.31<BR>\n" );
document.write( "*   **B. Marginal Probability Distribution of Y:**<BR>\n" );
document.write( "    *   1: 0.36<BR>\n" );
document.write( "    *   2: 0.64<BR>\n" );
document.write( "*   **C. Mean and Variance of X:**<BR>\n" );
document.write( "    *   Mean = 1.31<BR>\n" );
document.write( "    *   Variance = 0.2139<BR>\n" );
document.write( "*   **D. Mean and Variance of Y:**<BR>\n" );
document.write( "    *   Mean = 1.64<BR>\n" );
document.write( "    *   Variance = 0.2304<BR>\n" );
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