document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=248090>Question 248090</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A random variable X can assume two values 1 and 3. If its mean is 8/3, compute: the probabilities for those two points and the variance of X </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #180826 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=180826\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A random variable X can assume two values 1 and 3. If its mean is 8/3, compute: the probabilities for those two points<BR>\n" );
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document.write( "Let probability of \"1\" be x<BR>\n" );
document.write( "Then probability of \"3\" is 1-x<BR>\n" );
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document.write( "Mean = 1*x + 3(1-x) = 8/3<BR>\n" );
document.write( "x + 3 - 3x = 8/3<BR>\n" );
document.write( "-2x = 8/3 - 9/3<BR>\n" );
document.write( "-2x = -1/3<BR>\n" );
document.write( "x = 1/6 (Probability of \"1\")<BR>\n" );
document.write( "1-x = 5/6 (Probability of \"3\")<BR>\n" );
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document.write( "I'll let you figure the variance.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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