document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1081995>Question 1081995</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve the following probability question\r<BR>\n" );
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document.write( "The average length of stay in a hospital is useful for planning purposes. Suppose that the following is the distribution of the length of stay in a hospital after a minor operation.\r<BR>\n" );
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document.write( "   x =  1,   2,   3,    4,    5<BR>\n" );
document.write( "P(x) =  0.6, 0.1, 0.04, 0.02, c\r<BR>\n" );
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document.write( "Find the variance and standard deviation </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #696047 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=696047\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> E(X)=0.6+0.2+0.12+0.08+1.2; the 1.2 comes from5*0.24, which is what c has to be to have the probability =1<BR>\n" );
document.write( "E(X)=2.2<BR>\n" );
document.write( "V(X)= sum (x-E(X))^2*p(x)<BR>\n" );
document.write( "-(1.2)^2*0.6=0.864<BR>\n" );
document.write( "-(0.2)^2*0.1=0.004<BR>\n" );
document.write( "0.8^2*0.04=0.0256<BR>\n" );
document.write( "1.8^2*0.02=0.0648<BR>\n" );
document.write( "2.8^2*0.24=2.0256<BR>\n" );
document.write( "Adds to 2.984<BR>\n" );
document.write( "SD=1.73, sqrt of the variance </SPAN>\n" );
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