document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170287>Question 1170287</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Coffee with Meals A researcher wishes to determine the number of cups of coffee a customer drinks with an evening meal at a restaurant. Find the mean, vari- ance, and standard deviation for the distribution. 0 1 2 34 P(X) 0.31 0.42 0.21 0.04 0.02 \r<BR>\n" );
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document.write( "What is the expectation of the information provided above ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #795168 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=795168\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 0--0.31---product 0--------------squared deviation*probability=1.12^2*0.31=0.3889<BR>\n" );
document.write( "1--0.42-----------0.42-----------0.0144*0.42=0.006048<BR>\n" );
document.write( "2--0.21-----------0.42-----------0.88*0.21=0.1848<BR>\n" );
document.write( "3--0.04----------0.12------------1.88^2*0.04=0.1414<BR>\n" );
document.write( "4--0.04----------0.16------------2.88^2*0.04=0.3318<BR>\n" );
document.write( "sum is expected value, and that is 1.12 cups, which is the mean<BR>\n" );
document.write( "variance is sum of squared deviations * p(x)=1.0529 cups^2<BR>\n" );
document.write( "sd is sqrt(V)=1.026 cups\r<BR>\n" );
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