document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=924195>Question 924195</A>: <I><SPAN STYLE=\"word-break: break-all;\"> According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours.\r<BR>\n" );
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document.write( "Find the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1900 kilowatt-hours </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #560680 by </B><A HREF=/tutors/aboutme.mpl?userid=ewatrrr><B>ewatrrr(24785)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ewatrrr><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=ewatrrr><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=560680\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. <BR>\n" );
document.write( "Note: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z+=+blue%28x+-+1650%29%2Fblue%28320%29 ALIGN=MIDDLE  ALT=\"z+=+blue%28x+-+1650%29%2Fblue%28320%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >    <BR>\n" );
document.write( "P( x < 1900) = 9( z < 250/320) = P( z < .7813) = P(-100, .7813) = .7827<BR>\n" );
document.write( "Using a TI or Similarly a Casio fx-115 ES plus  </SPAN>\n" );
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