document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=268464>Question 268464</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 100?\r<BR>\n" );
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document.write( "I have tried to solve this using central limit therom and end up with <BR>\n" );
document.write( "-1/(3/6) = -1/0.5 = -0.5 however the book states that \r<BR>\n" );
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document.write( "SE with n=36 is 3/6 The critical ratio for a mean equal to 99 is (99-100)/0.5=-2.00 and for 101 is +2.00\r<BR>\n" );
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document.write( "I do not understand the critical ration portion...if you solve the equation the answer is -0.5 not -2.00...where am I going wrong?\r<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 #196777 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=196777\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 101?<BR>\n" );
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document.write( "t(99) = (99-100/(3/sqrt(36)) = -1/(1/2) = -2<BR>\n" );
document.write( "t(101) = (101-100)/(3/sqrt(36)) = 1/(1/2) = +2<BR>\n" );
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document.write( "P(xbar< 99) = P(t< -2 when df = 35) = 0.0267<BR>\n" );
document.write( "P(xbar> 101) = P(t > 2 when df = 35) = 0.0267<BR>\n" );
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document.write( "Comment: You just made an arithmetic error.  Statistics<BR>\n" );
document.write( "is full of these traps.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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document.write( "Given that you have a mean of 100 and standard deviation of 3; if repeated samples of 36 were selected what proportion of them would have means less than 99 and greater than 100? <BR>\n" );
document.write( "I have tried to solve this using central limit therom and end up with <BR>\n" );
document.write( "-1/(3/6) = -1/0.5 = (-0.5 WRONG) however the book states that <BR>\n" );
document.write( "SE with n=36 is 3/6 The critical ratio for a mean equal to 99 is (99-100)/0.5=-2.00 and for 101 is +2.00 (CORRECT)<BR>\n" );
document.write( "I do not understand the critical ration portion...if you solve the equation the answer is -0.5 not -2.00...where am I going wrong? \r<BR>\n" );
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