document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161607>Question 1161607</A>: <I><SPAN STYLE=\"word-break: break-all;\"> fruits weight is normally distributed with mean of 480 and standard deviation of 37. if you pick 23 at random,then 3% of their mean weight will be greater than how many grams<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 #785211 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=785211\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> z for 97th percentile is 1.881<BR>\n" );
document.write( "z for this type of sampling distribution is (x bar-mean)/sigma/sqrt(n)<BR>\n" );
document.write( "so <BR>\n" );
document.write( "1.881=(x bar-480)*sqrt(23)/37<BR>\n" );
document.write( "69.6/sqrt(23)=xbar-480=14.51<BR>\n" );
document.write( "the mean weight will be greater than 494.5 grams in 3% of all possible samples of 23.<BR>\n" );
document.write( "That's how I interpreted the problem.\r<BR>\n" );
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document.write( "If the author means 3% of the mean weight, or 14.835 grams is the answer.  I've never seen that type of question before. I am assuming one wants the upper 3% of the distribution of mean weights.\r<BR>\n" );
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