document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161984>Question 1161984</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose in fact 60% of all Filipino adults between the ages of 30 and 39 exercises at least 3 days per week. If we take a random sample of 100 Filipino adults in their 30s, what is the probability that at least 70% of them exercise at least 3 days per week? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #785753 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=785753\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> We would expect 60 to exercise 3 days a week, n*p=100*0.60<BR>\n" );
document.write( "the variance would be np(1-p)=60*0.4=24<BR>\n" );
document.write( "sd is sqrt (V)=4.9<BR>\n" );
document.write( "normal approximation would be z>(69.5-60)/4.9, with continuity correction factor, and that is z>1.939 with probability of 0.0263\r<BR>\n" );
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document.write( "can do a binomial using 1-binomcdf(100,0.6,69), because the binomial looks from the left and we want what is to the right.<BR>\n" );
document.write( "This is exact and the answer is 0.0248 </SPAN>\n" );
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