document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1186996>Question 1186996</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Q.3 An increasing number of consumers believe they have to look out for themselves in the marketplace. According to a<BR>\n" );
document.write( "survey conducted by the Yankelovich Partners for USA Weekend magazine, 60% of all consumers have called an 800<BR>\n" );
document.write( "or 900 telephone numbers for information about some product. Suppose a random sample of 20 consumers is<BR>\n" );
document.write( "contacted and interviewed about their buying habits. [Answers: a) 41.50%, b) 0.3%, c) 0%]<BR>\n" );
document.write( "Required:<BR>\n" );
document.write( "a) What is the probability that 13 or more of these consumers have called an 800 or 900 telephone numbers for<BR>\n" );
document.write( "information about some product?<BR>\n" );
document.write( "b) What is the probability that more than 17 of these consumers have called an 800 or 900 telephone numbers<BR>\n" );
document.write( "for information about some product?<BR>\n" );
document.write( "c) What is the probability that fewer than 5 of these consumers have called an 800 or 900 telephone numbers<BR>\n" );
document.write( "for information about some product?\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #820394 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=820394\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> want binomial n=12 p=0.6<BR>\n" );
document.write( "Very straight forward with c. You can look at the table with n=20 p=0.6 and 4. The probability both pdf and cdf is 0.<BR>\n" );
document.write( "For b, More than 17 is 18-20. For 18 it is 0.003, and 19 and 20 don't increase that significantly.<BR>\n" );
document.write( "-<BR>\n" );
document.write( "for 13: the calculator will show that for 14 or more (1-biniomcdf(20,0.6,13)=0.2500<BR>\n" );
document.write( "Can check that for 13: 0.1659.<BR>\n" );
document.write( "for 14: 0.1244<BR>\n" );
document.write( "for 15: 0.0746<BR>\n" );
document.write( "for 16: 0.0350<BR>\n" );
document.write( "for 17: 0.0123<BR>\n" );
document.write( "and the rest is 0.003 from above<BR>\n" );
document.write( "That is 0.416 or 41.6%<BR>\n" );
document.write( "also 1-binomcdf(20,0.6,12), since want 12 but not 13 to get rid of the left side. </SPAN>\n" );
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