document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1107808>Question 1107808</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The television show Found has been successful for many years. That show recently had a share of 19, meaning that among the TV sets in use, 19% were tuned to Found. Assume that an advertiser wants to verify that 19% share value by conducting its own survey, and a pilot survey begins with 14 households have TV sets in use at the time of a Found broadcast.\r<BR>\n" );
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document.write( "Find the probability that none of the households are tuned to Found. <BR>\n" );
document.write( "P(none) = <BR>\n" );
document.write( " (round answer to 4 decimal places)\r<BR>\n" );
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document.write( "Find the probability that at least one household is tuned to Found. <BR>\n" );
document.write( "P(at least one) = <BR>\n" );
document.write( " (round answer to 4 decimal places)\r<BR>\n" );
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document.write( "Find the probability that at most one household is tuned to Found. <BR>\n" );
document.write( "P(at most one) = <BR>\n" );
document.write( " (round answer to 4 decimal places)\r<BR>\n" );
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document.write( "If at most one household is tuned to Found, does it appear that the 19% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Found unusual?) <BR>\n" );
document.write( "yes, it is wrong<BR>\n" );
document.write( "no, it is not wrong </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #722951 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=722951\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  The television show Found has been successful for many years. That show recently had a share of 19, meaning that among the TV sets in use, 19% were tuned to Found. Assume that an advertiser wants to verify that 19% share value by conducting its own survey, and a pilot survey begins with 14 households have TV sets in use at the time of a Found broadcast.<BR>\n" );
document.write( "P(watch) = 0.19 ; p(don't watch) = 0.81<BR>\n" );
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document.write( "Find the probability that none of the households are tuned to Found. <BR>\n" );
document.write( "P(none) = (0.81)^14 = 0.0523<BR>\n" );
document.write( "(round answer to 4 decimal places)<BR>\n" );
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document.write( "Find the probability that at least one household is tuned to Found. <BR>\n" );
document.write( "P(at least one) = 1-P(none) = 0.9477<BR>\n" );
document.write( "(round answer to 4 decimal places)<BR>\n" );
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document.write( "Find the probability that at most one household is tuned to Found. <BR>\n" );
document.write( "P(at most one) = binomcdf(14,0.19,1) = 0.2242<BR>\n" );
document.write( "(round answer to 4 decimal places)<BR>\n" );
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document.write( "If at most one household is tuned to Found, does it appear that the 19% share value is wrong? (Hint: Is the occurrence of at most one household tuned to Found unusual?)<BR>\n" );
document.write( "Not unusual.<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( "yes, it is wrong<BR>\n" );
document.write( "no, it is not wrong  </SPAN>\n" );
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