document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1086151>Question 1086151</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The AIRLINE PILOTS AND PASSENGERS ASSOCIATION It has determined that 13.7% of all passengers leave their cell phones on when instructed to turn them off.  This was a recent problem when the Galaxy batteries were exploding.  From their survey of 61,000 passengers  find the variance of the number of passengers who left their phones on.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #700341 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=700341\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=\"black\" face=\"times\" size=\"3\">This is a Binomial Distribution problem because for any person, there are two outcomes: their phone is left on or it is turned off. Each trial (asking a person whether their phone is on or not) is independent from one another. \r<BR>\n" );
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document.write( "n = Sample size = 61000<BR>\n" );
document.write( "p = Probability of success = probability of person leaving phone on = 13.7% = 0.137\r<BR>\n" );
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document.write( "Variance = n*p*(1-p)<BR>\n" );
document.write( "Variance = 61000*0.137*(1-0.137)<BR>\n" );
document.write( "Variance = <font color=red>7212.091</font></font> </SPAN>\n" );
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