document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1189419>Question 1189419</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A large retail chain sends out an ad about a new product to 15 million users on Facebook and asks them to like the message. Assume the probability that a user will like the message (instead of ignoring it) is 0.00001.<BR>\n" );
document.write( "A. State assumptions for a binomial distribution to apply for X = the number of likes the message will receive. Identify n and p for that distribution.<BR>\n" );
document.write( "B. Suppose the assumptions are met. Find the mean and standard deviation of X.<BR>\n" );
document.write( "C. Is it reasonable to approximate the binomial distribution with the normal one? Why? (Assume the assumptions are met.) <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 #820897 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=820897\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> two outcomes, independent, unchanging probability, fixed number of trials.<BR>\n" );
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document.write( "15000000*0.00001=150<BR>\n" );
document.write( "so n=15,000,000 and p=0.00001<BR>\n" );
document.write( "the mean is np=150<BR>\n" );
document.write( "the variance is np(1-p)=150*0.99999=149.9985<BR>\n" );
document.write( "sd= sqrt (V)=12.25<BR>\n" );
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document.write( "np should be greater than 10, so yes, a normal approximation will work. </SPAN>\n" );
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