document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=973393>Question 973393</A>: <I><SPAN STYLE=\"word-break: break-all;\"> According to a survey in a country, 38% of adults do not have any credit cards. Suppose a simple random sample of 900 adults is obtained. \r<BR>\n" );
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document.write( "A) Determine the mean of the sampling distribuiton of p.\r<BR>\n" );
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document.write( "B) Determine the standard deviation of the sampling distribution of p. \r<BR>\n" );
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document.write( "C) In a random sample of 900 adults, what is the probability that less than 36% have no credit cards? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #595575 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=595575\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The mean of the sampling distribution is the point estimate, which is 0.38 or 342 adults from 900.\r<BR>\n" );
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document.write( "The standard deviation is<BR>\n" );
document.write( "sqrt {[(p)(1-p)]/(900)}\r<BR>\n" );
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document.write( "This is sqrt {(0.38)*(0.62)/900}\r<BR>\n" );
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document.write( "=sqrt (0.0002618)=0.0162.  This is the standard deviation of the sampling distribution.\r<BR>\n" );
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document.write( "The z-value for the random sample is {0.36-0.38)/0.0162; it is the sample-the postulated mean all divided by the standard error of the sampling distribution.\r<BR>\n" );
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document.write( "This is (-.02/0.0162), and we want the area on the normal distribution to the left of z< -1.234\r<BR>\n" );
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document.write( "We would expect with a large sample to have a fairly small probability of being 2% away.  As a rough guide, the error is 1/sqrt (n), here about 3%.  But when the probability differs from 50%, which it does, the error is less.  \r<BR>\n" );
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document.write( "Probability is 0.1085.   A z-score of -1.28 is the 10th percentile, so this is reasonable. </SPAN>\n" );
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