document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1079953>Question 1079953</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The distribution of heights of adult men in the U.S. is approximately normal with mean 69 inches and standard deviation 2.5 inches. Use what you know about a normal distribution and the 68-95-99.7 rule to answer the following. \r<BR>\n" );
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document.write( "NOTE: If your answer is a percent, such as 25 percent, enter: \"25 PERCENT\" (without the quotes). If your answer is in inches, such as 10 inches, enter: \"10 INCHES\" (without the quotes and with a space between the number and the INCHES). If your answer is an interval, such as 14 to 15 inches, then enter: \"14 TO 15 INCHES\" (without the quotes). Do not use extra zeros and do not include a decimal point unless your answer is not a whole number. Your answer must be entered in the correct format.\r<BR>\n" );
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document.write( "(a) About what percent of men are between 69 and 74 inches? \r<BR>\n" );
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document.write( "(b) Fill in the blank: About 2.5 percent of all men are shorter than ________. \r<BR>\n" );
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document.write( "(c) Between what approximate heights do the middle 95 percent of men fall? <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 #694224 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=694224\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 68–95–99.7 rule tells us the percentage of values that lie around the mean in a normal distribution with a width of one, two and three standard deviations<BR>\n" );
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document.write( "a)  74 is two standard deviations from the mean, therefore 34 percent + 13.5 percent = 47.5 percent.  Note that we had to take half of 68 percent and half of (95 percent - 68 percent).<BR>\n" );
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document.write( "b)  2.5 percent is approximately 2 standard deviations from the mean, therefore 2 * 2.5 = 5 and 69 - 5 = 64 inches<BR>\n" );
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document.write( "c)  the middle 95 percent of men fall between 2 standard deviations of the mean,  therefore the middle 95 percent is 64 to 74 inches<BR>\n" );
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