document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=331603>Question 331603</A>: <I><SPAN STYLE=\"word-break: break-all;\"> When designing the placement of a CD player in a new model car, engineers must consider the forward grip reach of the driver. Women have forward grip reaches that are normally distributed with a mean of 27.0 inches and a standard deviation of 1.3 inches.Use the 68-95-99.7 rule to find:<BR>\n" );
document.write( "A) The percentage of women with forward grip reaches between 24.4 inches and 29.6 inches.<BR>\n" );
document.write( "B) Find the percentage of women with forward grip reaches less than 30.9 inches.<BR>\n" );
document.write( "C) Find the percentage of women with forward grip reaches between 27.0 inches and 28.3 inches.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #237747 by </B><A HREF=/tutors/aboutme.mpl?userid=jrfrunner><B>jrfrunner(365)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jrfrunner><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=jrfrunner><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=237747\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given Normally distributed data with mean=27.0 and standard deviation =1.3<BR>\n" );
document.write( "1 standard deviation = 1.3<BR>\n" );
document.write( "2 standard deviations=2.6<BR>\n" );
document.write( "3 standard deviations =3.9<BR>\n" );
document.write( "-------------------<BR>\n" );
document.write( "A) The percentage of women with forward grip reaches between 24.4 inches and 29.6 inches.\r<BR>\n" );
document.write( "\n" );
document.write( "compute the mid-range<BR>\n" );
document.write( "Upper value-Xbar =29.6-27.0=2.6  <BR>\n" );
document.write( "Xbar - Lower value =27.0-24.4=2.6<BR>\n" );
document.write( "This means that 24.4 to 29.6 is 2 standard deviations from the mean<BR>\n" );
document.write( "approximately 95% of the data will lie within 2 standard dev from the mean<BR>\n" );
document.write( "===================<BR>\n" );
document.write( "B) Find the percentage of women with forward grip reaches less than 30.9 inches.<BR>\n" );
document.write( "----<BR>\n" );
document.write( "compute the distance in standard deviations from the mean to the largest value\r<BR>\n" );
document.write( "\n" );
document.write( "30.9-27.0 = 3.9  this is 3 standard deviations above the mean<BR>\n" );
document.write( "1/2 of 99.7% of the data will lie within the mean and 3 standard deviations <BR>\n" );
document.write( "50% of the data will lie below the mean<BR>\n" );
document.write( "so, the answer is  50% + 1/2*99.7% =99.85%<BR>\n" );
document.write( "================<BR>\n" );
document.write( "C) Find the percentage of women with forward grip reaches between 27.0 inches and 28.3 inches.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "compute the distance which starts at the mean 27.0 and goes to 28.3<BR>\n" );
document.write( "28.3-27.0=1.3  this is one standard deviation above the mean\r<BR>\n" );
document.write( "\n" );
document.write( "Since approx 68% of the data lies within 1 standard deviation on either side of the maan, the answer is 1/2* 68% = 34%\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );