document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1164888>Question 1164888</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 11. Suppose that 47 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.\r<BR>\n" );
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document.write( "What is the distribution of  <BR>\n" );
document.write( "X?  X~ N(________,________)\r<BR>\n" );
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document.write( "What is the distribution of  <BR>\n" );
document.write( "¯x?¯x~ N(_______,__________)\r<BR>\n" );
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document.write( "If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 69.2. \r<BR>\n" );
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document.write( "For the 47 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 67.3 and 69.2. \r<BR>\n" );
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document.write( "For part d), is the assumption that the distribution is normal necessary? Yes or No<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 #789314 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=789314\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> X ~ N(66, 121) <BR>\n" );
document.write( "x bar  is ~ N (66, 121/47) that can be converted to a decimal 2.57\r<BR>\n" );
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document.write( "for the first, the probability is <BR>\n" );
document.write( "z=(67.3-66)/11 and z=(69.2-66)/11 or z between 0.118 and 0.291<BR>\n" );
document.write( "=0.0675 probability (0.0674 not rounded on the calculator)\r<BR>\n" );
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document.write( "for a sample n=47, the sd is 11/sqrt (47) or 1.605, z between 0.810 and 2.00<BR>\n" );
document.write( "probability is 0.1859\r<BR>\n" );
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document.write( "With a sample size of 47 and a distribution that is likely not skewed, normal assumption can be used. The sample size >30 is used by many, but tire wear is likely not greatly skewed. </SPAN>\n" );
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