document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=316002>Question 316002</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland researchers).\r<BR>\n" );
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document.write( "a.	If you have a body temperature of 99.00 °F, what is your percentile score?<BR>\n" );
document.write( "b.	Convert 99.00 °F to a standard score (or a z-score).<BR>\n" );
document.write( "c.	Is a body temperature of 99.00 °F unusual? Why or why not?<BR>\n" );
document.write( "d.	Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?<BR>\n" );
document.write( "e.	A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?<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 #226093 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=226093\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assume the body temperatures of healthy adults are normally distributed with a mean of 98.20 °F and a standard deviation of 0.62 °F (based on data from the University of Maryland researchers). \r<BR>\n" );
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document.write( "a. If you have a body temperature of 99.00 °F, what is your percentile score?<BR>\n" );
document.write( "z(99) = (99-98.2)/0.62 = 1.2903<BR>\n" );
document.write( "P(z < 1.2903) = normalcdf(-100,1.2903) = 0.9015<BR>\n" );
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document.write( "99 degrees is at the 91%ile<BR>\n" );
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document.write( "b. Convert 99.00 °F to a standard score (or a z-score).<BR>\n" );
document.write( "See the problem above.<BR>\n" );
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document.write( "c. Is a body temperature of 99.00 °F unusual? Why or why not?<BR>\n" );
document.write( "Not too unusual.  It is 1.29 standard deviations above the mean.<BR>\n" );
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document.write( "d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98 °F or lower?<BR>\n" );
document.write( "z(97.98-98.2)/[0.62/sqrt(50)] = -2.5091<BR>\n" );
document.write( "P(z< -2.5091) = 0.0061<BR>\n" );
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document.write( "e. A person’s body temperature is found to be 101.00 °F. Is the result unusual? Why or why not? What should you conclude?<BR>\n" );
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document.write( "Find the z-value of 101 degrees.<BR>\n" );
document.write( "Based on the z-value decide if the result is \"unusual\".<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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