document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=974217>Question 974217</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Assume body temps of healthy adults are normally distributed with a mean of 98.20 degrees F and a standard deviation of 0.62 degrees F.<BR>\n" );
document.write( "1. If you have a body temp of 99.0 F, what is your percentile score?<BR>\n" );
document.write( "2. Convert 99.0 F to a standard score (or z score)<BR>\n" );
document.write( "3. Is a body temp of 99.0 F unusual?  Why or why not?<BR>\n" );
document.write( "4. 50 adults are randomly selected. What is the likelihood that the mean of their body temps is 97.98 F or lower?<BR>\n" );
document.write( "5. A person's body temp is found to be 101. F. Is the result unusual- why?<BR>\n" );
document.write( "6. What body temp is in the 95th percentile?<BR>\n" );
document.write( "7. What body type is in the 5th percentile? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #596176 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=596176\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean is 98.2;;;;;z=(x-mean)/sd   ;; For a sample, (x-mean)/[sd/sqrt(n)]<BR>\n" );
document.write( "sd is 0.62<BR>\n" );
document.write( "1.  z=(99-98.2)/0.62=  +1.29  90th percentile (slightly more)<BR>\n" );
document.write( "2.  see 1.<BR>\n" );
document.write( "3.  No.  I am a retired physician, and this may be found for a variety of reasons.  It can be normal, but in of itself, it is usually ignored.  Statistical and clinical significance are two different entities.<BR>\n" );
document.write( "4.z=(-0.22)*sqrt(50)/0.62   the SE is 0.62/sqrt(50).  I inverted to divide and put the sqrt in the numerator.  It is the same thing.  z=-2.52;;  Probability of this being random is 0.006.<BR>\n" );
document.write( "5. Yes.  It is more than 4 standard deviations above the mean.  This is highly unlikely to be normal.<BR>\n" );
document.write( "6. z=1.645  so 0.62*1.645 + Mean =99.22<BR>\n" );
document.write( "(x-mean)/sd =1.645, the 95th percentile. It is +1.02 degrees.<BR>\n" );
document.write( "7.It is the same amount in the other direction, - 1.02 degrees or 97.18.  Note that for one person, a low temperature is possible whereas for a sample of 50, even 20% of this decrease would be considered significant. </SPAN>\n" );
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