document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210598>Question 1210598</A>: <I><SPAN STYLE=\"word-break: break-all;\">   assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°c and a standard deviation of 1.00°c. a single thermometer is randomly selected and tested. find the probability of obtaining a reading less than  -0.903°c<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 #854332 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(2264)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=854332\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To find the probability of obtaining a reading less than -0.903°C in a normal distribution with a mean of 0°C and a standard deviation of 1.00°C, we can use the Z-score formula:\r<BR>\n" );
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document.write( "Z = (X - μ) / σ\r<BR>\n" );
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document.write( "Z = (-0.903 - 0) / 1 = -0.903\r<BR>\n" );
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document.write( "Using a Z-table or calculator, the probability for Z = -0.903 is approximately 0.1029. Thus, the probability of obtaining a reading less than -0.903°C is about 10.29%. </SPAN>\n" );
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