document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1110831>Question 1110831</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Oxygen demand is a term biologists use to describe the oxygen needed by fish and other aquatic organisms for survival. The Environmental Protection Agency conducted a study of a wetland area. In this wetland environment, the mean oxygen demand was <BR>\n" );
document.write( "&#956; = 9.3 mg/L<BR>\n" );
document.write( " with 95% of the data ranging from 6.7 mg/L to 11.9 mg/L. Let x be a random variable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal.\r<BR>\n" );
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document.write( "(a) Use the 95% data range to estimate the standard deviation for oxygen demand. <BR>\n" );
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document.write( "(b) An oxygen demand below 8 indicates that some organisms in the wetland environment may be dying. What is the probability that the oxygen demand will fall below 8 mg/L? (Round your answer to four decimal places.) <BR>\n" );
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document.write( "(c) A high oxygen demand can also indicate trouble. An oxygen demand above 12 may indicate an overabundance of organisms that endanger some types of plant life. What is the probability that the oxygen demand will exceed 12 mg/L? (Round your answer to four decimal places.) <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 #725830 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=725830\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> mean(mu) = 9.3 mg/L<BR>\n" );
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document.write( "(a)the 95% confidence interval = 9.3 + or - 2.6<BR>\n" );
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document.write( "margin of error(me) = cv * standard deviation<BR>\n" );
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document.write( "the me is 2.6, so we have<BR>\n" );
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document.write( "cv * standard deviation = 2.6<BR>\n" );
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document.write( "the cv for a 95% confidence interval is the z-score associated with a critical probability(p*) of 0.975 which is 1.96<BR>\n" );
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document.write( "Note p* is calculated this way, p* = 1 - (a/2), alpha(a) = 1 - (95/100)<BR>\n" );
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document.write( "standard deviation = 2.6 / 1.96 = 1.3265 <BR>\n" );
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document.write( "(b)  z-score = (8 - 9.3) / 1.3265 = -0.9800 <BR>\n" );
document.write( "Probability (P) ( X < 8 ) = 0.1635<BR>\n" );
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document.write( "(c) P( X > 12 ) = 1 - P ( X < 12 )<BR>\n" );
document.write( "z-score = (12 - 9.3) / 1.3265 = 2.0354<BR>\n" );
document.write( "P ( X < 12 ) = 0.9791<BR>\n" );
document.write( "P ( X > 12 ) = 1 - 0.9791 = 0.0209<BR>\n" );
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