document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1192344>Question 1192344</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 189.8 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 44.5 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 21 passengers.\r<BR>\n" );
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document.write( "What is the approximate probability (±0.001) that the total weight of the passengers exceeds 4375 pounds?   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849038 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.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=849038\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> **1. Calculate the Mean and Standard Deviation of the Total Weight**\r<BR>\n" );
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document.write( "* **Mean:**<BR>\n" );
document.write( "    * Mean weight of a single passenger = 189.8 pounds<BR>\n" );
document.write( "    * Mean weight of 21 passengers = 189.8 pounds/passenger * 21 passengers = 3985.8 pounds\r<BR>\n" );
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document.write( "* **Standard Deviation:**<BR>\n" );
document.write( "    * Standard deviation of a single passenger = 44.5 pounds<BR>\n" );
document.write( "    * Standard deviation of the total weight of 21 passengers = 44.5 pounds/passenger * √21 passengers ≈ 204.4 pounds\r<BR>\n" );
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document.write( "**2. Standardize the Total Weight Limit**\r<BR>\n" );
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document.write( "* We'll use the z-score to standardize the total weight limit:\r<BR>\n" );
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document.write( "   z = (Total Weight Limit - Mean Total Weight) / Standard Deviation of Total Weight<BR>\n" );
document.write( "   z = (4375 pounds - 3985.8 pounds) / 204.4 pounds<BR>\n" );
document.write( "   z ≈ 1.90\r<BR>\n" );
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document.write( "**3. Find the Probability**\r<BR>\n" );
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document.write( "* We're interested in the probability that the total weight exceeds 4375 pounds, which corresponds to the area to the right of z = 1.90 under the standard normal distribution curve.\r<BR>\n" );
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document.write( "* Using a standard normal distribution table or a calculator, we find:<BR>\n" );
document.write( "   P(Z > 1.90) ≈ 0.0287\r<BR>\n" );
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document.write( "**Therefore, the approximate probability that the total weight of the 21 passengers exceeds 4375 pounds is 0.029.**\r<BR>\n" );
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document.write( "**Note:**\r<BR>\n" );
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document.write( "* We've used the normal distribution as an approximation, even though passenger weights might not be perfectly normally distributed. <BR>\n" );
document.write( "* This approximation should be reasonably accurate due to the Central Limit Theorem, which states that the distribution of the sum or average of a large number of independent random variables tends to be approximately normal, regardless of the underlying distribution of the individual variables.<BR>\n" );
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