document.write( "Question 1142604: the average weight of a suitcase for an airline passenger is 50 pounds. the standard deviation is 2 pounds. if 20% of the suitcases are overweight, find the maximum weight allowed by the airline. \r
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Algebra.Com's Answer #763314 by Theo(13342)\"\" \"About 
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average weight of a suitcase is 50 pounds.
\n" ); document.write( "standard deviation is 2 pounds.
\n" ); document.write( "20% of the suitcases are overweight.
\n" ); document.write( "what is the maximum weight?\r
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\n" ); document.write( "\n" ); document.write( "normal distribution is assumed, i believe.
\n" ); document.write( "z-score for 20% of the area under the normal distribution curve being to the right of it would be z-score for 80% of the area under the normal distribution curve being to the left of it.
\n" ); document.write( "that z-score would be .8416 rounded to 4 decimal places.\r
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\n" ); document.write( "\n" ); document.write( "z-score formula is z = (x-m)/s\r
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\n" ); document.write( "\n" ); document.write( "z is the z-score.
\n" ); document.write( "x is the raw score.
\n" ); document.write( "m is the mean
\n" ); document.write( "s is the standard deviation\r
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\n" ); document.write( "\n" ); document.write( "formula becomes .8416 = (x-50)/2
\n" ); document.write( "solve for x to get x = .8416 * 2 +50 = 51.6832.
\n" ); document.write( "that should be your answer.\r
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\n" ); document.write( "\n" ); document.write( "you can see this in the following graphs.\r
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\n" ); document.write( "\n" ); document.write( "the first 2 graphs derive the z-score and the raw score from .2 of the area under the normal distribution curve to the right of that z-scire.\r
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\n" ); document.write( "\n" ); document.write( "the second 2 graphs derive the area to the right of the indicated z-score and raw score.\r
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