document.write( "Question 1158362: 2.38 Baggage fees: An airline charges the following baggage fees: $25 for the first bag and $30 for the second. Suppose 53% of passengers have no checked luggage, 29% have only one piece of checked luggage and 18% have two pieces. We suppose a negligible portion of people check more than two bags.\r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "a) The average baggage-related revenue per passenger is: $ (please round to the nearest cent)
\n" );
document.write( "b) The standard deviation of baggage-related revenue is: $ (please round to the nearest cent)
\n" );
document.write( "c) About how much revenue should the airline expect for a flight of 130 passengers? $ (please round to the nearest dollar) \n" );
document.write( "
Algebra.Com's Answer #854579 by KMST(5380) You can put this solution on YOUR website! 53%+29%+18%=100%, so the portion of people who check more than two bags must have been indeed negligible. \n" ); document.write( "The passenger who check two bags pay $25+$30=$55 \n" ); document.write( " \n" ); document.write( "a) The average baggage-related revenue per passenger (rounded to the nearest cent): \n" ); document.write( "The fractions of people who check 0, 1, and 2 bags are 0.53, 0.29, and 0.18 respectively, and add to 1.00, \n" ); document.write( "so the averaged revenue per passenger is the weighted average \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "b) The standard deviation of baggage-related revenue (per passenger, in $, rounded to the nearest cent): \n" ); document.write( "The standard deviation (in $) of baggage-related revenue per passenger can be calculated from the individuals deviations (in $) from the average: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "c) Revenue the airline should expect for a flight of 130 passengers (in $, rounded to the nearest dollar) \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |