document.write( "Question 1052171:
\n" );
document.write( "Shipping line W deals with transportation of goods in containers of various sizes. Charges for each container depend on its weight. A consignment of 800 containers in a particular vessel is in the high seas. Records indicate that the weights are normally distributed. The chances of a container weighing more than 20 tonnes are 97,72% while the chances of a container weighing less than 42,55 tonnes are 99,4%.
\n" );
document.write( "(1) Determine the mean µ and the standard deviation of the weights of the containers.
\n" );
document.write( "(2) Charges of one tonne are £2,50. Determine the number of containers whose weights are more than the mean µ but not exceeding 32,2 tonnes. Hence or otherwise, determine the total charges on these containers. \n" );
document.write( "
Algebra.Com's Answer #667603 by rothauserc(4718) You can put this solution on YOUR website! We have two equations in two unknowns to consider \n" ); document.write( ": \n" ); document.write( "let P be probability, u be the mean and std.dev. be the standard deviation, then \n" ); document.write( ": \n" ); document.write( "z-value associated with a P of (1 - 0.9772) is -2.00 \n" ); document.write( ": \n" ); document.write( "Note that we must subtract 0.9772 from 1 to get P( X < 20 ) in order to use z-tables \n" ); document.write( ": \n" ); document.write( "z-value associated with a P of 0.9940 is 2.51 \n" ); document.write( ": \n" ); document.write( "1) -2.00 = (20 - u) / std.dev. \n" ); document.write( ": \n" ); document.write( "2) 2.51 = (42.55 - u) / std.dev. \n" ); document.write( ": \n" ); document.write( "solve equation 1) for std.dev. and substitute in equation 2) \n" ); document.write( ": \n" ); document.write( "2.51 = (42.55 - u) / (20 -u) / -2.00 = ((42.55 -u) * -2.00) / (20 - u) = \n" ); document.write( ": \n" ); document.write( "2.51 * (20 - u) = ((42.55 - u) * -2.00) \n" ); document.write( ": \n" ); document.write( "50.2 - 2.51u = -85.1 + 2.00u \n" ); document.write( ": \n" ); document.write( "4.51u = 135.3 \n" ); document.write( ": \n" ); document.write( "u = 30 \n" ); document.write( ": \n" ); document.write( "std.dev. = (20 - 30) / -2.00 = 5 \n" ); document.write( ": \n" ); document.write( "******************************************************* \n" ); document.write( "1) mean is 30 tonnes and standard deviation is 5 tonnes \n" ); document.write( "******************************************************* \n" ); document.write( ": \n" ); document.write( "P ( X > 32.2 ) = 1 - P ( X < 32.2 ) \n" ); document.write( ": \n" ); document.write( "z-value for P ( X < 32.2 ) = (32.2 - 30) / 5 = 0.4400 \n" ); document.write( ": \n" ); document.write( "associated P is 0.6700, therefore \n" ); document.write( ": \n" ); document.write( "P ( X > 32.2 ) = 1 - 0.6700 = 0.3300 \n" ); document.write( ": \n" ); document.write( "******************************************************* \n" ); document.write( "The number of containers greater than the mean but less \n" ); document.write( "than 32.2 tonnes is 400 - (800 * 0.3300) = 136 \n" ); document.write( ": \n" ); document.write( "30 * 400 = 12000 tonnes of containers \n" ); document.write( ": \n" ); document.write( "12000 * (136/400) = 4080 tonnes \n" ); document.write( ": \n" ); document.write( "4080 * 2.50 = 10200 (British pounds?) \n" ); document.write( "******************************************************* \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |