Question 1052171:
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%.
(1) Determine the mean µ and the standard deviation of the weights of the containers.
(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.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We have two equations in two unknowns to consider
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let P be probability, u be the mean and std.dev. be the standard deviation, then
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z-value associated with a P of (1 - 0.9772) is -2.00
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Note that we must subtract 0.9772 from 1 to get P( X < 20 ) in order to use z-tables
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z-value associated with a P of 0.9940 is 2.51
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1) -2.00 = (20 - u) / std.dev.
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2) 2.51 = (42.55 - u) / std.dev.
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solve equation 1) for std.dev. and substitute in equation 2)
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2.51 = (42.55 - u) / (20 -u) / -2.00 = ((42.55 -u) * -2.00) / (20 - u) =
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2.51 * (20 - u) = ((42.55 - u) * -2.00)
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50.2 - 2.51u = -85.1 + 2.00u
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4.51u = 135.3
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u = 30
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std.dev. = (20 - 30) / -2.00 = 5
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1) mean is 30 tonnes and standard deviation is 5 tonnes
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P ( X > 32.2 ) = 1 - P ( X < 32.2 )
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z-value for P ( X < 32.2 ) = (32.2 - 30) / 5 = 0.4400
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associated P is 0.6700, therefore
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P ( X > 32.2 ) = 1 - 0.6700 = 0.3300
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The number of containers greater than the mean but less
than 32.2 tonnes is 400 - (800 * 0.3300) = 136
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30 * 400 = 12000 tonnes of containers
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12000 * (136/400) = 4080 tonnes
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4080 * 2.50 = 10200 (British pounds?)
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