Question 1052174: 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 stanbon(75887)   (Show Source):
You can put this solution on YOUR website! : 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.
z-value with a left tail of 1-0.9772 = 0.0228:: z = invNorm(0.0228)= -2
z-value with a left tail of 0.994 :: z = invNorm(0.994) = 2.5
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Use x = z*s + u
20 = -2*s + u
42.55 = 2.5s + u
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Subtract and solve for "s"::
22.55 = 4.5s
s = 5
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Solve for "u"::
20 = -2*5 + u
u = 30
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(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.
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z(32.2) = (32.2-30)/5 = 2.2/5 = 0.44
P(30< x < 32.2) = P(0< z <0.44) = normalcdf(0,0.44) = 17%
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Charge = 0.17*800*2.5 = L340.06
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Cheers,
Stan H.
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