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From past experience, an airline has found the luggage weight for individual air travelers on its trans-Atlantic route to have...Asked by JOSH
From past experience, an airline has found the luggage weight for individual air travelers on its trans-Atlantic route to have a mean of 80 pounds and a standard deviation of 20 pounds. The plane is consistently fully booked and holds 100 passengers. The pilot insists on loading an extra 500 pounds of fuel whenever the total luggage weight exceeds 8300 pounds, On what percentage of the flights will she end up having the extra fuel loaded?
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Answered by
jim
Since the flight is always fully booked, we can say that the luggage weight's mean is 8000 and the SD is 2000.
We're interested in the number of flights that is more than (300 / 2000 =) .15 SD above the mean.
Now we go to a z-score table - Google will provide, if you don' have one handy - and look up the value. The one I found says that 0.5596 of flights will be below the +0.15 SD mark, so that leaves .4404, or 44%, above it.
We're interested in the number of flights that is more than (300 / 2000 =) .15 SD above the mean.
Now we go to a z-score table - Google will provide, if you don' have one handy - and look up the value. The one I found says that 0.5596 of flights will be below the +0.15 SD mark, so that leaves .4404, or 44%, above it.
Answered by
economyst
I don't think jim is correct.
The SD of X+Y = sqrt[var(X)+var(Y)+2cov(XY) ]
Assuming all air-travelers are independent of each other, the last term goes away. Var(X)=SD(X)^2 = 400
so SD = sqrt(100*400) = 200
take it from here.
The SD of X+Y = sqrt[var(X)+var(Y)+2cov(XY) ]
Assuming all air-travelers are independent of each other, the last term goes away. Var(X)=SD(X)^2 = 400
so SD = sqrt(100*400) = 200
take it from here.
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