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?
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.
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.
To find the percentage of flights where the extra fuel will be loaded, we need to calculate the probability that the total luggage weight exceeds 8300 pounds.
We can use the concept of the standard normal distribution to solve this problem.
1. Convert the given mean and standard deviation into a standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. To convert the given mean and standard deviation, we use the formula:
Z = (X - μ) / σ
where Z is the standard score, X is the observed value, μ is the mean, and σ is the standard deviation.
In this case, X = 8300 pounds, μ = 80 pounds, and σ = 20 pounds.
Z = (8300 - 80) / 20 = 8220 / 20 = 411
2. Find the probability of the total luggage weight exceeding 8300 pounds using the standard normal distribution table or a calculator with a cumulative distribution function (CDF).
The standard normal distribution table gives the area under the curve to the left of a given Z-score. Since we need the area to the right of 411, we subtract the area to the left of 411 from 1.
P(Z > 411) = 1 - P(Z < 411)
Using a standard normal distribution table or calculator, we can find the corresponding probability for Z = 411 or use an appropriate software/tool.
3. Calculate the percentage of flights where the extra fuel will be loaded.
Multiply the probability by 100 to get the percentage.
Percentage = P(Z > 411) * 100
By following these steps and calculating the value, we can determine the percentage of flights where the pilot will end up having the extra fuel loaded.