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?
Mean luggage weight = 80*100=8000 pounds
Standard deviation = 20*100 = 2000 pounds
8300 pounds exceeds the mean by 300/2000=0.15σ
Assuming a normal distribution, look up standard tables where Z=0.15 to get a tail probability f(α) of 0.4404 (for the weight to be over 8300 pounds).
Hence calculate the percentage of flights which require extra fuels.
To solve this problem, we need to find the probability that the total luggage weight exceeds 8300 pounds. We can use the concept of the normal distribution and z-scores to calculate this probability.
First, let's calculate the average luggage weight for the total number of passengers on the flight, which is 100 passengers. The mean luggage weight for a single passenger is given as 80 pounds. Therefore, the average luggage weight for 100 passengers is 80 * 100 = 8000 pounds.
Next, we need to calculate the standard deviation of the total luggage weight for 100 passengers. Since the luggage weights are independent variables, we can use the property that the standard deviation of the sum of independent random variables is equal to the square root of the sum of the variances.
The standard deviation for a single passenger is given as 20 pounds. Therefore, the standard deviation for 100 passengers is 20 * √100 = 200 pounds.
Now, to find the probability that the total luggage weight exceeds 8300 pounds, we need to calculate the z-score for this value. The z-score is calculated using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
Using the above formula, the z-score is z = (8300 - 8000) / 200 = 1.5.
We can then look up the corresponding probability in the standard normal distribution table or use a calculator to find that the probability of z > 1.5 is approximately 0.0668.
Therefore, the percentage of flights where the pilot will end up having the extra fuel loaded is 0.0668 * 100 = 6.68%.