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?
Z = (score-mean)/SD = (83-80)/20 = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/percentage related to that Z score.
To find the percentage of flights where the pilot will end up loading extra fuel, we need to first find the probability of the total luggage weight exceeding 8300 pounds.
We are given that the luggage weight for individual air travelers on this route follows a normal distribution with a mean of 80 pounds and a standard deviation of 20 pounds.
Since the plane holds 100 passengers, the total luggage weight for the passengers can be modeled as a normal distribution with a mean of 80 pounds * 100 passengers = 8000 pounds and a standard deviation of 20 pounds * sqrt(100 passengers) = 200 pounds. Here, we are assuming that the individual luggage weights are independent.
Now, we can calculate the probability using the standard normal distribution.
Step 1: Calculate the z-score. The z-score (standard score) measures how many standard deviations an element is from the mean.
z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
For the given problem, we want to find the probability of the total luggage weight exceeding 8300 pounds. So, x = 8300, μ = 8000, and σ = 200.
z = (8300 - 8000) / 200 = 300 / 200 = 1.5
Step 2: Look up the z-score in the standard normal distribution table or use a calculator to find the corresponding cumulative probability. The cumulative probability represents the probability of getting a value less than or equal to the z-score.
Looking up the z-score of 1.5 in the standard normal distribution table, we find that the cumulative probability is approximately 0.9332.
However, we are interested in the probability of the total luggage weight exceeding 8300 pounds, so we need to subtract this cumulative probability from 1 to get the probability of the total luggage weight being greater than 8300 pounds.
Probability = 1 - 0.9332 = 0.0668
Multiply by 100 to get the percentage.
Percentage = 0.0668 * 100 = 6.68%
Therefore, the pilot will end up having the extra fuel loaded on approximately 6.68% of the flights.