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?

To find the percentage of flights where the pilot will end up loading extra fuel, we need to calculate the probability that the total luggage weight exceeds 8300 pounds.

Given that the luggage weight for individual air travelers follows a normal distribution with a mean of 80 pounds and a standard deviation of 20 pounds, we can use a normal distribution to estimate the probability.

First, we need to calculate the z-score for the total luggage weight of 8300 pounds using the formula:

z = (x - μ) / σ

where
x = 8300 pounds
μ = 80 pounds (mean)
σ = 20 pounds (standard deviation)

z = (8300 - 80) / 20
z = 8220 / 20
z = 411

Next, we can use a standard normal distribution table or calculator to find the cumulative probability corresponding to the z-score of 411. Let's assume the cumulative probability is P.

The percentage of flights where the pilot will end up loading extra fuel can then be calculated as:

Percentage = (1 - P) * 100

This is done because we want to find the percentage of flights where the total luggage weight exceeds 8300 pounds, which is equivalent to finding the percentage of flights where the cumulative probability is less than or equal to P.

Please note that a z-score of 411 is extremely large, and it is beyond the range of most standard normal distribution tables or calculators. This indicates that the likelihood of the total luggage weight exceeding 8300 pounds is very low.

To calculate the exact percentage, you may need to use more advanced statistical techniques or consult specialized software that can handle extreme values like 411.

To calculate the answer, we need to find the probability that the total luggage weight exceeds 8300 pounds.

We can use the concept of the normal distribution to solve this problem. The total luggage weight can be considered as a sum of 100 independent random variables (individual luggage weights) with a mean of 80 pounds and a standard deviation of 20 pounds.

First, we need to convert the problem into a standardized normal distribution by calculating the z-score. The z-score can be calculated using the formula:

z = (x - μ) / σ

Where:
x = the desired value (8300 pounds)
μ = the mean (80 pounds)
σ = the standard deviation (20 pounds)

z = (8300 - 100 * 80) / (sqrt(100) * 20)
z = (8300 - 8000) / (sqrt(100) * 20)
z = 300 / (10 * 20)
z = 300 / 200
z = 1.5

Now, we need to find the probability of having a standardized z-score larger than 1.5. We can look up this probability in the z-table or use a calculator.

Using the z-table, we find that the probability for a z-score larger than 1.5 is approximately 0.0668.

Therefore, the pilot will end up having the extra fuel loaded on approximately 6.68% of the flights.