Questions: Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The aircraft can carry 35 passengers, and a flight has fuel and baggage that allows for a total passenger load of 5,775 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than 5,775 lb / 35 = 165 lb. What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 179.3 lb and a standard deviation of 35.2. The probability is approximately (Round to four decimal places as needed.)

Solution Steps

The aircraft can carry a maximum passenger load of \( 5,775 \, \text{lb} \) with \( 35 \) passengers. The mean weight limit per passenger is calculated as:

\[ \text{Mean weight limit} = \frac{5,775 \, \text{lb}}{35} = 165 \, \text{lb} \]

The weights of men are normally distributed with a mean \( \mu = 179.3 \, \text{lb} \) and a standard deviation \( \sigma = 35.2 \, \text{lb} \). We need to find the probability that the mean weight of the passengers exceeds \( 165 \, \text{lb} \).

To find the probability that the mean weight exceeds \( 165 \, \text{lb} \), we first calculate the Z-score for \( 165 \, \text{lb} \):

\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} = \frac{165 - 179.3}{35.2 / \sqrt{35}} \approx -2.4034 \]

Using the Z-score, we can find the probability that the mean weight is greater than \( 165 \, \text{lb} \):

\[ P(X > 165) = P(Z > -2.4034) = 1 - P(Z < -2.4034) = \Phi(\infty) - \Phi(-2.4034) \]

From the calculations, we find:

\[ P(X > 165) \approx 0.9919 \]

Since the probability that the aircraft is overloaded is \( 0.9919 \), which is significantly greater than \( 0.5 \), the pilot should take action to correct for an overloaded aircraft.

Final Answer