Question

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 43 passengers and a flight has fuel and baggage that allows for a total passenger load of 7052 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 $\frac{7,052}{43}approx164$ 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 lb and a standard deviation of 30.
the probability is approximately
(round to four decimal places as needed.)
should the pilot take any action to correct for an overloaded aircraft?
a. no. because the probability is high the aircraft is safe to fly with its current load
b. yes. because the probability is high, the pilot should take action by somehow reducing the weight of the aircraft.

Explanation:

Step1: Identify the sampling - distribution parameters

We know that the population mean $\mu = 179$ lb, the population standard deviation $\sigma=30$ lb, and the sample size $n = 43$. The mean of the sampling - distribution of the sample mean is $\mu_{\bar{x}}=\mu = 179$ lb, and the standard deviation of the sampling - distribution of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{30}{\sqrt{43}}\approx4.55$. The critical value for the sample - mean weight that would cause an overload is $\bar{x}=164$ lb.

Step2: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. Substituting the values, we get $z=\frac{164 - 179}{4.55}=\frac{- 15}{4.55}\approx - 3.30$.

Step3: Find the probability

We want to find $P(\bar{X}>164)$. Since $P(\bar{X}>164)=1 - P(\bar{X}\leq164)$, and looking up the z - score of $z=-3.30$ in the standard normal distribution table, $P(Z\leq - 3.30)=0.0005$. So $P(\bar{X}>164)=1 - 0.0005 = 0.9995$.

Answer:

The probability is approximately $0.9995$.
Yes. Because the probability is high, the pilot should take action by somehow reducing the weight of the aircraft. (Option B)