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 40 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,640 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 6,640 / 40 = 166 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 177.3 lb and a standard deviation of 35.3. the probability is approximately □. (round to four decimal places as needed.)

Explanation:

Step1: Calculate the sample mean

The sample size \(n = 40\). The population mean \(\mu=177.3\) lb and the population standard deviation \(\sigma = 35.3\) lb. The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}=\mu = 177.3\) lb. The standard deviation of the sampling distribution of the sample mean (standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{35.3}{\sqrt{40}}\approx5.58\) lb.

Step2: Calculate the z - score

We want to find \(P(\bar{X}>166)\). The z - score is calculated using the formula \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\). Substitute \(\bar{x} = 166\), \(\mu_{\bar{x}}=177.3\) and \(\sigma_{\bar{x}}\approx5.58\) into the formula: \(z=\frac{166 - 177.3}{5.58}=\frac{- 11.3}{5.58}\approx - 2.03\)

Step3: Find the probability

We know that \(P(\bar{X}>166)=P(Z>-2.03)\). Since \(P(Z > z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq - 2.03)=0.0212\). So \(P(Z>-2.03)=1 - 0.0212 = 0.9788\)

Answer:

\(0.9788\)