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 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6,253 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 { 6,253 \mathrm { lb } } { 37 } = 169 \mathrm { 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 175.1 lb and a standard deviation of 35.9. the probability is approximately \\( \square \\). (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \), where \( \sigma = 35.9 \) and \( n = 37 \).

$$ \sigma_{\bar{x}}=\frac{35.9}{\sqrt{37}}\approx\frac{35.9}{6.0828}\approx5.902 $$

Step2: Calculate the z - score

The formula for the z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \), where \( \bar{x}=169 \), \( \mu = 175.1 \), and \( \sigma_{\bar{x}}\approx5.902 \).

$$ z=\frac{169 - 175.1}{5.902}=\frac{- 6.1}{5.902}\approx - 1.03 $$

Step3: Find the probability

We want \( P(\bar{X}>169) \). Since \( P(\bar{X}>169)=1 - P(\bar{X}\leq169) \), and from the standard normal table \( P(Z\leq - 1.03)=0.1515 \).

$$ P(\bar{X}>169)=1 - 0.1515 = 0.8485 $$

Answer:

\(0.8485\)