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Loading Aircraft Before every flight, the pilot must
Chapter 6, Problem 22BSC(choose chapter or problem)
Loading Aircraft 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 Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 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 \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). 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 182.9 lb and a standard deviation of 40.8 lb (based on Data Set 1 in Appendix B).
Questions & Answers
QUESTION:
Loading Aircraft 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 Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 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 \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). 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 182.9 lb and a standard deviation of 40.8 lb (based on Data Set 1 in Appendix B).
ANSWER:Step 1 of 2
We are dealing with mean of sample 37 passengers not an individual so we need to find sample mean and sample standard deviation which is given below.
The Sample Mean is given by
\(\begin{aligned}
\mu_{\bar{x}} & =\mu \\
& =182.9 \mathrm{lb}
\end{aligned}\)
The Standard deviation of the distribution of the sample mean is
\(\begin{aligned}
\sigma_{\bar{x}} & =\frac{\sigma}{\sqrt{n}} \\
& =\frac{40.8}{\sqrt{37}} \\
& =6.7074
\end{aligned}\)
By Central limit theorem, the sample mean is normally distributed with mean \(\mu\) and standard deviation \(\sigma / \sqrt{n}\)