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Doorway Height The Boeing 757-200 ER airliner carries 200
Chapter 6, Problem 21BSC(choose chapter or problem)
Doorway Height The Boeing 757200 ER airliner carries 200 passengers and has doors with a height of 72 in. Heights of men are normally distributed with a mean of 69.5 in. and a standard deviation of 2.4 in. (based on Data Set 1 in Appendix B).
a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.
b. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 in.
c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?
d. When considering the comfort and safety of passengers, why are women ignored in this case?
Questions & Answers
QUESTION:
Doorway Height The Boeing 757200 ER airliner carries 200 passengers and has doors with a height of 72 in. Heights of men are normally distributed with a mean of 69.5 in. and a standard deviation of 2.4 in. (based on Data Set 1 in Appendix B).
a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.
b. If half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 72 in.
c. When considering the comfort and safety of passengers, which result is more relevant: the probability from part (a) or the probability from part (b)? Why?
d. When considering the comfort and safety of passengers, why are women ignored in this case?
ANSWER:Step 1 of 2
Given, the Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 in. Heights of men are normally distributed with a mean of 69.5 in. and a standard deviation of 2.4 in.
a. If a male passenger is randomly selected, the probability that he can fit through the doorway without bending.
We are dealing with individual value from a normally distributed population so mean is 69.5 in. and standard deviation is 2.4 in.
The Test Statistic can be expressed as
\(\begin{aligned}
Z & =\frac{x-\mu}{\sigma} \\
Z & =\frac{72-69.5}{2.4} \\
Z & =1.04
\end{aligned}\)
To find the associated probability, we have
\(\mathrm{P}\left(Z \leq \frac{x-\mu}{\sigma}\right)\)
\(\mathrm{P}(\mathrm{Z} \leq 1.04)=0.8508\) (From Area Under Normal Curve table)