SOLUTION: Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a sta

Question 1191531: Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. If an airliner is filled with 110 randomly selected adult men, what is the probability that any one of those adult male will have a hip width greater than 16.2 inches? What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches? (Round your answers to three decimal places; add trailing zeros as needed.)

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The first uses z=(x-mean)/sd, so z(>16.2)=>(16.2-14.4)/1.1=1.8/1.1=>1.64
probability is 0.0509.
The second has a sd=sigma/sqrt(110)=0.1049.
this makes z>1.8/0.1049 or 17.159 and that has probability 0.
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