Respuesta :
Answer:
Step-by-step explanation:
Given that an airline uses a seat width of 16.2 in. Assume men have hip breadths that are normally distributed with a mean of 14.5 in. and a standard deviation of 1 in.
X is the hip breadth of men
X is N(14.5, 1)
We can convert X into std normal variate Z as
[tex]z=\frac{x-14.5}{1}[/tex]
a) the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.2 in.
= [tex]P(X>16.2) = P(Z>1.7) = 0.0446[/tex]
b) Here n =128 and each person is independent of the other
Hence for any person to have hip greater than 16.2 inches is binomial with constant prob = 0.0446
the probability that these men have a mean hip breadth greater than 16.2 in
[tex]\bar X :[/tex] N(14.5, 1/sqrt 128)
[tex]P(\bar X >16.2) = 0.0000[/tex]
i.e. very unusual event.
