5.
Suppose that an airline uses a seat width of
16.3
in. Assume men have hip breadths that are normally distributed with a mean of
14.4
in. and a standard deviation of
0.9
in. Complete parts (a) through (c) below.
(a) Find the probability that if an individual man is randomly selected, his hip breadth will be greater than
16.3
in.The probability is
nothing.
(Round to four decimal places as needed.)
solution
given that
P(x >16.3 ) = 1 - P(x< 16.3)
= 1 - P[(x -) / < (16.3-14.4) / 0.9]
= 1 - P(z < 2.11)
Using z table
= 1 - 0.9826
= 0.0174
probability= 0.0174
5. Suppose that an airline uses a seat width of 16.3 in. Assume men have hip...
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