Question

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.)

Answer 1

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