Suppose that an airline uses a seat width of 16.516.5 in. assume men have hip breadths that are normally distributed with a mean of 14.414.4 in. and a standard deviation of 0.90.9 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 16.516.5 in. the probability is nothing. (round to four decimal places as needed.)

it would be 6xy and -16xy.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

i hope this is what you were looking for.

see attached graph