Question

(6.4.15) Suppose that an airline uses a seat width of 16.4 in. Assume men have hip breadths that are normally distributed with a mean of 14.1 in. and a standard deviation of 1.1 in.
Find the probability that if an individual man is randomly​ selected, his hip breadth will be greater than 16.4 in.
The probability is?

Answer 1

$P(X\ge16.4) = P(Z \ge \frac {16.4 - \mu} {\sigma}) = P(Z \ge \frac {16.4 - 14.1} {1.1}) = P(Z\ge 2.09) = \\ 1-P(Z<2.09)=1- 0.9817= 0.0183$