Question

Suppose 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. 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 16.2 in. The probability is 0.0446 (Round to four decimal places as needed.) (b) If a plane is filled with 128 randomly selected men, find the probability that these men have a mean hip breadth greater than 16.2 in The probability is (Round to four decimal places as needed.)

Answer 1

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

$z=\frac{x-14.5}{1}$

a) the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.2 in.

= $P(X>16.2) = P(Z>1.7) = 0.0446$

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

$\bar X :$ N(14.5, 1/sqrt 128)

$P(\bar X >16.2) = 0.0000$

i.e. very unusual event.