Question

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of 14.4 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 16.5 in. The probability is 0.0179. ​(Round to four decimal places as​ needed.) ​(b) If a plane is filled with 123 randomly selected​ men, find the probability that these men have a mean hip breadth greater than 16.5 in.

Answer 1

Answer:

a) 0.018

b) 0            

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,  

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly​ selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected​ men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$  

P(x > 16.5)  

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$  

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,  

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in