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.

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:

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

P(x > 16.5)

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

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:

P(x > 16.5)  

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

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

blue marble is 15 times out of 50 = 15/50 reduces to 3/10

landing on tails is 24 out of 50 = 24/50 reduces to 12/25

to find the probability of both multiply together:

3/10 x 12/25 = 3 x 12 / 10 x 25 = 36/250

if you multiplied the numerator and denominator by 10 you get 360/2500.

the answer is 360/2500

25

step-by-step explanation: