Suppose that an airline uses a seat width of 17.3 inches. Assume men have hip breadths that are normally distributed with a mean of 14.8 inches and a standard deviation of 0.9 inches complete parts a though c

A. Find the probability that if an individual man is randomly selected his hip breadth will be greater than 17.3 in
B. If a plane is filled with 126 randomly selected meant find the probability that these men have a mean hip breadth greater than 17.3 inches
C. Which results should be considered for any changes in seat design the result from part a or part B?

a) 0.00272

b) 0.00

c) To satisfy every customer that comes in, the result From part a should be considered for any changes in seat design.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 14.8 inches

Standard deviation = σ = 0.9 inches

A) The probability that if an individual man is randomly selected his hip breadth will be greater than 17.3 in = P(x > 17.3)

We first need to standardize/normalize/obtain the z-scores of 17.3 inches

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (17.3 - 14.8)/0.9 = 2.78

To determine the probability that if an individual man is randomly selected his hip breadth will be greater than 17.3 in

P(x > 17.3) = P(z > 2.78)

We'll use data from the normal probability table for these probabilities

P(x > 17.3) = P(z > 2.78) = 1 - P(z ≤ 2.78) = 1 - 0.99728 = 0.00272

b) If a plane is filled with 126 randomly selected meant find the probability that these men have a mean hip breadth greater than 17.3 inches

The sample mean = population mean = 14.8 inches

But

Sample standard deviation = σₓ = (σ/√n)

where n = sample size = 126

σₓ = (0.9/√126)

σₓ = 0.080

z = (x - μ)/σ = (17.3 - 14.8)/0.08 = 31.25

P(x > 17.3) = P(z > 31.25) = 0.00

c) To satisfy every customer that comes in, the result From part a should be considered for any changes in seat design.

Part a shows that there are still about 0.272% of customers that the maximum hip breadth of 17.3 inches cannot contain. Although, a small number, it is still important enough to make sure everyone is satisfied.

The part B shows that the average of passengers' hip breadth can never exceed 17.3 inches. This would be appropriate if the company's thinking is that not all customers can be satisfied and the very few customers with hip breadth more than 17.3 inches would survive.

Hope this Helps!!!

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !