Question

Suppose that an airline uses a seat width of 16.7 in. Assume men have hip breadths that are normally distributed with a mean of 14.2 in. and a standard deviation of 0.9 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 16.7 in. B) If the plane is filled with 126 randomly selected men find the the probability that these men have a mean hip breadth greater than 16.7. C) Which results should be considered for any changes in seat design: the result from part a or part b. TI84 use pleaese

Answer 1

Answer:

Step-by-step explanation:

Given that X, the hip width of men is N(14.2, 0.9)

i.e. we have $\frac{x-14.2}{0.9}$ is N(0,1)

Seat width an airline uses = 16.7 inches.

a) the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.7 in

$=P(X>16.7 inches)\\= P(Z>\frac{16.7-14.2}{0.9} =P(Z>2.77)\\\\\\=P(Z>0)-(0$

b) Here sample size = 126

Each person is independent of the other

the probability that these men have a mean hip breadth greater than 16.7

=$0.0028^{126}$ <0.00001

c) part a is important since even a single person not fitting will cause embarassment and leads to customer dissatisfaction.

Answer 2

Answer:

a) 0.003

b) 0.00001

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.2

Standard Deviation, σ = 0.9

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

Formula:

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

a) P(hip breadth will be greater than 16.7)

P(x > 16.7)

$P( x > 16.7) = P( z > \displaystyle\frac{16.7 - 14.2}{0.9}) = P(z > 2.77)$

$= 1 - P(z \leq 2.77)$

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

$P(x > 16.7) = 1 - 0.997 = 0.003$

b) Standard error due to sampling

$=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{0.9}{\sqrt{126}} = 0.0802$

a) P(hip breadth will be greater than 16.7 for the sample)

P(x > 16.7)

$P( x > 16.7) = P( z > \displaystyle\frac{16.7 - 14.2}{0.0802}) = P(z > 31.17)$

$= 1 - P(z \leq 31.17)$

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

$P(x > 16.7) \approx 0.000001$

c)Result in a) should be considered for any changes in seat design because we need to consider the whole population and not the result for a sample.