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:

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.

Answer 2

Using the normal distribution and the central limit theorem, it is found that:

a) 0.0027 = 0.27% probability that if an individual man is randomly selected, his hip breadth will be greater than 16.7 in.

b) 0% probability that these men have a mean hip breadth greater than 16.7.

c) The results of part a should be considered, as there would be a problem if many of the passengers could not fit in the seats.

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Normal Probability Distribution  

In a normally distributed set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is from the mean.
• Each z-score has a respective p-value.
• This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
• Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

• The mean is of 14.2 in, thus $\mu = 14.2$
• The standard deviation is of 0.9 in, thus $\sigma = 0.9$

Item a:

• This probability is 1 subtracted by the p-value of Z when X = 16.7, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{16.7 - 14.2}{0.9}$

$Z = 2.78$

$Z = 2.78$ has a p-value of 0.9973.

1 - 0.9973 = 0.0027.

0.0027 = 0.27% probability that if an individual man is randomly selected, his hip breadth will be greater than 16.7 in.

Item b:

• By the Central Limit Theorem, the standard deviation of the sampling distributions of sample means of size n is: $s = \frac{\sigma}{\sqrt{n}}$.
• Sample of 126, thus $n = 126, s = \frac{0.9}{\sqrt{126}}$.

$Z = \frac{X - \mu}{s}$

$Z = \frac{16.7 - 14.2}{\frac{0.9}{\sqrt{126}}}$

$Z = 31.2$

$Z = 31.2$ has a p-value of 1.

1 - 1 = 0

0% probability that these men have a mean hip breadth greater than 16.7.

Item c:

The results of part a should be considered, as there would be a problem if many of the passengers could not fit in the seats.

A similar problem is given at https://brainly.com/question/24342706