Question

Tatooine is a fictional desert planet that appears in the Star Wars franchise. It is home to many settlers, including humans. Let X = weight of a human inhabitant on Tatooine. You know from prior research that X is approximately Normally distributed with population mean u = 96kg and population standard deviation o = 23.3kg. A. Specify the sampling distribution of statistic X, including stating what (5pts) type of distribution, its sampling mean uy, and its standard deviation Oy, for samples of 9 humans living on Tatooine. B. What is the Probability that a randomly selected group of 9 humans on (10pts) Tatooine would have a mean weight between 95 and 100kg? Include a rough sketch and Probability statement.

Answer 1

Answer:

a) For this case we select a sample size of n =9. And the distribution for the sample mean is given by:

$\bar X \sim N (\mu , \sqrt{\frac{\sigma}{\sqrt{n}}})$

With the following parameters:

$\mu_{\bar X}= 96$

$\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767$

b) $P(95< \bar X < 100)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limit we got:

$z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129$

$z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515$

$P( -0.129 < Z< 0.515) = P(Z$

The sketch is on the figure attached

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(96,23.3)$  

Where $\mu=96$ and $\sigma=23.3$

Part a

For this case we select a sample size of n =9. And the distribution for the sample mean is given by:

$\bar X \sim N (\mu , \sqrt{\frac{\sigma}{\sqrt{n}}})$

With the following parameters:

$\mu_{\bar X}= 96$

$\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767$

Part b

For this case we want this probability:

$P(95< \bar X < 100)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limit we got:

$z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129$

$z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515$

$P( -0.129 < Z< 0.515) = P(Z$

The sketch is on the figure attached