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

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{(\sigma)/(√(n))})`

With the following parameters:

`\mu_(\bar X)= 96`

`\sigma_(\bar X) = (23.3)/(√(9)) =7.767`

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

And we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

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

`z = (95-96)/((23.3)/(√(9)))= -0.129`

`z = (100-96)/((23.3)/(√(9)))= 0.515`

`P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248`

The sketch is on the figure attached

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{(\sigma)/(√(n))})`

With the following parameters:

`\mu_(\bar X)= 96`

`\sigma_(\bar X) = (23.3)/(√(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 = (\bar X -\mu)/((\sigma)/(√(n)))`

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

`z = (95-96)/((23.3)/(√(9)))= -0.129`

`z = (100-96)/((23.3)/(√(9)))= 0.515`

`P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248`

The sketch is on the figure attached