The weights of people in a certain population are normally distributed with a mean of 159 l b and a standard deviation of 25 l b. For samples of size 7​, determine whether the distribution of x overbar is normal or approximately normal and give its mean and standard deviation.

Respuesta :

Answer:

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:

[tex]X \sim N(159,25)[/tex]  

Where [tex]\mu=159[/tex] and [tex]\sigma=25[/tex]

Since the distribution of X is normal than we can conclude that we know the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

With the following parameters:

[tex] \mu_{\bar X} = 159 [/tex]

[tex]\sigma_{\bar X} =\frac{25}{\sqrt{7}} = 9.449[/tex]

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:

[tex]X \sim N(159,25)[/tex]  

Where [tex]\mu=159[/tex] and [tex]\sigma=25[/tex]

Since the distribution of X is normal than we can conclude that we know the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

With the following parameters:

[tex] \mu_{\bar X} = 159 [/tex]

[tex]\sigma_{\bar X} =\frac{25}{\sqrt{7}} = 9.449[/tex]