Respuesta :
Answer:
a) For this case we select a sample size of n =9. And the distribution for the sample mean is given by:
[tex]\bar X \sim N (\mu , \sqrt{\frac{\sigma}{\sqrt{n}}}) [/tex]
With the following parameters:
[tex]\mu_{\bar X}= 96[/tex]
[tex]\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767[/tex]
b) [tex] P(95< \bar X < 100)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for the limit we got:
[tex] z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129[/tex]
[tex] z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515[/tex]
[tex] P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248[/tex]
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". Â
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(96,23.3)[/tex] Â
Where [tex]\mu=96[/tex] and [tex]\sigma=23.3[/tex]
Part a
For this case we select a sample size of n =9. And the distribution for the sample mean is given by:
[tex]\bar X \sim N (\mu , \sqrt{\frac{\sigma}{\sqrt{n}}}) [/tex]
With the following parameters:
[tex]\mu_{\bar X}= 96[/tex]
[tex]\sigma_{\bar X} = \frac{23.3}{\sqrt{9}} =7.767[/tex]
Part b
For this case we want this probability:
[tex] P(95< \bar X < 100)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for the limit we got:
[tex] z = \frac{95-96}{\frac{23.3}{\sqrt{9}}}= -0.129[/tex]
[tex] z = \frac{100-96}{\frac{23.3}{\sqrt{9}}}= 0.515[/tex]
[tex] P( -0.129 < Z< 0.515) = P(Z<0.515) -P(Z<-0.129) = 0.697-0.449= 0.248[/tex]
The sketch is on the figure attached
