Respuesta :
Answer:
a) We are interested on this probability
[tex]P(X>13.17)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367 [/tex]
b) [tex] P(\bar X >13.17)[/tex]
The new z score is given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207 [/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".
Part a
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(13,0.5)[/tex]
Where [tex]\mu=13[/tex] and [tex]\sigma=0.5[/tex]
We are interested on this probability
[tex]P(X>13.17)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367 [/tex]
Part b
Since the distribution for X is normal then we can conclude that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we want this probability:
[tex] P(\bar X >13.17)[/tex]
The new z score is given by:
[tex] z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207 [/tex]
