Respuesta :
Answer:
a) [tex]\mu_{\bar X}=80[/tex]
[tex]\sigma_{\bar X}=\frac{3}{\sqrt{36}}[/tex]
b) [tex]z=\frac{81 -80}{\frac{3}{\sqrt{36}}}=2[/tex]
c) [tex]P(\bar X <81)=P(\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}<2)=P(z<2)=0.977[/tex]
d) No, it would not be unusual because more than 5% of all such samples have means less than 81
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 a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(80,3)[/tex]
Where [tex]\mu=80[/tex] and [tex]\sigma=3[/tex]
And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(80,\frac{3}{\sqrt{36}})[/tex]
[tex]\mu_{\bar X}=80[/tex]
[tex]\sigma_{\bar X}=\frac{3}{\sqrt{36}}[/tex]
Part b
In order to find the z score we can use this formula:
[tex]z=\frac{\bar X -\mu_{\bar X}}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we replace on the before formula we got:
[tex]z=\frac{81 -80}{\frac{3}{\sqrt{36}}}=2[/tex]
Part c
We want this probability:
[tex]P(\bar X <81)=P(\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}<2)=P(z<2)=0.977[/tex]
Part d
If we see the probability obtained on part c we have that more than 90% of the samples have samples means less than 81. So the best answer is:
No, it would not be unusual because more than 5% of all such samples have means less than 81
