Respuesta :
Answer:
a)[tex]P(X>300)=P(\frac{X-\mu}{\sigma}>\frac{300-\mu}{\sigma})=P(Z>\frac{300-300}{25})=P(z>0)= 0.5[/tex]
[tex]P(X>335)=P(\frac{X-\mu}{\sigma}>\frac{335-\mu}{\sigma})=P(Z>\frac{335-300}{25})=P(z>1.4)=0.0808[/tex]
b)[tex]P(\bar X>300)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{300-\mu}{\sigma_{\bar x}})=P(Z>\frac{300-300}{17.5})=P(z>0)= 0.5[/tex]
[tex]P(\bar X>335)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{335-\mu}{\sigma_{\bar x}})=P(Z>\frac{335-300}{17.5})=P(z>2)=0.0228[/tex]
Step-by-step explanation:
Assuming the following questions:
a) Choose one twelfth-grader at random. What is the probability that his or her score is higher than 300? Higher than 335?
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".
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(300,35)[/tex]
Where [tex]\mu=300[/tex] and [tex]\sigma=35[/tex]
We are interested on this probability
[tex]P(X>300)[/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>300)=P(\frac{X-\mu}{\sigma}>\frac{300-\mu}{\sigma})=P(Z>\frac{300-300}{25})=P(z>0)= 0.5[/tex]
We find the probabilities with the normal standard table or with excel.
And for the other case:
[tex]P(X>335)=P(\frac{X-\mu}{\sigma}>\frac{335-\mu}{\sigma})=P(Z>\frac{335-300}{25})=P(z>1.4)=0.0808[/tex]
b) Now choose an SRS of four twelfth-graders. What is the probability that his or her mean score is higher than 300? Higher than 335?
For this case since the distribution for X is normal then the distribution for the sample mean is also normal and given by:
[tex] \bar X = \sim N(\mu = 300, \sigma_{\bar x} = \frac{35}{\sqrt{4}}=17.5)[/tex]
The new z score is given by:
[tex]z=\frac{\bar X -\mu}{\sigma_{\bar x}}[/tex]
And using the formula we got:
[tex]P(\bar X>300)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{300-\mu}{\sigma_{\bar x}})=P(Z>\frac{300-300}{17.5})=P(z>0)= 0.5[/tex]
We find the probabilities with the normal standard table or with excel.
And for the other case:
[tex]P(\bar X>335)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{335-\mu}{\sigma_{\bar x}})=P(Z>\frac{335-300}{17.5})=P(z>2)=0.0228[/tex]
