Respuesta :
Answer:
[tex]P(\bar X >86)=P(Z>\frac{86-80}{\frac{12}{\sqrt{9}}}=1.5)[/tex]
And using a calculator, excel ir the normal standard table we have that:
[tex]P(Z>1.5)=0.0668[/tex]
TRUE.
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 variable of interest of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(80,12)[/tex]
Where [tex]\mu=80[/tex] and [tex]\sigma=12[/tex]
Since S follows a normal distribution then we know 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]
We can find the probability of interest with the following expression:
[tex]P(\bar X >86)=P(Z>\frac{86-80}{\frac{12}{\sqrt{9}}}=1.5)[/tex]
And using a calculator, excel ir the normal standard table we have that:
[tex]P(Z>1.5)=0.0668[/tex]
TRUE.
