Respuesta :
Answer:
[tex]P(\bar X <2.2)=P(Z<\frac{2.2-2.25}{\frac{0.2}{\sqrt{12}}})=P(Z<-0.866)[/tex]
[tex]P(\bar X<2.2)= P(Z<-0.866)=0.193[/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".
Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,\sigma)[/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(2.25,\frac{0.2}{\sqrt{12}})[/tex]
Solution to the problem
We want this probability:
[tex]P(\bar X<2.2)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X <2.2)=P(Z<\frac{2.2-2.25}{\frac{0.2}{\sqrt{12}}})=P(Z<-0.866)[/tex]
[tex]P(\bar X<2.2)= P(Z<-0.866)=0.193[/tex]
