Respuesta :
Answer:
[tex]P(38<\bar X<44)=0.498[/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=41,\sigma=34)[/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(41,\frac{34}{\sqrt{58}})[/tex]
We are interested on this probability
[tex]P(38<\bar X<44)[/tex]
If we apply the Z score formula to our probability we got this:
[tex]P(38<\bar X<44)=P(\frac{38-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{44-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]
[tex]=P(\frac{38-41}{\frac{34}{\sqrt{58}}}<Z<\frac{44-41}{\frac{34}{\sqrt{58}}})=P(-0.672<z<0.672)[/tex]
And we can find this probability on this way:
[tex]P(-0.672<z<0.672)=P(z<0.672)-P(z<-0.672)[/tex]
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
[tex]P(-1.2<z<2.35)=P(z<2.35)-P(z<-1.2)=0.749-0.251=0.498[/tex]
