Respuesta :
Answer:
[tex] z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028[/tex]
[tex] z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441[/tex]
An we can use the normal standard table and the following difference and we got this result:
[tex] P(-1.028<z<0.441)= P(z<0.441) -P(z<-1.028) = 0.670 -0.152 =0.518[/tex]
Step-by-step explanation:
Assuming this statement to complete the problem "with a standard deviation 5 mpg"
We have the following info given:
[tex]\mu = 34[/tex] represent the mean
[tex]\sigma= 5[/tex] represent the deviation
We have a sample size of n = 54 and we want to find this probability:
[tex] P(33.3 < \bar X< 34.3)[/tex]
And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:
[tex]\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]
And we can use the z score formula given by:
[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And replacing we got:
[tex] z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028[/tex]
[tex] z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441[/tex]
An we can use the normal standard table and the following difference and we got this result:
[tex] P(-1.028<z<0.441)= P(z<0.441) -P(z<-1.028) = 0.670 -0.152 =0.518[/tex]
