Respuesta :
Answer:
[tex]z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51[/tex]
[tex]z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51[/tex]
Then we can find the probability of interest with this difference:
[tex]P(-0.51<z<0.51)=P(z<0.51)-P(z<-0.51)[/tex]
And using the normal standard distribution or excel we got:
[tex]P(-0.51<z<0.51)=P(z<0.51)-P(z<-0.51)=0.6950 -0.3050= 0.390[/tex]
So then the probability that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390
Step-by-step explanation:
We define the variable of interest as the per capita income and we know the following properties for this variable:
[tex]\mu=20908[/tex] and [tex]\sigma=407[/tex]
We want to find this probability:
[tex]P(20908-28<\bar X<20908+28) = P(20880< \bar X< 20936)[/tex]
We select a sample size of n=55 and we define the z score formula given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
We can find the z score then for 20880 and 20936 and we got:
[tex]z=\frac{20880-20908}{\frac{407}{\sqrt{55}}}= -0.51[/tex]
[tex]z=\frac{20936-20908}{\frac{407}{\sqrt{55}}}= 0.51[/tex]
Then we can find the probability of interest with this difference:
[tex]P(-0.51<z<0.51)=P(z<0.51)-P(z<-0.51)[/tex]
And using the normal standard distribution or excel we got:
[tex]P(-0.51<z<0.51)=P(z<0.51)-P(z<-0.51)=0.6950 -0.3050= 0.390[/tex]
So then the probability that the sample mean would differ from the true mean by less than 28 dollars from the sample of 55 is approximately 0.390
