Respuesta :
Answer:
[tex] 25.75 = z_{\alpha/2} \frac{100}{\sqrt{100}}[/tex]
And solving for the critical value we got:
[tex] z_{\alpha/2}= \frac{25.75*10}{100} = 2.575[/tex]
Now we need to find the confidence level and for this case we can use find this probability:
[tex] P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)[/tex]
And using the normal standard distribution or excel we got:
[tex] P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)= 0.9950-0.0050= 0.99[/tex]
So then the confidence interval for this case is 99%
Step-by-step explanation:
For this case the random variable X is the scores for the SAT math scores and we know that the distribution for X is normal:
[tex] X\sim N(\mu , \sigma =100)[/tex]
They select a random sample of n =100 and they construc a confidence interval for the true population mean of interest and they got:
[tex]512.00 \pm 25.75[/tex]
for this problem we need know that the confidence interval for the true mean when the deviation is known is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]
The margin of error is given by:
[tex] ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And the margin of error for this interval is [tex] ME = 25.75[/tex] then we can solve for the critical value in order to find the confidence level:
[tex] 25.75 = z_{\alpha/2} \frac{100}{\sqrt{100}}[/tex]
And solving for the critical value we got:
[tex] z_{\alpha/2}= \frac{25.75*10}{100} = 2.575[/tex]
Now we need to find the confidence level and for this case we can use find this probability:
[tex] P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)[/tex]
And using the normal standard distribution or excel we got:
[tex] P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)= 0.9950-0.0050= 0.99[/tex]
So then the confidence interval for this case is 99%
