Suppose that the population of the scores of all high school seniors that took the SAT-M (SAT math) test this year follows a normal distribution with unknown population mean and…

Question

asked Oct 2, 2021 222k views

1 Answer

Daren Thomas · Answer 1 · 2021-10-09T04:09:14+0000

Answer:

$25.75 = z_(\alpha/2) (100)/(√(100))$

And solving for the critical value we got:

$z_(\alpha/2)= (25.75*10)/(100) = 2.575$

Now we need to find the confidence level and for this case we can use find this probability:

P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)

And using the normal standard distribution or excel we got:

P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)= 0.9950-0.0050= 0.99

So then the confidence interval for this case is 99%

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:

$X\sim N(\mu , \sigma =100)$

They select a random sample of n =100 and they construc a confidence interval for the true population mean of interest and they got:

$512.00 \pm 25.75$

for this problem we need know that the confidence interval for the true mean when the deviation is known is given by:

$\bar X \pm z_(\alpha/2)(\sigma)/(√(n))$

The margin of error is given by:

$ME =z_(\alpha/2) (\sigma)/(√(n))$

And the margin of error for this interval is
ME = 25.75 then we can solve for the critical value in order to find the confidence level: