Suppose that the population of the scores of all high school seniors who took the sat math (sat-m) test this year follows a normal distribution with standard deviation σ = 100.
you read a report that says, "on the basis of a simple random sample of 100 high school seniors that took the sat-m test this year, a confidence interval for μ is found to be 512.00 ± 25.76."

what was the confidence level used to calculate this confidence interval?

Confidence =

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

Assuming the X follows a normal distribution

The distribution for the sample mean is given by:

represent the sample mean

represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

  (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

And we can find the critical value like this:

And we know that on the right tail of the z score =2.576 we have of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998=, so then , and then we can find the confidence like this:

Confidence =

And then the confidence level would be given by 99%

umumumumum

im not sure