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 known standard deviation 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 population mean is 512.00 ± 25.75." The confidence level for this interval is

And solving for the critical value we got:

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

And using the normal standard distribution or excel we got:

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:

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

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

The margin of error is given by:

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

And solving for the critical value we got:

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

And using the normal standard distribution or excel we got:

So then the confidence interval for this case is 99%

answer4 to the second power is 16

and 4 to the third power is 64,

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)