A simple random sample of 35 colleges and universities in the United States has a mean tuition of $17,700 with a standard deviation of $10,300. Construct a 95% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.

Step-by-step explanation:

Sample size greater than 30, which means that we use the normal distribution to find the interval.

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 17700 - 3412 = $14,288

The upper end of the interval is the sample mean added to M. So it is 17700 + 3412 = $21,112.

The 95% confidence interval for the mean tuition for all colleges and universities in the United States is between $14,288 and $21,112.

10/60

step-by-step explanation:

if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6

u bout ugly as   a mf

step-by-step explanation: lol