A) for 30 randomly selected rolling stones concerts, the mean gross earnings is 2.79 million dollars. assuming a population standard deviation gross earnings of 0.47 million dollars, obtain a 99% confidence interval for the mean gross earnings of all rolling stones concerts (in millions). confidence interval: ( , )

The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).

Step-by-step explanation:

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 mean subtracted by M. So it is 2.79 - 0.2210 = 2.5690 million dollars.

The upper end of the interval is the mean added to M. So it is So it is 2.79 + 0.2210 = 3.0110 million dollars.

The 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions) is (2.5690, 3.0110).

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest