Answer:
The 95% confidence interval for the mean amount of the increase is ($541.6, $588.4). This means that we are 95% sure that the mean amount of increase of all customers who charge at least $3,000 in a year is between these two values.
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 Z-table as such z has a p-value of
.
That is z with a pvalue of
, so Z = 1.96.
Now, find the margin of error 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 565 - 23.4 = $541.6
The upper end of the interval is the sample mean added to M. So it is 565 + 23.4 = $588.4
The 95% confidence interval for the mean amount of the increase is ($541.6, $588.4). This means that we are 95% sure that the mean amount of increase of all customers who charge at least $3,000 in a year is between these two values.