Answer:
A sample size of 554 is needed.
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
.
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.
Standard deviation is known to be $12,000
This means that 
What sample size do you need to have a margin of error equal to $1000, with 95% confidence?
This is n for which M = 1000. So



Dividing both sides by 1000:



Rounding up:
A sample size of 554 is needed.