Answer:
The 99% confidence interval for the average score in the population is between 99.24 and 112.44.
The margin of error is 6.60.
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 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 mean subtracted by M. So it is 105.84 - 6.60 = 99.24
The upper end of the interval is the mean added to M. So it is 105.84 + 6.60 = 112.44
The 99% confidence interval for the average score in the population is between 99.24 and 112.44.