Answer:
The 90% confidence interval would be between 73.80 and 82.20.
Explanation:
95% confidence 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 the margin of errorM as such

In which
is the standard deviation of the population and n is the size of the sample.
We have that M = 5, n = 64. We have to find the standard deviation of the population. So





90% confidence 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 78 - 4.20 = 73.80.
The upper end of the interval is the sample mean added to M. So it is 78 + 4.20 = 82.20
The 90% confidence interval would be between 73.80 and 82.20.