Answer:
We need to survey at least 217 students.
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.
How many randomly selected Foothill College students must be surveyed?
We need to survey at least n students.
n is found when
. So






Rounding up
We need to survey at least 217 students.