Answer:
We need a sample size of at least 101
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.
What sample size is needed to estimate the true average speed to within 2 mph at 99% confidence?
We need a sample of at least n.
n is found when M = 2. So






Rounding up
We need a sample size of at least 101