the average number of miles that 90 truckers drove in a day was 540
with a standard deviation of 40 miles. What is the 99% confidence
interval of the true mean number of miles driven by all truckers?

Hi!!


We want that the confidence level be 0.99 . So, for a standard normal
variable we need to find Q that:
P(Z > Q) = 0.005 or 0.5% 

Due the symmetry of the normal distribution around 0, if Q is such that 
P(Z > Q) = 0.005 , 
then we will have that
P(Z < -Q) = 0.005. 

Now we can conclude that 
P(-Q < Z < Q) = 1 - P(Z > Q) - P(Z < -Q) =
              = 1 - 0.005 -0.005 =
              = 0.99
This means that the interval (-Q,Q) defines a 99% confidence interval.

Using the table after the following link we find that the value 0,005
correspond to a Z equal to -2.58, this is our -Q, then Q is +2.58:
"Normal Curve Table"
http://www.stat.psu.edu/~rho/stat200/normtable.html


Then for a standard normal variable (-2.58, +2.58) defines a 99%
confidence interval.

Having a large number n of samples from the population (say n > 30)
and knowing Mean and STD (standard deviation) we can calculate the
extremes of the 95% confidence interval by the formula:

Extremes (99% confidence) = Mean +/- (2.58 * STD)/sqrt(n)

For this problem we have that:
n = 90 > 30
Mean = 540 miles
STD = 40 miles


Extreme 1 = 540 + (2.58 * 40)/sqrt(90) = 550.88 miles
Extreme 2 = 540 - (2.58 * 40)/sqrt(90) = 529.12 miles

The 99% confidence interval is (550.88, 529.12) .

I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards.
livioflores-ga