121 is big enough to assume normality and not worry about the t distribution. By the 68-95-99.7 rule a 95% confidence interval includes plus or minus two standard deviations. So 95% of the cars will be in the mph range
[tex](73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)[/tex]
The question is a bit vague, but it seems we're being asked for the 95% confidence interval on the average of 121 cars. The 121 is a hint of course.
The standard deviation of the average is in general the standard deviation of the individual samples divided by the square root of n:
[tex]\sigma = \dfrac{ 11}{\sqrt{121}} = 1[/tex]
So repeating our experiment of taking the average 121 cars over and over, we expect 95% of the averages to be in the mph range
[tex](73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75) [/tex]
That's probably the answer they're looking for.
