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

$(73 - 2 \cdot 11, 73 + 2 \cdot 11) = (51,95)$

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:

$\sigma = \dfrac{ 11}{\sqrt{121}} = 1$

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

$(73 - 2 \cdot 1, 73 + 2 \cdot 1) = (71,75)$

That's probably the answer they're looking for.

4 0

2 years ago

A bag contains 3 triangles, 4 squares, 5 pentagons, and 6 hexagons. If one shape is randomly chosen, what is the probability tha

e-lub [12.9K]

I’d say that it’s definitely random choice for the squares, pentagons, and hexagons. But, if we are looking at how common it would be between the 3 of them, then I would say that it would be one of the hexagons, because there are more hexagons than any of the other shapes in the bag.

3 0

2 years ago