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

While playing volleyball, Ralph drink 3/4 of a jug of water. If the jughead 64 fluid ounces of water, how many fluid ounces of w

MaRussiya [10]

Answer:

Step-by-step explanation:

4 0

2 years ago

Find the area for a trapezoid 9.85 m 15 m 11 m 9 m

igor_vitrenko [27]

Answer:

total area = 99m² + 39.4m²

= 138.4m²

7 0

2 years ago