In a random sample of 144 games of the 2014 NFL regular season, mean passing yards per game were 222 with a standard deviation of 36 yards per game. Recalling that the standard error of a sample mean is the standard deviation divided by the square root of the sample size, what is the 95% confidence interval for the mean passing yards per game

The 95% confidence interval for the mean passing yards per game is between 216.12 yards and 227.88 yards.

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 M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the mean subtracted by M. So it is 222 - 5.88 = 216.12 yards

The upper end of the interval is the mean added to M. So it is 222 + 5.88 = 227.88 yards

The 95% confidence interval for the mean passing yards per game is between 216.12 yards and 227.88 yards.

option b. 4.9%

Explanation:

1) The probability that a dart thrown at random within the dart board will hit the bull's eye is equal to the ratio of the areas:

probability = area of the bull's eye / aera of the dart board.

2) Area of the dart board = π × (radius)² = π × (18 in)²

3) Area of the bull's eye = π × (radius)² = π × (4 in)²

4) So, the probaility is:

  π (4 in)²           4²
= = 0.0494
 π (18 in)²         18²

5) The percent probability is that number times 100: 0.0494 × 100 = 4.94%