Two boxes labeled A and B are full of numbered tickets. We know that Box A has an average of 80 and an SD of 20. We also know that Box B has an average of 50 and an SD of 10. We will draw at random with replacement 100 times from Box A. Independently, we will draw at random with replacement 100 times from Box B. From Box A:
The sum of the 100 draws expected value equals SE equals
The average of the 100 draws expected value equals SE equals
From Box B:
The sum of the 100 draws expected value equals SE equals
The average of the 100 draws expected value equals SE equals
The difference between the sums of the draws: expected value equals SE equals (decimal form, three places) The difference between the averages of the draws: expected value equals SE equals (decimal form, three places).

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

a≈380.13mi² would be your answer if the radius was 11 miles

is there any answers choices? l can try to you if there is answer choices to this question,so sorry