Suppose that a baseball player has been a .250 hitter for his career, which means that his probability of a hit (success) has been 0.250. Then during one winter the player genuinely improves to the point that his probability of success improves to 0.333. You will investigate how likely the player is, in a sample of at-bats, to convince the manager that he has improve

366

step-by-step explanation:

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.

-1/4 > -1 2/3. -1/4 is greater than -1 2/3 because it is closer to 0 on the number line.

to be honest i don't know but i need points so