An experiment to determine the most effective way to teach safety principles applied four different teaching methods. some employees were given programmed instruction booklets and worked through the course at their own pace. other employees attended lectures. a third group watched a television presentation, and a fourth group was divided into small discussion groups. a high of 10 was possible. a sample of five tests was selected from each group. the test grade results were:

i belive those are the ones that are under 90 degreese.

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.