3.28 chickenpox, part ii. we learned in exercise 3.26 that about 90% of american adults had chickenpox before adulthood. we now consider a random sample of 120 american adults. (a) how many people in this sample would you expect to have had chickenpox in their childhood? and with what standard deviation? (b) would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) what is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? how does this probability relate to your answer to part (b)?

my answer got deleted but it's a

step-by-step explanation:

you would need to also lose 1 point for each of the other outcomes for it to be fair.

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.