4. (07.05)
a recent survey by the american cancer society shows that the probability that someone is a smoker is p(s) = 0.19. it
has also determined that the probability that someone has lung cancer, given that he or she is a smoker, is p(los) =
0.158. what is the probability (rounded to the nearest hundredth) that a random person is a smoker and has lung cancer
p(
s lc)? (4 points)
0.03
0.83
0.35
0.02

can you rephrase the question ? it is unclear : -)

step-by-step explanation:

step-by-step explanation:

the answer is 128

d. corresponding side lengths in similar triangles are proportional in length.

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.