The average annual precipitations (in inches) of random sample of 30 years in San Francisco, California, have a sample mean of 40.12 inches and sample standard deviation of 8.18 inches. Find an 80% confidence interval for the average annual precipitations. a. (39.13, 136.11) b. (38.16, 42.08) c. (8.18, 40.12) d. None of these

do you have a picture of the table

46

step-by-step explanation:

given

height = 168 cm

formula

height =2.47*f+54.13

solution

168 = 2.47f + 54.13             subtract 54.13 from both sides

168 - 54.13   = 2.47f             combine

113.87 = 2.47f                     divide by 2.47

113.87 / 2.47 = f                   combine

46.1 = f  

answer: 46 cm

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.