A random sample of 8 recent college graduates found that starting salaries for computer scientists had a mean of $85000 and a standard deviation of $15000. Construct a 95% confidence interval for the average starting salary of all computer scientists in Gainesville.

the answer is d. 75%

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.

y=35

x=5

step-by-step explanation:

x=-2

step-by-step explanation:

28-32x=92

-32x=92-28

-32x=64

x=64/-32

x=-2