The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. Suppose you want to calculate the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75. Which distribution would you use to solve the problem?
Accepted Solution
A:
Answer:7%Step-by-step explanation:Given Data:mean =$1000standard deviation s = $370number of employee n 80weekly salary at most of $75the value of z can be determined [tex]n = [z\times \frac{s}{E}]^2[/tex][tex]80 = [z \times \frac{370}{75}]^2[/tex][tex]\sqrt(80) = 4.93z[/tex]z = 1.81From standard table of z P(z > 1.81) = 0.035P(condition described) = 2*0.035 = 7%Z DISTRIBUTION