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?

7%

Step-by-step explanation:

Given Data:

mean =$1000

standard deviation s = $370

number of employee n 80

weekly salary at most of $75

the value of z can be determined

z = 1.81

From standard table of z

P(z > 1.81) = 0.035

P(condition described) = 2*0.035 = 7%

Z DISTRIBUTION

h

step-by-step explanation:

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10