SOLUTION: The salaries of the employees of a corporation are normally distributed with a mean of $58,750 and a standard deviation of $4700.
SHOW THE WORK TO GET YOUR ANSWER!
A. What is
Algebra.Com
Question 1197729: The salaries of the employees of a corporation are normally distributed with a mean of $58,750 and a standard deviation of $4700.
SHOW THE WORK TO GET YOUR ANSWER!
A. What is the probability that a randomly selected employee will have a salary of at least $63,000?
B. What is the probability that a randomly selected employee will have a salary less that $50,000?
C. What are the minimum and maximum salaries of the middle 95% of the employees?
D. What would be the salary of an employee have to be to be in the top 5% of all employees?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean = 58,750.
standard deviation = 4,700.
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation.
A. What is the probability that a randomly selected employee will have a salary of at least $63,000?
z = (63,000 - 58,750) / 4,700 = .9043 rounded to 4 decimal places.
using the ti-84 plus, i find that the area to the right of that z-score = .1829 rounded to 4 decimal places.
i confirmed with the online calculator at https://davidmlane.com/hyperstat/z_table.html that this is correct.
B. What is the probability that a randomly selected employee will have a salary less that $50,000?
z = (50,000 - 58750) / 4700 = -1.8617 rounded to 4 decimal places.
using the ti-84 plus, i find that the area to the left of that z-score = .0313 rounded to 4 decimal places.
i confirmed with the online calculator at https://davidmlane.com/hyperstat/z_table.html that this is correct.
C. What are the minimum and maximum salaries of the middle 95% of the employees?
middle 95% of employees would have z-score with .025 area to the left of it on the low side and .025 area to the right of it on the high side.
using the ti-84 plus, i get the low side z-score equal to -1.9600 rounded to 4 decimal places, and i get the high side z-score equal to 1.9600 rounded to 4 decimal places.
the area to the left of z-score of -1.9600 is equal to .025.
the area to the left of z-score of 1.9600 is equal to .975.
the area in between is equal to .975 minus .025 = .95.
use the z-score formula to find the raw score.
z = (x - m) / s becomes -1.9600 = (x - 58750) / 4700 on the left side of the interval.
solve for x to get x = -1.9600 * 4700 + 58750 = 49538.
z = (x - m) / s becomes 1.9600 = (x - 58750) / 4700 on the right side of the interval.
solve for x to get x = 1.9600 * 4700 + 58750 = 67962.
the middle 95% of the employees will make between 49538 and 67962 salary per year.
i confirmed with the online calculator at https://davidmlane.com/hyperstat/z_table.html that this is correct.
D. What would be the salary of an employee have to be to be in the top 5% of all employees?
using the t-84 plus, i get the z-score that has .05 area to the right of it equal to 1449.
the raw score will be based on the z-score formula as shown below:
4.6449 = (x - 58750) / 4700.
solve for x to get x = 1.6449 * 4700 + 58750 = 66481.03.
i confirmed with the online calculator at https://davidmlane.com/hyperstat/z_table.html that this is correct.
here are the results from using the calculator.
using the online calculator, you can work directly with the raw scores instead of the z-scores.
if you want to work with the z-scores, you enter a mean of 0 and a sandard deviation of 1.
if y ou want to work with the raw scores, you enter the mean and the standad deviation required by the problem.
you can also get the z-scores and the areas using the z-score table, but the calculators make it easier and they are accurate to more decimal places.
if you need to know how to use the tables, let me know and i'll give you a tutorial on that.
theo
RELATED QUESTIONS
I hope this is in the right category.
The monthly salaries in a company with 5000... (answered by Theo)
XYZ Corporation gives each of its employees an aptitude test. The scores on the test are... (answered by Boreal)
Suppose that, in a company with 5000 employees the monthly salaries are normally... (answered by oscargut)
starting salaries of marketing majors are normally distributed with a mean of 42,000 and... (answered by ewatrrr)
Assume that the salaries of elementary school teachers in the united states are normally... (answered by ewatrrr)
Assume that the salaries of elementary school teachers in the United States are normally... (answered by Boreal)
A set of data with a mean of 58 and a standard deviation of 6.3 is normally distributed. (answered by jim_thompson5910)
The mean salary of 350 workers at company A is $ 1050 and the standard deviation is $... (answered by Boreal)
A large corporation employs 26744 individuals. the average income several years ago for... (answered by Boreal)