Question

Sample annual salaries​ (in thousands of​ dollars) for employees at a company are listed. 44 33 49 50 41 41 44 33 49 34 50 44 54 ​.

(a) Find the sample mean and sample standard deviation.
​(b) Each employee in the sample is given a ​$3000 raise. Find the sample mean and sample standard deviation for the revised data set.
​(c) Each employee in the sample takes a pay cut of ​$4000 from their original salary. Find the sample mean and the sample standard deviation for the revised data set.
​(d) What can you conclude from the results of​ (a), (b), and​ (c)?

Answer 1

The sample mean is 43.538 and the sample standard deviation is 7.956.

Step-by-step explanation:

(a) To find the sample mean, you add up all the salaries and divide by the number of salaries in the sample. In this case, you add up all the salaries: 44 + 33 + 49 + 50 + 41 + 41 + 44 + 33 + 49 + 34 + 50 + 44 + 54. which equals 566. Then, you divide by the number of salaries, which is 13. So, the sample mean is 566/13 = 43.538.

To find the sample standard deviation, you first need to find the mean of the data set. You have already found it in the previous step which is 43.538. Next, subtract the mean from each of the data points, square the result, and add up all the squared differences. In this case, subtract the mean from each of the salaries: 44 - 43.538, 33 - 43.538, 49 - 43.538, 50 - 43.538, 41 - 43.538, 41 - 43.538, 44 - 43.538, 33 - 43.538, 49 - 43.538, 34 - 43.538, 50 - 43.538, 44 - 43.538, 54 - 43.538. Then, square the results: (-0.538)^2, (-10.538)^2, (5.462)^2, (6.462)^2, (-2.538)^2, (-2.538)^2, (-0.538)^2, (-10.538)^2, (5.462)^2, (-9.538)^2, (6.462)^2, (-0.538)^2, (10.462)^2. Finally, add up all the squared differences and divide by the number of salaries minus 1, then take the square root of the result. You should get a sample standard deviation of approximately 7.956.