Question

Sample annual salaries (in thousands of dollars) for employees at a company are listed. 42 36 48 51 39 39 42 36 48 33 39 42 45 (a) Find the sample mean and the sample standard deviation. (b) Each employee in the sample receives a 5% raise. Find the sample mean and the sample standard deviation for the revised data set. (c) To calculate the monthly salary, divide each original salary by 12. 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

Answer:

a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

$\bar X = 41.538$

$s = 5.317$

b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

$\bar X = 43.615$

$s= 5.583$

c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

$\bar X= 3.462$

$s = 0.443$

d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.

Step-by-step explanation:

For this case we can use the following formulas for the mean and standard deviation:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Part a

Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45

And if we calculate the mean we got:

$\bar X = 41.538$

$s = 5.317$

Part b

For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:

44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25

And if we calculate the new mean and deviation we got:

$\bar X = 43.615$

$s= 5.583$

Part c

The new dataset would be each value divided by 12 so we have:

3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75

And the new mean and deviation would be:

$\bar X= 3.462$

$s = 0.443$

Part d

As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).

And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.

And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.