Question

Sample annual salaries​ (in thousands of​ dollars) for employees at a company are listed. ​(a) Find the sample mean and sample standard deviation. ​(b) Each employee in the sample is given a ​% raise. Find the sample mean and 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 sample standard deviation for the revised data set. ​(d) What can you conclude from the results of​ (a), (b), and​ (c)?

Answer 1

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

The value is attached in the image file please find it.

In point a:

First, we calculate the find the mean,

Formula:

$\to Mean (\bar{x}) = \frac{( \sum x)}{n}$

                   $=\frac{540}{13}\\\\ = 41.54$

To calculate the standard deviation, subtract the mean value from all observations then square its value:

$SD = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$   $= \sqrt{ \frac{339.2308}{13-1}}$

                             $= \sqrt{ \frac{339.2308}{12}}\\\\= \sqrt{28.269}\\\\=5.31$

please find attached file

In point b:

New $x_i = 1 x_i + 0.05 x_i = 1.05 x_i$

Calculate new mean:

$\to \bar{x} = \frac{\sum x}{n}$

       $=\frac{567}{13} \\\\= 43.62$

calculating the standard deviation:

$SD = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$   $= \sqrt{ \frac{374.0022}{13-1}}$

                             $= \sqrt{ \frac{374.0022}{12}}\\\\= \sqrt{31.16}\\\\=5.58$

please find attached file

In point C:

Calculate new mean:

$\to \bar{x} = \frac{\sum x}{n}$

       $=\frac{47.25}{13} \\\\= 3.46$

calculating the standard deviation:

$SD = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$   $= \sqrt{ \frac{2.5975}{13-1}}$

                             $= \sqrt{ \frac{2.5975}{12}}\\\\= \sqrt{0.216}\\\\=0.46$

please find attached file

In point d:

for b,

New $\bar{x}= 1.05 \bar{x}= 1.05(41.54) =43.62$

New$s= 1.05s= 1.05(5.31)=5.57$

for c,

New$\bar{x}= \frac{ \bar{x}}{12}= \frac{41.54}{12} = 3.46$

New $s = \frac{s}{ 12} = \frac{5.31}{12}= 0.44$