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

Follows are the solution to the given points:

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}) = (( \sum x))/(n)`

`=(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{ (339.2308)/(13-1)}`

`= \sqrt{ (339.2308)/(12)}\\\\= √(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} = (\sum x)/(n) \\`

`=(567)/(13) \\\\= 43.62`

calculating the standard deviation:

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

`= \sqrt{ (374.0022)/(12)}\\\\= √(31.16)\\\\=5.58`

please find attached file

In point C:

Calculate new mean:

`\to \bar{x} = (\sum x)/(n) \\`

`=(47.25)/(13) \\\\= 3.46`

calculating the standard deviation:

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

`= \sqrt{ (2.5975)/(12)}\\\\= √(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}= (41.54)/(12) = 3.46`

New
`s = (s)/( 12) = (5.31)/(12)= 0.44`