Questions: Part 5 of 7 Points: 0.43 of 1 Sample annual salaries (in thousands of dollars) for employees at a company are listed. 52, 38, 50, 58, 40, 40, 52, 38, 50, 27, 58, 52, 43 (a) Find the sample mean and sample standard deviation. (b) Each employee in the sample is given a 6% 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)? (a) The sample mean is x̄=46 thousand dollars. The sample standard deviation is s=9.1 thousand dollars. (b) The sample mean is x̄=48.8 thousand dollars. The sample standard deviation is s=9.7 thousand dollars. (c) The sample mean is x̄= thousand dollars.

Part 5 of 7
Points: 0.43 of 1
Sample annual salaries (in thousands of dollars) for employees at a company are listed.
52, 38, 50, 58, 40, 40, 52, 38, 50, 27, 58, 52, 43
(a) Find the sample mean and sample standard deviation.
(b) Each employee in the sample is given a 6% 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)?
(a) The sample mean is x̄=46 thousand dollars.
The sample standard deviation is s=9.1 thousand dollars.
(b) The sample mean is x̄=48.8 thousand dollars.
The sample standard deviation is s=9.7 thousand dollars.
(c) The sample mean is x̄= thousand dollars.

Transcript text: Part 5 of 7 Points: 0.43 of 1 Sample annual salaries (in thousands of dollars) for employees at a company are listed. $\begin{array}{lllllllllllll}52 & 38 & 50 & 58 & 40 & 40 & 52 & 38 & 50 & 27 & 58 & 52 & 43\end{array}$ (a) Find the sample mean and sample standard deviation. (b) Each employee in the sample is given a $6 \%$ 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)? (a) The sample mean is $\bar{x}=46$ thousand dollars. The sample standard deviation is $s=9.1$ thousand dollars. (b) The sample mean is $\bar{x}=48.8$ thousand dollars. The sample standard deviation is $\mathrm{s}=9.7^{7}$ thousand dollars. (c) The sample mean is $\bar{x}=\square$ thousand dollars.

Solution

Solution Steps

To solve the given problem, we need to follow these steps:

(a) Calculate the sample mean and sample standard deviation for the given salaries. (b) Apply a 6% raise to each salary and then calculate the new sample mean and sample standard deviation. (c) Convert each salary to a monthly salary by dividing by 12, and then calculate the sample mean and sample standard deviation for the revised data set.

Step 1: Calculate the Sample Mean and Sample Standard Deviation for the Given Salaries

Given the salaries in thousands of dollars: \[ \{52, 38, 50, 58, 40, 40, 52, 38, 50, 27, 58, 52, 43\} \]

The sample mean ($\bar{x}$) is calculated as: \[ \bar{x} = 46.0 \]

The sample standard deviation ($s$) is calculated as: \[ s = 9.1196 \]

Step 2: Apply a 6% Raise and Calculate the New Sample Mean and Sample Standard Deviation

Each salary is increased by 6%, resulting in the revised salaries: \[ \{55.12, 40.28, 53.00, 61.48, 42.40, 42.40, 55.12, 40.28, 53.00, 28.62, 61.48, 55.12, 45.58\} \]

The new sample mean ($\bar{x}$) is: \[ \bar{x} = 48.76 \]

The new sample standard deviation ($s$) is: \[ s = 9.6668 \]

Step 3: Convert to Monthly Salaries and Calculate the Sample Mean and Sample Standard Deviation

Each original salary is divided by 12 to convert to monthly salaries: \[ \{4.3333, 3.1667, 4.1667, 4.8333, 3.3333, 3.3333, 4.3333, 3.1667, 4.1667, 2.2500, 4.8333, 4.3333, 3.5833\} \]

The sample mean ($\bar{x}$) for the monthly salaries is: \[ \bar{x} = 3.8333 \]

The sample standard deviation ($s$) for the monthly salaries is: \[ s = 0.7600 \]