Question

sample annual salaries (in thousands of dollars) for employees at a company are listed.
37 32 49 63 30 30 37 32 49 26 63 37 46
(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 \\( \overline { x } = 40.8 \\) thousand dollars.
(round to one decimal place as needed.)
the sample standard deviation is \\( s = 12.2 \\) thousand dollars.
(round to one decimal place as needed.)
(b) the sample mean is \\( \overline { x } = 41.7 \\) thousand dollars,
(round to one decimal place as needed.)

Explanation:

Step1: Calculate sample mean for part (a)

The formula for sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Given data \(x=\{37,32,49,63,30,30,37,32,49,26,63,37,46\}\), \(n = 13\).
\(\sum_{i=1}^{13}x_{i}=37+32 + 49+63+30+30+37+32+49+26+63+37+46=530\)
\(\bar{x}=\frac{530}{13}\approx40.8\)

Step2: Calculate sample variance for part (a)

The formula for sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\((37 - 40.8)^{2}=(- 3.8)^{2}=14.44\), \((32-40.8)^{2}=(-8.8)^{2}=77.44\), \((49 - 40.8)^{2}=(8.2)^{2}=67.24\), \((63-40.8)^{2}=(22.2)^{2}=492.84\), \((30 - 40.8)^{2}=(-10.8)^{2}=116.64\), \((30 - 40.8)^{2}=(-10.8)^{2}=116.64\), \((37 - 40.8)^{2}=(-3.8)^{2}=14.44\), \((32-40.8)^{2}=(-8.8)^{2}=77.44\), \((49 - 40.8)^{2}=(8.2)^{2}=67.24\), \((26-40.8)^{2}=(-14.8)^{2}=219.04\), \((63-40.8)^{2}=(22.2)^{2}=492.84\), \((37 - 40.8)^{2}=(-3.8)^{2}=14.44\), \((46-40.8)^{2}=(5.2)^{2}=27.04\)
\(\sum_{i = 1}^{13}(x_{i}-\bar{x})^{2}=14.44+77.44+67.24+492.84+116.64+116.64+14.44+77.44+67.24+219.04+492.84+14.44+27.04 = 1900.4\)
\(s^{2}=\frac{1900.4}{12}\approx158.4\)
\(s=\sqrt{158.4}\approx12.2\)

Step3: For part (b)

If each data point \(x_{i}\) is multiplied by \(k = 1.06\) (a \(6\%\) raise), the new mean \(\bar{y}=k\bar{x}\) and new standard deviation \(s_{y}=ks\)
\(\bar{y}=1.06\times40.8 = 43.248\approx43.2\) (There seems to be a calculation error in the provided answer. Using the formula: if original \(\bar{x}=40.8\), after \(6\%\) raise \(\bar{x}_{new}=(1 + 0.06)\times40.8=43.248\approx43.2\)). The standard deviation \(s_{new}=1.06\times12.2 = 12.932\approx12.9\)

Step4: For part (c)

If each data point \(x_{i}\) is divided by \(k = 12\), the new mean \(\bar{z}=\frac{\bar{x}}{12}\) and new standard deviation \(s_{z}=\frac{s}{12}\)
\(\bar{z}=\frac{40.8}{12}=3.4\), \(s_{z}=\frac{12.2}{12}\approx1.0\)

Answer:

(a) The sample mean \(\bar{x}\approx40.8\) thousand dollars and sample standard deviation \(s\approx12.2\) thousand dollars.
(b) The sample mean \(\bar{x}\approx43.2\) thousand dollars and sample standard deviation \(s\approx12.9\) thousand dollars.
(c) The sample mean \(\bar{x}\approx3.4\) thousand dollars and sample standard deviation \(s\approx1.0\) thousand dollars.
(d) When data is multiplied by a constant \(k\), the mean and standard deviation are multiplied by \(k\). When data is divided by a constant \(k\), the mean and standard deviation are divided by \(k\).