# The monthly incomes for 12 randomly selected​ people, each with a​ bachelor's de

The monthly incomes for 12 randomly selected​ people, each with a​ bachelor's degree in​ economics, are shown on the right. Complete parts​ (a) through​ (c) below.
Assume the population is normally distributed.
a) Find the sample mean.
b) Find the sample standard deviation
c) Construct a $$\displaystyle{95}\%$$ confidence interval for the population mean $$\displaystyle\mu$$. A $$\displaystyle{95}\%$$ confidence interval for the population mean is
$$\begin{array}{|c|c|}\hline 4450.42 & 4596.96 & 4366.46 \\ \hline 4455.62 & 4151.52 & 3727.77 \\ \hline 4283.26 & 4527.94 & 4407.68 \\ \hline 3946.49 & 4023.61 & 4221.73\\ \hline \end{array}$$

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Step 1
$$\begin{array}{|c|c|}\hline X & X-\bar{X} & (X-\bar{X})^{2} \\ \hline 4450.42 & 187.1292 & 35017.33 \\ \hline 4596.96 & 33.6692 & 111335.1 \\ \hline 4366.46 & 103.1692 & 10643.88 \\ \hline 4455.65 & 192.3592 & 37002.05 \\ \hline 4151.52 & -111.771 & 12492.72 \\ \hline 3727.77 & -535.521 & 286782.6 \\ \hline 4283.26 & 19.96917 & 398.7676 \\ \hline 4527.94 & 264.6492 & 70039.18 \\ \hline 4407.68 & 144.3892 & 20848.23 \\ \hline 3946.49 & -316.801 & 100362.8 \\ \hline 4023.61 & -239.681 & 57446.9 \\ \hline 4221.73 & -41.5608 & 1727.303 \\ Total=51159.49 & & Total=744096.8 \\ \hline \end{array}$$
Step 2
a) Sample mean:
$$\displaystyle\overline{{{x}}}={\frac{{\sum{X}}}{{{n}}}}$$
$$\displaystyle={\frac{{{51159.49}}}{{{12}}}}$$
$$\displaystyle={4263.291}$$
The sample mean is 4263.291
Step 3
b) Sample Standard Deviation
$$\displaystyle{s}{d}=\sqrt{{{\frac{{\sum{\left({X}-\overline{{{X}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}$$
$$\displaystyle=\sqrt{{{\frac{{{744096.8}}}{{{12}-{1}}}}}}$$
$$\displaystyle={260.087}$$
The sample standard deviation is 260.087
Step 4
c) $$\displaystyle{95}\%$$ Confidence Interval for Population mean:
$$\displaystyle\overline{{{x}}}\pm{t}_{{{n}-{1}}}\times{\frac{{{s}{d}}}{{\sqrt{{{n}}}}}}$$
$$\displaystyle{4263.291}\pm{2.201}\times{\frac{{{260.087}}}{{\sqrt{{{12}}}}}}$$
$$\displaystyle{4263.291}\pm{165.253}$$
$$\displaystyle{\left({4098.038},\ {4428.544}\right)}$$
$$\displaystyle{95}\%$$ confidence interval for the population mean is $$\displaystyle{\left({4098.038},\ {4428.544}\right)}$$