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Find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution of sample means and compare them with the mean and standard deviation of the population. The minutes of overtime reported by each of the three executives at a corporation are 90, 120, and 210. Use a sample size of 3.

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Given: 90, 120, 210

n=Sample size=3N=Population size=3\begin{align*} n&=\text{Sample size}=3 \\ N&=\text{Population size}=3 \end{align*}

The mean\textbf{mean} is the sum of all values divided by the number of values:

x=i=1nxiN=90+120+2103=4203=140\begin{align*} \overline{x}&=\dfrac{\sum_{i=1}^n x_i}{N} \\ &=\dfrac{\begin{matrix}90+120+210\end{matrix}}{3} \\ &=\dfrac{420}{3} \\ &=140 \end{align*}

The population variance\textbf{population variance} is the sum of squared deviations from the mean divided by NN:

σ2=(xμ)2N=(90140)2+(120140)2+(210140)23=104003=2600\begin{align*} \sigma^2&=\dfrac{\sum (x-\mu)^2}{N} \\ &=\dfrac{\begin{matrix}(90-140)^2+(120-140)^2+(210-140)^2\end{matrix}}{3} \\ &=\dfrac{10400}{3} \\ &=2600 \end{align*}

The population standard deviation\textbf{population standard deviation} is the square root of the population variance:

σ=σ2=260050.9902\sigma=\sqrt{\sigma^2}=\sqrt{2600}\approx 50.9902

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