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The Probability Distribution Shown Below Describes a Population of Measurements x5101520p(x)14141414\begin{array}{l|cccc}\hline x & 5 & 10 & 15 & 20 \\\hline p(x) & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\hline\end{array}

question 20

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The probability distribution shown below describes a population of measurements that
can assume values of 5, 10, 15, and 20, each of which occurs with the same frequency: x5101520p(x)14141414\begin{array}{l|cccc}\hline x & 5 & 10 & 15 & 20 \\\hline p(x) & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\hline\end{array}

Find E(x)=μE ( x ) = \mu . Then consider taking samples of n=2n = 2 measurements and calculating xˉ\bar { x } for each sample. Find the expected value, E(xˉ)E ( \bar { x } ) , of xˉ\bar { x } .


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