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The Probability Distribution Shown Below Describes a Population of Measurements x25811p(x)14141414\begin{array}{l|rrrr}\hline x & 2 & 5 & 8 & 11 \\\hline p(x) & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\hline\end{array}

question 22

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The probability distribution shown below describes a population of measurements that can assume values of 2, 5, 8, and 11, each of which occurs with the same frequency: x25811p(x)14141414\begin{array}{l|rrrr}\hline x & 2 & 5 & 8 & 11 \\\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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