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A Discrete Random Variable X Can Assume Five Possible Values x235810p(x)0.100.200.300.300.10\begin{array}{c|ccccc}x & 2 & 3 & 5 & 8 & 10 \\\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10\end{array}

Question 116

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A discrete random variable x can assume five possible values: 2, 3, 5, 8, 10. Its probability distribution is shown below. Find the standard deviation of the distribution. x235810p(x) 0.100.200.300.300.10\begin{array}{c|ccccc}x & 2 & 3 & 5 & 8 & 10 \\\hline p(x) & 0.10 & 0.20 & 0.30 & 0.30 & 0.10\end{array}


A) 1.8451.845
B) 5.75.7
C) 6.416.41
D) 2.5322.532

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