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xP(x)00.0510.1520.130.7Find the standard deviation of this probability distribution. Give your answer to at least 2 decimal places

User Liutong Chen
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1 Answer

18 votes
18 votes

We have to find the standard deviation of this probability distribution.

We have to start by calculating the mean of this distribution.

It can be calculated as:


\begin{gathered} \bar{x}=\sum^nP(x_i)\cdot x_i \\ \bar{x}=0.05\cdot0+0.15\cdot1+0.1\cdot2+0.7\cdot3 \\ \bar{x}=0+0.15+0.2+2.1 \\ \end{gathered}

We now can calculate the standard deviation using the last result as:


\begin{gathered} \sigma=\sqrt{\sum_^P(x_i)\cdot(x_i-\bar{x})^2} \\ \sigma=√(0.05(0-2.1)^2+0.15(1-2.1)^2+0.1(2-2.1)^2+0.7(3-2.1)^2) \\ \sigma=√(0.05(-2.1)^2+0.15(-1.1)^2+0.1(-0.1)^2+0.7(0.9)^2) \\ \sigma=√(0.05(4.41)+0.15(1.21)+0.1(0.01)+0.7(0.81)) \\ \sigma=√(0.2205+0.1815+0.001+0.567) \\ \sigma=√(0.97) \\ \sigma\approx0.98 \end{gathered}

Answer: the standard deviation is approximately 0.98.

User Jammer
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