Answer:
Explanation:
We'll go ahead and complete the 3rd column(xPr(x)) as seen below;
[tex]\begin{gathered} 0\times0.35=0 \\ 1\times0.25=0.25 \\ 2\times0.22=0.44 \\ 5\times0.18=0.90 \end{gathered}[/tex]
The mean can be determined using the below formula;
[tex]\begin{gathered} \text{Mean(}\mu)=\sum ^{}_{}x\cdot Pr(x)_{} \\ \mu=0+0.25+0.44+0.90=1.59 \\ \mu=1.59 \end{gathered}[/tex]
Let's now complete the 4th column(x^2P(x)) as seen below;
[tex]\begin{gathered} 0^2\times0.35=0 \\ 1^2\times0.25=0.25 \\ 2^2\times0.22=0.88 \\ 5^2\times0.18=4.5 \\ \sum ^{}_{}x^2.P(x)=0+0.25+0.88+4.5=5.63 \end{gathered}[/tex]
We'll use the below formula to determine the standard deviation;
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