91Ó°ÊÓ

Given in the question is to refer exercise 13.

The histogram depicts the probability distribution.

The provided distribution is symmetric (the graph to the left of $5$is a mirror image of the graph to the right of $5$), and the mean is located in the middle, which is $5$.

The expected value of the squared variation from the mean is the variance:

$$\begin{gathered} {\mathrm{σ}}^2=∑(x-\mathrm{μ}{)}^{2}P(x) \\ =(1-5{)}^2×\frac{1}{9}+(2-5{)}^2×\frac{1}{9}+(3-5{)}^2×\frac{1}{9}+(4-5{)}^2×\frac{1}{9}+(5−5{)}^2×\frac{1}{9}+(6−5{)}^2×\frac{1}{9}+(7−5{)}^2×\frac{1}{9}+(8−5{)}^2×\frac{1}{9}+(9−5{)}^2×\frac{1}{9} \\ =\frac{20}{3} \\ ≈6.6667 \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} \mathrm{σ}=\sqrt{{\mathrm{σ}}^2} \\ =\sqrt{6.6667} \\ ≈2.5820 \end{gathered}$$

The following histogram shows the probability distribution

The expected value is calculated by multiplying each possibility by its probability:

$$\begin{gathered} E\left(X\right)=∑xP\left(x\right) \\ =1×0.301+2×0.176+3×0.125+4×0.097+5×0.079+6×0.067+7×0.058+8×0.051+9×0.046 \\ =3.441 \end{gathered}$$

The expected value of the squared variation from the mean is the variance:

$$\begin{gathered} {\mathrm{σ}}^2=∑(x-\mathrm{μ}{)}^{2}P(x) \\ =(1-3.441{)}^2×0.301+(2-3.441{)}^2×0.176+(3-3.441{)}^2×0.125+(4-3.441)+(5−3.441{)}^2×0.079+(6−3.441{)}^2×0.067+(7−3.441{)}^2×0.058+(8−3.441{)}^2×0.051+(9−3.441) \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} \mathrm{σ}=\sqrt{{\mathrm{σ}}^2} \\ =\sqrt{6.060519} \\ ≈2.4618 \end{gathered}$$