91影视

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:

$$\begin{gathered} {蟽}^2=鈭�(x-渭{)}^{2}P(x) \\ =(1-5{)}^2脳\frac{1}{9}+(2-5{)}^2脳\frac{1}{9}+(3-5{)}^2脳\frac{1}{9}+(4-5{)}^2脳\frac{1}{9}+(5鈭�5{)}^2脳\frac{1}{9}+(6鈭�5{)}^2脳\frac{1}{9}+(7鈭�5{)}^2脳\frac{1}{9}+(8鈭�5{)}^2脳\frac{1}{9}+(9鈭�5{)}^2脳\frac{1}{9} \\ =\frac{20}{3} \\ 鈮�6.6667 \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} 蟽=\sqrt{{蟽}^2} \\ =\sqrt{6.6667} \\ 鈮�2.5820 \end{gathered}$$

The following histogram shows the probability distribution

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

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

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

$$\begin{gathered} {蟽}^2=鈭�(x-渭{)}^{2}P(x) \\ =(1-3.441{)}^2脳0.301+(2-3.441{)}^2脳0.176+(3-3.441{)}^2脳0.125+(4-3.441)+(5鈭�3.441{)}^2脳0.079+(6鈭�3.441{)}^2脳0.067+(7鈭�3.441{)}^2脳0.058+(8鈭�3.441{)}^2脳0.051+(9鈭�3.441) \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} 蟽=\sqrt{{蟽}^2} \\ =\sqrt{6.060519} \\ 鈮�2.4618 \end{gathered}$$