91影视

Properties of discrete probability distribution:

1) for all values of x,$\mathrm{P}\left(x\right)猢�0$

2)$鈭�\mathrm{P}\left(x\right)=1$

The probability distribution of the capacity for Arc 1 is given below:

From the given table, it is clear that $\mathrm{P}\left(x\right)猢�0$, for all values of x.

Now

$\begin{array}{rcl}鈭�\mathrm{P}\left(x\right)& =& 0.05+0.10+0.25+0.60\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 1.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 2.

The probability distribution of the capacity for Arc 2 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)猢�0$, for all values of x.

Now

$\begin{array}{rcl}鈭�\mathrm{P}\left(x\right)& =& 0.10+0.30+0.60+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 2.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 3

The probability distribution of the capacity for Arc 3 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)猢�0$, for all values of x.

Now

$\begin{array}{rcl}鈭�\mathrm{P}\left(x\right)& =& 0.05+0.25+0.70+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 3.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 4

The probability distribution of the capacity for Arc 4 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)猢�0$for all values of x.

Now

$\begin{array}{rcl}鈭�\mathrm{P}\left(x\right)& =& 0.90+0.10+0+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 4.

To obtain the probability that the capacity for arc 1 exceeds 1, it means to find$\mathrm{P}\left(x>1\right)$

$\begin{array}{rcl}P\left(x>1\right)& =& 1-\mathrm{P}\left(x猢�1\right)\\ & =& 1-\left(0.05+0.10\right)\\ & =& 1-0.15\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.85$

Hence the probability that the capacity for arc 1 exceeds 1 is 0.85.

To obtain the probability that the capacity for arc 2 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x猢�1\right)\\ & =& 1-\left(0.10+0.30\right)\\ & =& 1-0.40\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.60$

Therefore, the probability that the capacity for arc 2 exceeds 1 is 0.60.

Calculating the probability that the capacity for arc 3 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x猢�1\right)\\ & =& 1-\left(0.05+0.25\right)\\ & =& 1-0.30\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.70$

Therefore, the probability that the capacity for arc 3 exceeds 1 is 0.70.

Calculating the probability that the capacity for arc 4 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x猢�1\right)\\ & =& 1-\left(0.90+0.10\right)\\ & =& 1-1\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0$

Therefore, the probability that the capacity for arc 4 exceeds 1 is 0

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0脳0.05+1脳0.10脳2脳0.25+3脳0.60\\ & =& 2.4\\ & & \end{array}$

Thus, the mean capacity for arc 1 is 2.4.

Calculating the mean for arc 2

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0脳0.10+1脳0.30脳2脳0.60+3脳0\\ & =& 1.5\\ & & \end{array}$

Thus the mean capacity for arc 2 is 21.5

Calculating the mean for arc 3

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0脳0.05+1脳0.25脳2脳0.70+3脳0\\ & =& 1.65\\ & & \end{array}$

Thus,the mean capacity for arc 3 is$\begin{array}{rcl}& & 1.65\end{array}$

Calculating the mean for arc 3

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0脳0.90+1脳0.10脳2脳0+3脳0\\ & =& 0.10\\ & & \end{array}$

Thus,the mean capacity for arc 4 is$\begin{array}{rcl}& & 0.10\end{array}$

${蟽}^2=E\left[{\left(x-渭\right)}^2\right]$

${蟽}^2={\left(x-渭\right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{蟽}^2& =& {\left(0-2.4\right)}^{2}0.05+{\left(1-2.4\right)}^{2}0.10+{\left(2-2.4\right)}^{2}0.25+{\left(3-2.4\right)}^{2}0.60\\ & =& 0.288+0.196+0.04+0.216\\ & =& 0.74\\ & & \end{array}$

$\begin{array}{rcl}蟽& =& \sqrt{0.74}\\ & =& 0.86\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}渭卤2蟽& =& 2.4卤2脳0.86\\ & =& 2.4卤1.72\\ & =& \left(2.4-1.72,2.4+1.72\right)\\ & & \end{array}$

$渭卤2蟽=\left(0.68,4.12\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${蟽}^2=E\left[{\left(x-渭\right)}^2\right]$

${蟽}^2={\left(x-渭\right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{蟽}^2& =& {\left(0-1.5\right)}^{2}0.10+{\left(1-1.5\right)}^{2}0.30+{\left(2-1.5\right)}^{2}0.60+{\left(3-1.5\right)}^{2}0\\ & =& 0.225+0.075+0.15+0\\ & =& 0.45\\ & & \end{array}$

$\begin{array}{rcl}蟽& =& \sqrt{0.45}\\ & =& 0.67\\ & & \end{array}$

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}渭卤2蟽& =& 1.5卤2脳0.67\\ & =& 1.5卤1.34\\ & =& \left(1.5-1.34,1.5+1.34\right)\\ & & \end{array}$

$渭卤2蟽=\left(0.16,2.84\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${蟽}^2=E\left[{\left(x-渭\right)}^2\right]$

${蟽}^2={\left(x-渭\right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{蟽}^2& =& {\left(0-1.65\right)}^{2}0.05+{\left(1-1.65\right)}^{2}0.25+{\left(2-1.65\right)}^{2}0.70+{\left(3-1.65\right)}^{2}0\\ & =& 0.1316+0.1056+0.0857+0\\ & =& 0.327\\ & & \end{array}$

$\begin{array}{rcl}蟽& =& \sqrt{0.327}\\ & =& 0.57\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}渭卤2蟽& =& 1.65卤2脳0.57\\ & =& 1.65卤1.14\\ & =& \left(1.65-1.14,1.65+1.14\right)\\ & & \end{array}$

$渭卤2蟽=\left(0.51,2.79\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${蟽}^2=E\left[{\left(x-渭\right)}^2\right]$

${蟽}^2={\left(x-渭\right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{蟽}^2& =& {\left(0-0.1\right)}^{2}0.90+{\left(1-0.1\right)}^{2}0.10+{\left(2-0.1\right)}^{2}0+{\left(3-0.1\right)}^{2}0\\ & =& 0.009+0.081+0+0\\ & =& 0.09\\ & & \end{array}$

$\begin{array}{rcl}蟽& =& \sqrt{0.327}\\ & =& 0.3\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}渭卤2蟽& =& 0.1卤2脳0.3\\ & =& 0.1卤0.6\\ & =& \left(0.1-0.6,0.1+0.6\right)\\ & & \end{array}$

$渭卤2蟽=\left(-0.5,0.7\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean