Question

Suppose that Martian dice are 4-sided (tetrahedra) with points labeled $1-4$. When a pair of these dice is tossed, let x be the product of the two numbers at the tops of the dice if the product is odd; otherwise$x=0$.

Short answer

The required values are mentioned below.

$\begin{array}{c}\mu =1\\ \mathrm{var}\left(x\right)=\frac{21}{4}\\ \sigma =\frac{\sqrt{21}}{2}\end{array}$

Step-by-step solution

Given Information

Martian dice is tossed

Definition of the cumulative distribution function

The likelihood that a comparable continuous random variable has a value less than or equal to the function's argument is the value of the function.

Find the values

The random variables are given below.

$\begin{array}{c}{x}_1=1\\ p\left(x_1\right)=1/16\\ x_2=0\\ p\left(x_2\right)=2/16\end{array}$

Solve further.

$\begin{array}{c}{x}_3=3\\ p\left(x_3\right)=2/16\\ x_4=0\\ p\left(x_4\right)=3/16\end{array}$

Solve further.

$\begin{array}{c}{x}_6=0\\ p\left(x_6\right)=2/16\\ x_8=0\\ p\left(x_8\right)=2/16\end{array}$

Solve further.

$\begin{array}{c}{x}_9=9\\ p\left(x_9\right)=1/16\\ x_{12}=0\\ p\left(x_{12}\right)=2/16\end{array}$

Solve further.

$\begin{array}{c}{x}_{16}=0\\ p\left(x_{16}\right)=1/16\end{array}$

The mean is given below.

$\begin{array}{c}\mu =\sum x_{i}p\left(x_i\right)\\ \mu =\frac{1}{16}+\frac{6}{19}+\frac{9}{16}\\ \mu =1\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{var}\left(x\right)=\sum {\left(x_i-\mu \right)}^{2}p\left(x_i\right)\\ \mathrm{var}\left(x\right)=\frac{2}{16}+\frac{3}{16}+\frac{2}{16}+\frac{2}{16}+\frac{1}{16}+\frac{4\times 2}{16}+\frac{64}{16}\\ \mathrm{var}\left(x\right)=\frac{21}{4}\end{array}$

The standard deviation is given below.

$\begin{array}{c}\sigma =\sqrt{\mathrm{var}\left(x\right)}\\ \sigma =\sqrt{\frac{21}{4}}\\ \sigma =\frac{\sqrt{21}}{2}\end{array}$

The graph is shown below.

Hence, the required values are mentioned below.

.

$\begin{array}{c}\mu =1\\ \mathrm{var}\left(x\right)=\frac{21}{4}\\ \sigma =\frac{\sqrt{21}}{2}\end{array}$