Two dice are thrown; x = sum of the numbers on the dice

Three coins are tossed

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

Let S be the sample space.

$S=\{2,3,4,5,6,7,8,9,10,11,12\}$

The probability of each number in S is given below.

$\begin{array}{c}{x}_1=2\\ p\left(x_1\right)=1/36\\ x_2=3\\ p\left(x_2\right)=2/36\end{array}$

Solve further.

$\begin{array}{c}{x}_3=4\\ p\left(x_3\right)=3/36\\ x_4=5\\ p\left(x_4\right)=4/36\end{array}$

Solve further.

$\begin{array}{c}{x}_5=6\\ p\left(x_5\right)=5/36\\ x_6=7\\ p\left(x_6\right)=6/36\end{array}$

Solve further.

$\begin{array}{c}{x}_7=8\\ p\left(x_7\right)=5/36\\ x_8=9\\ p\left(x_8\right)=4/36\end{array}$

Solve further.

$\begin{array}{c}{x}_9=10\\ p\left(x_9\right)=3/36\\ x_{10}=11\\ p\left(x_{10}\right)=2/36\end{array}$

Solve further.

$\begin{array}{c}{x}_{11}=12\\ p\left(x_{11}\right)=1/36\end{array}$

The mean is given below.

$\begin{array}{c}\mu =\sum _2{x}_{i}p\left(x_i\right)\\ =\frac{2}{36}+\frac{6}{36}+\frac{12}{36}+\frac{20}{36}+\frac{30}{36}+\frac{49}{36}+\frac{40}{36}+\frac{36}{36}+\frac{30}{36}+\frac{22}{36}+\frac{12}{36}\\ \approx 7\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{var}\left(x_i\right)=\sum {\left(x_i-\mu \right)}^{2}p\left(x_i\right)\\ \mathrm{var}\left(x_i\right)=\sum {\left(x_i-7\right)}^{2}p\left(x_i\right)\\ \mathrm{var}\left(x_i\right)=\frac{5^2}{36}+\frac{4^2\left(2\right)}{36}+\frac{3^2\left(3\right)}{36}+\frac{2^2\left(4\right)}{36}+\frac{\left(5\right)}{36}+\frac{\left(5\right)}{36}+\frac{2^2\left(4\right)}{36}+\frac{3^2\left(3\right)}{36}+\frac{4^2\left(2\right)}{36}+\frac{5^2}{36}\\ \mathrm{var}\left(x_i\right)=\frac{35}{6}\end{array}$

The standard deviation is given below.

$\begin{array}{c}\sigma =\sqrt{\mathrm{var}\left(x\right)}\\ \sigma =\sqrt{\frac{35}{6}}\end{array}$

Cumulative function is given below.

$\begin{array}{c}{x}_1=2\\ F\left(x_1\right)=1/36\\ x_2=3\\ F\left(x_2\right)=3/36\end{array}$

Solve further.

$\begin{array}{c}{x}_3=4\\ F\left(x_3\right)=6/36\\ x_4=5\\ F\left(x_4\right)=10/36\end{array}$

Solve further.

$\begin{array}{c}{x}_5=6\\ F\left(x_5\right)=15/36\\ x_6=7\\ F\left(x_6\right)=21/36\end{array}$

Solve further.

$\begin{array}{c}{x}_7=8\\ F\left(x_7\right)=26/36\\ x_8=9\\ F\left(x_8\right)=30/36\end{array}$

Solve further.

$\begin{array}{c}{x}_9=10\\ F\left(x_9\right)=33/36\\ x_{10}=11\\ F\left(x_{10}\right)=35/36\end{array}$

Solve further.

$\begin{array}{c}{x}_9=10\\ F\left(x_9\right)=33/36\\ x_{10}=11\\ F\left(x_{10}\right)=35/36\end{array}$

Thegraph is shown below.