91Ó°ÊÓ

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.

$S=\left\{p^3,3p^2(1−p),3p{(1−p)}^2,{(1−p)}^3\right\}$

Find the random variable.

$\begin{array}{c}{x}_1=3\\ p_1=p^3\\ x_2=1\\ p_2=3p^2(1−p)\end{array}$

Solve further.

$\begin{array}{c}{x}_3=−1\\ p_3=3p{\left(1-p\right)}^2\\ x_4=−3\\ p_4={\left(1-p\right)}^3\end{array}$

The mean is given below.

$\begin{array}{c}\mathrm{μ}\left(x\right)=∑x_i{p}_i\\ =3p^3+3p^2(1−p)−3p−3{\left(1-p\right)}^2-3{\left(1-p\right)}^3\\ =3[−p−1+3p]\\ =3[2p−1]\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{var}\left(x\right)=E\left(x^2\right)−{\mathrm{μ}}^2\\ \mathrm{var}\left(x\right)=9p^3+3p^2\left(1-p\right)+3p{\left(1-p\right)}^2+9{\left(1-p\right)}^3-{\left[3\left(2p-1\right)\right]}^2\\ \mathrm{var}\left(x\right)=9p^3+3p^2−3p^3+3p\left(1−2p+p^2\right)+9\left(1−3p+3p^2−p^3\right)−9\left[4p^2−4p+1\right]\\ \mathrm{var}\left(x\right)=12p−12p^2\end{array}$

The standard deviation is given below

$\begin{array}{c}\mathrm{σ}=\sqrt{\mathrm{var}\left(x\right)}\\ \mathrm{σ}=\sqrt{12p−12p^2}\\ \mathrm{σ}=2\sqrt{3p(1−p)}\end{array}$

The graph is shown below.

Hence, the required values are mentioned below.
$\begin{array}{c}\mathrm{μ}\left(x\right)=3[2p−1]\\ \mathrm{var}\left(x\right)=12p−12p^2\\ \mathrm{σ}=2\sqrt{3p(1−p)}\end{array}$