Question

A weighted coin with probability p of coming down heads is tossed three times; x = number of heads minus number of tails.

Short answer

The required values are mentioned below.

$\begin{array}{c}\mu \left(x\right)=3[2p-1]\\ \mathrm{var}\left(x\right)=12p-12p^2\\ \sigma =2\sqrt{3p(1-p)}\end{array}$

Step-by-step solution

 Step 1: Given Information 

Three coins are 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.

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.

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

The variance is given below.

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

The standard deviation is given below

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

The graph is shown below.

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