Question

A coin is tossed repeatedly; x = number of the toss at which a head first appears.

Short answer

The required values are mentioned below.

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

Step-by-step solution

Given Information

A coin is tossed repeatedly.

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

$\begin{array}{c}x=1\\ p\left(x_1\right)=p\\ x=2\\ p\left(x_2\right)=pq\end{array}$

Solve further.

$\begin{array}{c}x=3\\ p\left(x_3\right)=p{q}^2\\ x=4\\ p\left(x_4\right)=p{q}^3\end{array}$

Solve further.

$\begin{array}{c}x=5\\ p\left(x_5\right)=p{q}^4\\ x=n\\ p\left(x_n\right)=p{q}^{n-1}\end{array}$

The mean is given below.

$\begin{array}{c}\mu =\sum _{x=1}^{\infty }xp{q}^{x-1}\\ =p\left(1+2q+3q^2+4q^3+5q^4+\dots \right)\end{array}$

Let$S=\left(1+2q+3q^2+4q^3+5q^4+\cdots +\right)$

$\begin{array}{c}qS=\left(q+2q^2+3q^3+4q^4+5q^5+\dots \right)\\ S-qS=\left(1+q+q^2+q^3+q^4+\cdots +\right)\\ S(1-q)=\frac{1}{1-q}\\ S=\frac{1}{{(1-q)}^2}\end{array}$

The mean becomes as follows.

$\begin{array}{c}\mu =\frac{p}{{(1-q)}^2}\\ \mu =\frac{1/2}{{(1/2)}^2}\\ \mu =2\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{var}\left(x\right)=\sum {(x-2)}^{2}p{q}^{x-1}\\ =\sum (x-1)q^x\\ =q+q^3+4q^4+9q^5+\dots \\ =q+q^3\left(1+4q+9q^2+16q^3+\dots \right)\end{array}$

Let$S_1=1+4q+9q^2+16q^3+\dots$

$S_1=\frac{1+q}{{(1-q)}^3}$

The value of variance becomes as follows.

$\begin{array}{c}\mathrm{var}\left(x\right)=q+q^3\left[\frac{1+q}{{(1-q)}^3}\right]\\ \mathrm{var}\left(x\right)=\frac{1}{2}+\times \frac{3/2}{{\left(0.5\right)}^3}\\ \mathrm{var}\left(x\right)=2\end{array}$

The standard deviation is given below.

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

Cumulative function is given below.

$\begin{array}{c}x=1\\ F\left(x_1\right)=1/2\\ x=2\\ F\left(x_2\right)=3/4\end{array}$

Solve further.

$\begin{array}{c}x=3\\ F\left(x_3\right)=7/8\\ x=4\\ F\left(x_4\right)=15/16\end{array}$

Solve further.

$\begin{array}{c}x=5\\ F\left(x_5\right)=31/32\end{array}$

The graph is shown below

Hence, the required values are mentioned be

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