Chapter 10.4, Problem 54E

Solution

a.

To determine: the probability model for X.

X	0	1	2	3	4
P(X)	0.0625	0.25	0.375	0.25	0.0625

Given information:

The possible values of X is $0,1,2,3$ and $4$ .

Concept Used:

Binomial probability Distribution:

Consider a simple event with these properties:

Each trial has two possible outcomes, called success and failure.

The probability of success on each trial is the same. (We denote the probability of success as $p$ and the probability of failure as $q$ . Note that $q=1-p$ .)

The trials are independent.

If the binomial random variable $X$ counts the number of successes in $n$ trials with probability of success $p$ and the probability of getting $k$ successes in $n$ trials, then $P\left(X=k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)p^k{q}^{n-k}$ .

Calculation:

The probability of getting a head on a coin toss is $\frac{1}{2}$ or $0.5$ .

Substitute $0$ for $k$ , $n=4,p=0.5$ in the formula.

  $\begin{array}{c}$ P\left(X=0\right)=\left(\begin{array}{c}4\\ 0\end{array}\right){(0.5)}^0{(0.5)}^{4-0}$=0.0625\end{array}$

Similarly, calculate the probability of getting $1,2,3,4$ heads.

  $\begin{array}{l}P\left(1\right)=0.25\\ P\left(2\right)=0.375\\ P\left(3\right)=0.25\\ P\left(4\right)=0.0625\end{array}$

X	0	1	2	3	4
P(X)	0.0625	0.25	0.375	0.25	0.0625

b.

To determine: the expected value.

The expected value is 2.

Given information:

The possible values of X is $0,1,2,3$ and $4$ .

Concept Used:

Expected Value of a Discrete Random Variable:

The weighted mean of all possible values of a discrete random variable is called its expected values.

It is the sum of the products of all possible values and their probability.

  $E\left(X\right)={\displaystyle \sum \left|X\cdot P\left(X\right)\right|}$

Calculation:

X	0	1	2	3	4
P(X)	0.0625	0.25	0.375	0.25	0.0625

  $\begin{array}{c}E\left(X\right)=\left(0\cdot 0.0625\right)+\left(1\cdot 0.25\right)+\left(2\cdot 0.375\right)+\left(4\cdot 0.0625\right)\\ =2\end{array}$

c.

To explain: the standard deviation of the number of heads.

The standard deviation is 1.

Given information:

The possible values of X is $0,1,2,3$ and $4$ .

Calculation:

The square root of the expected value of squared deviations from the mean, $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$

Substitute all the values in $\sigma =\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2\cdot P\left(x_i\right)}}$ .

  $\begin{array}{c}\sigma =\sqrt{{\left(0-2\right)}^2\cdot 0.0625+{\left(1-2\right)}^2\cdot 0.25+{\left(2-2\right)}^2\cdot 0.375+{\left(3-2\right)}^2\cdot 0.25+{\left(4-2\right)}^2\cdot 0.0625}\\ =1\end{array}$

d.

To verify: the same value of $\mu =np$ and $\sigma =\sqrt{npq}$ .

The values of $\mu and\sigma$ are $2and1$ .

Given information:

X	0	1	2	3	4
P(X)	0.0625	0.25	0.375	0.25	0.0625

Concept Used:

Expected Value for a Binomial Distribution:

If the binomial random variable $X$ counts the number of successes in $n$ independent trials with probability of success $p$ , then $\mu =E(x)=np$ .

Standard Deviation for a Binomial Distribution:

If the binomial random variable $X$ counts the number of successes in $n$ independent trials with probability of success $p$ , then $\sigma =\sqrt{npq}$ where $q$ represents the probability of failure.

Calculation:

The sample size $n$ is 4, the probability of success $p$ is $0.5$ , and the probability of failure $q$ is $(1-0.5)=0.5$ .

  $\begin{array}{c}\mu =4(0.5)\\ =2\end{array}$

  $\begin{array}{c}\sigma =\sqrt{(4)(0.5)(0.5)}\\ =1\end{array}$