Chapter 14.4, Problem 15E

(a)

To determine

To find:

The each probability of the given binomial experiments .

Answer to Problem 15E

The number of possibilities by the total number of possible outcomes $=64$

Explanation of Solution

Given:

The probability is of success is $p=0.5$

The number of trials is $n=6$

Concept used:

The distribution is a binomial distribution

P (k-success)

  ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

The distribution is a binomial distribution

The probability is of success is $p=0.5$

The number of trials is $n=6$

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

The left side of the table

The probability of getting EXACTLY $1,2,3,4,5\text{and}6$

The right side of the table is cumulative distribution

Binomial distribution $\left(n=6,p=0.5\right)$

	$f\left(x\right)$	$F\left(x\right)$
$x$	$P\left[X=x\right]$	$P\left[X\le x\right]$
$0$	$0.0156$	$0.0156$
$1$	$0.0938$	$0.1094$
$2$	$0.2344$	$0.3438$
$3$	$0.3125$	$0.6563$
$4$	$0.2344$	$0.8906$
$5$	$0.0938$	$0.9844$
$6$	$0.0156$	$1.0000$

The number of possibilities by the total number of possible outcomes

  $\begin{array}{l}=2^6\\ =64\end{array}$

(b)

To determine

To draw:

A bar graph and the experiment was repeated $64$ times.

Answer to Problem 15E

The number of possibilities by the total number of possible outcomes $=64$

Explanation of Solution

Given:

The probability is of success is $p=0.5$

The number of trials is $n=6$

Concept used:

The distribution is a binomial distribution

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

The distribution is a binomial distribution

The probability is of success is $p=0.5$

The number of trials is $n=6$

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

The number of possibilities by the total number of possible outcomes

  $\begin{array}{l}=2^6\\ =64\end{array}$

  

(c)

To determine

A bar graph and the mean number of tails.

Answer to Problem 15E

The mean number of tails $=20$

Explanation of Solution

Given:

The probability is of success is $p=0.5$

The number of trials is $n=6$

Concept used:

The distribution is a binomial distribution

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

The distribution is a binomial distribution

The probability is of success is $p=0.5$

The number of trials is $n=6$

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

The number of possibilities by the total number of possible outcomes

  $\begin{array}{l}=2^6\\ =64\end{array}$

  

The average number is the center peak of the probability distribution.

This is $20$ tails in $64$ trials of flipping $6$ coins.

(d)

To determine

To find:

The standard deviation of the number of tails.

Answer to Problem 15E

  $\sigma =1.22$

Explanation of Solution

Given:

The probability is of success is $p=0.5$

The number of trials is $n=6$

Concept used:

The distribution is a binomial distribution

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

The distribution is a binomial distribution

The probability is of success is $p=0.5$

The number of trials is $n=6$

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

The number of possibilities by the total number of possible outcomes

  $\begin{array}{l}=2^6\\ =64\end{array}$

  

The equation for the standard deviation of a binomial distribution

  $\begin{array}{l}\sigma =\sqrt{npq}\mathrm{..................}\left(1\right)\\ q=1-p\\ p=q=0.5\\ n=6\end{array}$

Using these values in equation (1)

  $\begin{array}{l}\sigma =\sqrt{npq}\\ \sigma =\sqrt{6\left(0.5\right)\left(0.5\right)}\\ \sigma =1.22\end{array}$

(e)

To determine

To draw:

A bar graph toa normal distribution.

Explanation of Solution

Given:

The probability is of success is $p=0.5$

The number of trials is $n=6$

Concept used:

The distribution is a binomial distribution

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

Calculation:

The distribution is a binomial distribution

The probability is of success is $p=0.5$

The number of trials is $n=6$

P (k-success) ${=}^n{C}_k{p}^k{\left(1-p\right)}^{n-k}$

The number of possibilities by the total number of possible outcomes

  $\begin{array}{l}=2^6\\ =64\end{array}$

  

A continuous normal distribution curve is overlaid on top of the discrete distribution.