Chapter 4.4, Problem 87E

(a)

Section 1:

To determine

The mean of the number of heads.

Answer to Problem 87E

Solution: The mean of the number of heads $\left({\mu }_X\right)$ is 0.5.

Explanation of Solution

The chance of the occurrence of the number of heads in the tossing of single coin is either 0 or 1, which means each has probability of $\frac{1}{2}$. So, the formula for mean is equal to the expectation of chance. Thus, the mean can be calculated as shown below:

$\begin{array}{c}\text{Mean}\left({\mu }_X\right)={\displaystyle \sum _{i=1}^2{x}_i}\times p\left(x_i\right)\\ =0\times 0.5+1\times 0.5\\ =0.5\end{array}$

Section 2:

To determine

The standard deviation of number of heads.

Answer to Problem 87E

Solution: The standard deviation $\left({\sigma }_X\right)$ is 0.5.

Explanation of Solution

The chance of the occurrence of number of heads in the tossing of single coin is 0 or 1, which means each has probability of $\frac{1}{2}$. So, the formula for variance is as follows:

$\text{Variance}\left({\sigma }_X^2\right)={\left(x_i-{\mu }_X\right)}^2\times p_i$

And it is calculated as

$\begin{array}{c}\text{Variance}\left({\sigma }_X^2\right)={\left(x_i-{\mu }_X\right)}^2\times p_i\\ ={\left(0-0.5\right)}^2\times \frac{1}{2}+{\left(1-0.5\right)}^2\times \frac{1}{2}\\ =0.25\end{array}$

So, the standard deviation is calculated as

$\begin{array}{c}\text{Standard deviation }\left(\sigma \right)=\sqrt{\text{Variance}}\\ \text{ =}\sqrt{0.25}\\ =0.5\end{array}$

(b)

Section 1:

To determine

The mean of the number of heads.

Answer to Problem 87E

Solution: The mean is 2.

Explanation of Solution

The coin is tossed four times, so there will be five possible numbers of heads (0, 1, 2, 3, 4). So, the number of arrangements for the number of heads is as follows:

For 0 head, there will be ${}^4{C}_0=1$ arrangement that has no heads.

For 1 head, there will be ${}^4{C}_1=4$ arrangements that have 1 head.

For 2 heads, there will be ${}^4{C}_2=6$ arrangements that have 2 heads.

For 3 heads, there will be ${}^4{C}_3=4$ arrangements that have 3 heads.

For 4 heads, there will be ${}^4{C}_4=1$ arrangement that has 4 heads.

So, there will be total $16\left(1+4+6+4+1\right)$ arrangements. Thus, the corresponding probabilities are as follows:

Number of heads $\left(x_i\right)$	Favorable arrangements	Probability $\left(p_i\right)$
0	1	1/16
1	4	4/16
2	6	6/16
3	4	4/16
4	1	1/16

Now, the formula of mean is as follows:

$\text{Mean}\left({\mu }_X\right)={\displaystyle \sum _{i=1}^5{x}_i\times p_i}$

And it is calculated as

$\begin{array}{c}\text{Mean}\left({\mu }_X\right)={\displaystyle \sum _{i=1}^5{x}_i\times p_i}\\ =\left(0\times \frac{1}{16}\right)+\left(1\times \frac{4}{16}\right)+\left(2\times \frac{6}{16}\right)+\left(3\times \frac{4}{16}\right)+\left(4\times \frac{1}{16}\right)\\ =2\end{array}$

Section 2:

To determine

The standard deviation.

Answer to Problem 87E

Solution: The standard deviation is 1.

Explanation of Solution

The formula for standard deviation is as follows:

$\text{Standard}\text{deviation (}{\sigma }_X)=\sqrt{{\displaystyle \sum _{i=1}^5{\left(x_i-{\mu }_X\right)}^2\times p_i}}$

And it is calculated as

$\begin{array}{c}\text{Standard}\text{deviation (}{\sigma }_X)=\sqrt{{\displaystyle \sum _{i=1}^5{\left(x_i-{\mu }_X\right)}^2\times p_i}}\\ =\sqrt{\left(\begin{array}{l}{\left(0-2\right)}^2\times \frac{1}{16}+{\left(1-2\right)}^2\times \frac{4}{16}+{\left(2-2\right)}^2\times \frac{6}{16}\\ +{\left(3-2\right)}^2\times \frac{4}{16}+{\left(4-2\right)}^2\times \frac{1}{16}\end{array}\right)}\\ =1\end{array}$

(c)

Section 1:

To determine

The mean using the distribution provided in Example 4.23.

Answer to Problem 87E

Solution: The mean is 2.

Explanation of Solution

The mean for the referred distribution is calculated by using the formula:

$\begin{array}{c}\text{Mean}{\mu }_X={\displaystyle \sum _{i=1}^5{x}_i\times p_i}\\ =0\times 0.0625+1\times 0.25+2\times 0.375+3\times 0.25+4\times 0.0625\\ =2\end{array}$

Section 2:

To determine

The standard deviation using the distribution provided in Example 4.23.

Answer to Problem 87E

Solution: The standard deviation is 1.

Explanation of Solution

The standard deviation for the referred distribution is calculated by using the formula:

$\text{Standard deviation (}{\sigma }_X)=\sqrt{{\displaystyle \sum _{i=1}^5{\left(x_i-{\mu }_X\right)}^2\times p_i}}$

And it is calculated as

$\begin{array}{c}\text{Standard deviation (}{\sigma }_X)=\sqrt{{\displaystyle \sum _{i=1}^5{\left(x_i-{\mu }_X\right)}^2\times p_i}}\\ =\sqrt{\left(\begin{array}{l}{\left(0-2\right)}^2\times 0.0625+{\left(1-2\right)}^2\times 0.25+{\left(2-2\right)}^2\times 0.375\\ +{\left(3-2\right)}^2\times 0.25+{\left(4-2\right)}^2\times 0.0625\end{array}\right)}\\ =1\end{array}$

Section 3:

To determine

Whether the results obtained in part (b) and part (c) are the same.

Answer to Problem 87E

Solution: Yes, the results obtained in part (b) and parts (c) are same.

Explanation of Solution

The mean and standard deviation obtained in part (b) are 2 and 1, respectively. Similarly, the mean and standard deviation obtained in part (c) are 2 and 1, respectively. Hence, the values obtained for the mean and standard deviation is the same in both the parts.