Chapter PIV, Problem 12RE

(a)

To determine

To find: a probability model for the result on the spinner.

Explanation of Solution

Let X be the spinner's result.

The following table gives the model of probability.

X	5	10	20
P(X)	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{4}$

(b)

To determine

To find: the mean and standard deviation of the outcomes of the spinner.

Answer to Problem 12RE

Mean = 10

Standard Deviation = 6.12

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum XP\left(x\right)}\\ Var\left(X\right)={\displaystyle \sum {\left(X-\mu \right)}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

The mean of the spinner-is

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum XP\left(x\right)}\\ =5\left(\frac{1}{2}\right)+10\left(\frac{1}{4}\right)+20\left(\frac{1}{4}\right)\\ =2.5+2.5+5\\ =10\end{array}$

variance is

  $\begin{array}{c}Var\left(X\right)={\displaystyle \sum {\left(X-\mu \right)}^{2}P\left(x\right)}\\ ={\left(5-10\right)}^2\left(\frac{1}{2}\right)+{\left(10-10\right)}^2\left(\frac{1}{4}\right)+{\left(20-10\right)}^2\left(\frac{1}{4}\right)\\ =25\left(\frac{1}{2}\right)+0\left(\frac{1}{4}\right)+100\left(\frac{1}{4}\right)\\ =12.5+25\\ =37.5\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{37.5}\\ =6.12\end{array}$

(c)

To determine

To find: a probability model for the result on the die.

Explanation of Solution

Let Y be the result of a dying.

The following table gives the model of probability.

Y	0	1	2	3	4
P(y)	$\frac{2}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

(d)

To determine

To find: the mean and standard deviation of the findings of the death.

Answer to Problem 12RE

Mean = 10/6

Standard Deviation = 1.49

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum XP\left(x\right)}\\ Var\left(X\right)={\displaystyle \sum {\left(X-\mu \right)}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean is

  $\begin{array}{c}0\left(\frac{2}{6}\right)+1\left(\frac{1}{6}\right)+2\left(\frac{1}{6}\right)+3\left(\frac{1}{6}\right)+4\left(\frac{1}{6}\right)\\ =\frac{1+2+3+4}{6}\\ =\frac{10}{6}\end{array}$

Variance is

  $\begin{array}{c}Var\left(Y\right)={\displaystyle \sum {\left(Y-\mu \right)}^{2}P\left(y\right)}\\ ={\left(0-\frac{10}{6}\right)}^2\left(\frac{2}{6}\right)+{\left(1-\frac{10}{6}\right)}^2\left(\frac{1}{6}\right)+{\left(2-\frac{10}{6}\right)}^2\left(\frac{1}{6}\right)\\ +{\left(3-\frac{10}{6}\right)}^2\left(\frac{1}{6}\right)+{\left(4-\frac{10}{6}\right)}^2\left(\frac{1}{6}\right)\\ =2.78\left(\frac{2}{6}\right)+0.45\left(\frac{1}{6}\right)+0.12\left(\frac{1}{6}\right)+1.78\left(\frac{1}{6}\right)+5.45\left(\frac{1}{6}\right)\\ =2.23\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(Y\right)=\sqrt{\mathrm{var}\left(Y\right)}\\ =\sqrt{2.23}\\ =1.49\end{array}$

(e)

To determine

To find: the mean and standard deviation of the number of spaces that are allowed to pass.

Answer to Problem 12RE

Mean = 70/6

Standard Deviation = 6.31

Explanation of Solution

Formula used:

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum XP\left(x\right)}\\ Var\left(X\right)={\displaystyle \sum {\left(X-\mu \right)}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Let Z be the total number of spaces you get to move.

  $Z=X+Y$

Mean:

  $\begin{array}{c}E\left(Z\right)=E\left(X+Y\right)\\ =E\left(X\right)+E\left(Y\right)\\ =10+\frac{10}{6}\\ =\frac{70}{6}\end{array}$

Variance:

  $\begin{array}{c}Var\left(Z\right)=\mathrm{var}\left(X+Y\right)\\ =\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ =37.5+2.33\\ =39.83\end{array}$

Standard deviation:

  $\begin{array}{c}SD\left(Z\right)=\sqrt{\mathrm{var}\left(Z\right)}\\ =\sqrt{39.83}\\ =6.31\end{array}$