Chapter 16, Problem 14E

To determine

To Calculate: the standard deviation of the quantity may win rolling a die.

Answer to Problem 14E

42.002

Explanation of Solution

Given:

Pay five dollars to play and plan to pay 20 dollars. It means that the researcher is willing to play the 0.10 chance of hitting the ball 4 times.

Formula used:

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

Calculation:

Suppose X be number of darts. The next step is the X model of probability

Suppose H =hit the target

The probability that only one dart would take is 0.1. This is the first try possibility when hit that 2 darts are needed. when first miss and second hit

  $\begin{array}{c}P\left(H\text{'}H\right)=\left(0.9\right)\left(0.1\right)\\ =0.09\end{array}$

Will be 3 darts by losing first two before finding the third time

  $\begin{array}{c}P\left(H\text{'}H\text{'}H\right)=\left(0.9\right)\left(0.9\right)\left(0.1\right)\\ =0.081\end{array}$

4 darts are required if lose the first 3 and the fourth

  $\begin{array}{c}P\left(H\text{'}H\text{'}H\text{'}H\text{'}\right)+P\left(H\text{'}H\text{'}H\text{'}H\right)\\ =\left(0.9\right)\left(0.9\right)\left(0.9\right)\left(0.9\right)+\left(0.9\right)\left(0.9\right)\left(0.9\right)\left(0.1\right)\\ =0.729\end{array}$

Suppose X be paid to subtract the sum gained from the amount received to play is the probability model for X.

X	0 (H’ H’ H’ H’)	$95 (H)	90 (H’ H’)	85 (H’ H’ H’)	80 (H’ H’ H’ H’)
P(X = x)	${\left(0.9\right)}^4=0.6561$	0.1	0.009	${\left(0.9\right)}^2\left(0.1\right)=0.081$	${\left(0.9\right)}^3\left(0.1\right)=0.0729$

The expected winning number is

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.6561\right)+95\left(0.1\right)+90\left(0.09\right)+85\left(0.081\right)+80\left(0.0729\right)\\ =0+9.5+8.1+6.885+5.832\\ =\$30.32\end{array}$

Expected winning is $30.32

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ ={\left(0-30.32\right)}^2\left(0.6561\right)+{\left(95-30.32\right)}^2\left(0.1\right)+{\left(90-30.32\right)}^2\left(0.09\right)\\ +{\left(85-30.32\right)}^2\left(0.081\right)+{\left(80-30.32\right)}^2\left(0.0729\right)\\ =1764.165\end{array}$

The standard deviation

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{1764.165}\\ =42.002\end{array}$

Thus, the standard deviation is 42.002