Chapter 17, Problem 28E

(a)

To determine

To find: the mean and standard deviation of the number of researchers with bull 's eyes can be obtained.

Answer to Problem 28E

Mean = 160

Standard Deviation = 5.66

Explanation of Solution

Given:

  $\begin{array}{c}n=200\\ p=0.8\end{array}$

Formula used:

  $E\left(X\right)=np$

  $SD\left(X\right)=\sqrt{npq}$

Calculation:

The mean is

  $\begin{array}{c}E\left(x\right)=np\\ =200\left(0.8\right)\\ =160\end{array}$

The standard deviation is

  $\begin{array}{c}SD\left(x\right)=\sqrt{npq}\\ =\sqrt{200\left(0.8\right)\left(0.2\right)}\\ =5.66\end{array}$

(b)

To determine

To Explain: that normal model is an appropriate here.

Explanation of Solution

Given:

  $\begin{array}{c}n=200\\ p=0.8\end{array}$

Calculation:

  $\begin{array}{c}np=200\left(0.8\right)\\ =160\left(>10\right)\end{array}$

And,

  $\begin{array}{c}np=n\left(1-p\right)\\ =200\left(1-0.8\right)\\ =40\left(>10\right)\end{array}$

Here, it could be noticed that np > 10 and nq > 10. Therefore, the normality conditions are satisfied.

(c)

To determine

To Explain: the distribution of the number of bull’s-eyes researcher may get using the 68-95-99.7.

Explanation of Solution

Given:

  $\begin{array}{c}n=200\\ p=0.8\end{array}$

Calculation:

Using the 68-95-99.7 rule

68% of the data lies between the limits of one sigma.

  $\begin{array}{c}\mu \pm \sigma =\left(\mu -\sigma ,\mu +\sigma \right)\\ =\left(160-5.66,160+5.66\right)\\ =\left(154.34,165.66\right)\end{array}$

Using the 68-95-99.7 rule, researcher expected to obtain between 154.34 and 165.66 About 68 percent of the time, Bull's eyes.

95% of the data lies between the limits of two sigma

  $\begin{array}{c}\mu \pm 2\sigma =\left(\mu -2\sigma ,\mu +2\sigma \right)\\ =\left(160-2\left(5.66\right),160+2\left(5.66\right)\right)\\ =\left(148.69,171.31\right)\end{array}$

Using the 68-95-99.7 rule, researcher expected to obtain get between 148.69 and 171.31 About 95 percent of the time, Bull's eyes.

99.7% of the data lies between the limits of three sigma

  $\begin{array}{c}\mu \pm 3\sigma =\left(\mu -3\sigma ,\mu +3\sigma \right)\\ =\left(160-3\left(5.66\right),160+3\left(5.66\right)\right)\\ =\left(143.03,176.97\right)\end{array}$

Using the 68-95-99.7 rule, researcher expected to obtain between 143.03and 176.97 About 99.7 percent of the time, Bull's eyes.

(d)

To determine

To Explain: that it would be surprised if researcher made only 140 bull’s eyes.

Answer to Problem 28E

0.0002

Explanation of Solution

Given:

  $\begin{array}{c}n=200\\ p=0.8\end{array}$

Calculation:

The probability is,

  $\begin{array}{c}P\left(X\le 140\right)=P\left(\frac{X-\mu }{\sigma }\le \frac{140-160}{5.66}\right)\\ =P\left(Z\le -3.53\right)\\ =1-P\left(Z\le 3.53\right)\\ =1-0.9998\\ =0.0002\end{array}$

Here, the value of chance is very small. So, it is obvious that it is quite unlikely that only 140 bull's-eyes out of 200 will be struck by the archer.