Chapter PV, Problem 13RE

(a)

To determine

To find out what are the mean and standard deviation of the sampling distribution model for $\widehat{p}$ .

Answer to Problem 13RE

The mean is $0.8$ and standard deviation is $0.028$ .

Explanation of Solution

It is given in the question that a champion archer can generally hit the bull’s-eye $80\%$ of the time. Suppose she shoots $200$ arrows during competition. Thus, the mean and the standard deviation of the sampling distribution model for $\widehat{p}$ can be calculated as:

  $\begin{array}{l}E(x)=\mu \\ =p\\ =0.8\\ \sigma =SD=\sqrt{\frac{p(1-p)}{n}}\\ =\sqrt{\frac{0.8\times 0.2}{200}}\\ =0.028\end{array}$

(b)

To determine

To explain is normal model appropriate here.

Answer to Problem 13RE

Yes, the normal model is appropriate.

Explanation of Solution

It is given in the question that a champion archer can generally hit the bull’s-eye $80\%$ of the time. Suppose she shoots $200$ arrows during competition. The mean is $0.8$ and standard deviation is $0.028$ . Thus, a normal model is appropriate when the success or failure conditions are satisfied because both are greater than ten as,

  $\begin{array}{l}np=0.8\times 200>10\\ n(1-p)=0.2\times 200>10\end{array}$

(c)

To determine

To sketch the sampling model using the $68-95-99.7$ rule.

Explanation of Solution

It is given in the question that a champion archer can generally hit the bull’s-eye $80\%$ of the time. Suppose she shoots $200$ arrows during competition. The mean is $0.8$ and standard deviation is $0.028$ . Thus, the sketch of sampling model using the $68-95-99.7$ rule as,

  

(d)

To determine

To find out what is the probability that she gets at least $85\%$ bull’s-eyes.

Answer to Problem 13RE

The probability is $0.039$ .

Explanation of Solution

It is given in the question that a champion archer can generally hit the bull’s-eye $80\%$ of the time. Suppose she shoots $200$ arrows during competition. The mean is $0.8$ and standard deviation is $0.028$ . Thus, the probability that she gets at least $85\%$ bull’s-eyes can be calculated as,

  $\begin{array}{l}z=\frac{0.85-0.8}{0.028}\\ =1.77\end{array}$

And then from tables we get,

  $\begin{array}{l}P(z>1.77)=1-0.9616\\ =0.039\end{array}$