Chapter 17, Problem 27E

(a)

To determine

To find: the mean and standard deviation of the predicted number of successful first serves.

Answer to Problem 27E

Mean= 56

Standard Deviation = 4.099

Explanation of Solution

Given:

  $\begin{array}{l}n=80\\ p=0.7\end{array}$

Formula used:

  $\begin{array}{c}E\left(X\right)=np\\ SD\left(X\right)=\sqrt{npq}\end{array}$

Calculation:

The Mean is

  $\begin{array}{c}E\left(X\right)=np\\ =\left(80\right)\times \left(0.7\right)\\ =56\end{array}$

The Standard deviation is

  $\begin{array}{c}SD\left(X\right)=\sqrt{npq}\\ =\sqrt{80\left(0.7\right)\left(0.3\right)}\\ =\sqrt{16.8}\\ =4.099\end{array}$

(b)

To determine

To Verify: the normal model to approximate the number of successful 1^sts serves distribution.

Explanation of Solution

Given:

  $\begin{array}{l}n=80\\ p=0.7\end{array}$

Calculation:

From the mention information,

  $\begin{array}{c}nq=56\\ nq\text{=80}\left(0.3\right)\\ =24\end{array}$

Here $nq>10\text{and }nq\text{>10}$ these serves are <10% of all the serves.

Therefore, it could be approximated this binomial probability by using normal model.

(c)

To determine

To Explain: the distribution using the 68-95-99.7 rule.

Explanation of Solution

Given:

  $\begin{array}{l}n=80\\ p=0.7\end{array}$

Formula used:

Calculation:

Using the 68-95-99.7 rule

1. 68% of amount of time of serves needed is between

  $\begin{array}{c}=\left(\mu -\sigma ,\mu +\sigma \right)\\ =\left(56-4.099,56+4.099\right)\\ =\left(51.9,60.1\right)\end{array}$

2. There is 95% amount of time of serves needed is between

  $\begin{array}{c}=\left(\mu -2\sigma ,\mu +2\sigma \right)\\ =\left(56-2\left(4.099\right),56+2\left(4.099\right)\right)\\ =\left(47.8,64.2\right)\end{array}$

3. There is 99.7% amount of time of serves needed is between

  $\begin{array}{c}=\left(\mu -3\sigma ,\mu +3\sigma \right)\\ =\left(56-3\left(4.099\right),56+3\left(4.099\right)\right)\\ =\left(43.7,68.3\right)\end{array}$

(d)

To determine

To find: the probability that researcher makes minimum 65 1^st serves.

Answer to Problem 27E

0.0141

Explanation of Solution

Given:

  $\begin{array}{l}n=80\\ p=0.7\end{array}$

Calculation:

The probability that researcher makes minimum 65 first serves is,

  $\begin{array}{c}P\left(X\ge 65\right)=1-P\left(\frac{X-\mu }{\sigma }\ge \frac{65-56}{4.099}\right)\\ =1-P\left(Z\ge 2.1957\right)\\ =1-0.9859\left(\text{using excel function }\left(=\text{NORMSDIST}\left(0.9859\right)\right)\right)\\ =0.0141\end{array}$