Chapter 17, Problem 35E

(a)

To determine

To find: the number of cards that are supposed to stop before a driver whose seatbelt is not buckled is detected.

Answer to Problem 35E

5

Explanation of Solution

Given:

The probability of the driver wearing his seat belt is 0.80.

The probability of a driver not bucking is 0.20

Calculation:

X is geom. (0.2)

Expected number is

  $\begin{array}{c}\frac{1}{p}=\frac{1}{0.2}\\ =5\end{array}$

Before finding a driver, whose seat belt is not bucked up, 5 cars are likely to be stopped.

(b)

To determine

To find: the probability of stopping the first unbelted driver in the sixth vehicle.

Answer to Problem 35E

0.066

Explanation of Solution

Given:

The probability of the driver wearing his seat belt is 0.80.

The probability of a driver not bucking is 0.20

  $p=0.20$

Calculation:

Required probability is

  $\begin{array}{c}P\left(X=6\right)=q^{5}p\\ ={\left(0.8\right)}^5\times \left(0.2\right)\\ =0.066\end{array}$

There is, therefore, a 6.6 percent probability that the 6th car will be the first unbelted driver.

(c)

To determine

To Find: the probability that all first 10 drivers will wear their seat belts.

Answer to Problem 35E

0.1074

Explanation of Solution

Given:

The probability of the driver wearing his seat belt is 0.80.

The probability of a driver not bucking is 0.20

  $\begin{array}{l}p=0.20\\ n=10\end{array}$

Calculation:

Required probability is

  $\begin{array}{c}P\left(X=10\right)=C{}_{10}{\left(0.8\right)}^{10}{\left(0.2\right)}^0\\ =0.1074\end{array}$

There is a 10.74% probability that any of the first 10 drivers will wear seat belts.

(d)

To determine

To find: the mean and standard deviation of the number of seatbelts required to be worn by drivers.

Answer to Problem 35E

Mean = 2.4

Standard deviation = 2.19

Explanation of Solution

Given:

The probability of the driver wearing his seat belt is 0.80.

The probability of a driver not bucking is 0.20

  $\begin{array}{l}p=0.20\\ n=30\end{array}$

Formula used:

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

Calculation:

Expected number is

  $\begin{array}{c}np=30\left(0.8\right)\\ =2.4\end{array}$

Standard deviation is

  $\begin{array}{c}=\sqrt{npq}\\ =\sqrt{30\left(0.8\right)\left(0.2\right)}\\ =2.19\end{array}$

(e)

To determine

To find: the probability of finding that minimum 20 drivers do not wear their seat belts.

Explanation of Solution

Given:

The probability of the driver wearing his seat belt is 0.80.

The probability of a driver not bucking is 0.20

  $\begin{array}{l}p=0.20\\ n=120\end{array}$

Formula used:

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

Calculation:

  $\begin{array}{c}\text{Mean of }X\text{=120}\left(0.2\right)\\ =24\end{array}$

  $\begin{array}{c}\text{Standard deviation = }npq\\ =\sqrt{120\left(0.8\right)\left(0.2\right)}\\ =4.38\end{array}$

Less than 10% of all drivers are people who do not wear seat belts.

  $\begin{array}{c}np=24\text{ and}\\ npq=\left(120\right)\left(0.8\right)\\ =96\end{array}$

  $\text{ Both }npandnq\ge 10$

It is therefore possible to approximate binomial distribution with normal

  $X~\left(24,4.38\right)$