Chapter 17, Problem 6E

(a)

To determine

To explain: among the five cars, that would simulate the amount of seatbelt drivers wearing.

Explanation of Solution

Define how to simulate among the five cars the number of seatbelt-wearing drivers:

It is considered that 75% of all drivers wear their seat belts all the time.

It is also provided that at the z traffic light, 5 cars are waiting.

As provided below, the simulation process is:

It will use a random number generator to produce 5 numbers between 1 and 15 (for each car at the traffic light), where each number will appear equally. (In Excel, those who'll use the RANDBETWEEN(1, 15) function).

It would then assume that if there is a number between 1-10 (inclusive), it would assume that it is a driver wearing a non-seatbelt.

The number of seatbelt-wearing drivers would then be registered.

(b)

To determine

To conduct: at least 30 trials.

Explanation of Solution

Run the simulation for at least 30 times:

Trial Number	1st Car	2nd Car	3rd Car	4th Car	5th Car	Number of seat belt-wearing drivers
1	9	11	9	10	1	4
2	4	1	10	10	1	5
3	9	6	8	14	11	3
4	1	15	3	11	15	2
5	11	7	9	11	1	3
6	1	13	6	13	10	3
7	12	13	9	11	4	2
8	11	4	14	4	5	3
9	9	13	7	3	6	4
10	13	15	15	8	3	2
11	5	3	13	10	13	3
12	8	1	2	9	4	5
13	12	8	4	1	10	4
14	5	2	7	13	9	4
15	7	11	3	14	13	2
16	4	8	4	11	1	4
17	1	8	1	7	6	5
18	1	15	7	3	3	4
19	5	5	6	4	14	4
20	15	10	4	9	7	4
21	12	12	13	4	3	2
22	6	15	10	2	5	4
23	14	15	7	3	1	3
24	3	8	1	8	10	5
25	13	2	3	6	2	4
26	3	2	1	11	10	4
27	13	13	7	5	1	3
28	1	9	10	1	3	5
29	9	3	13	1	1	4
30	3	2	11	14	6	3

(c)

To determine

To calculate: the probabilities that there are not belted drivers, exactly one and two etc.

Explanation of Solution

Estimating the probabilities for not belted drivers, 1 belted driver etc:

The frequency table on the basis of the simulation is mention below:

Number of seatbelt-wearing drivers(X)	Frequency
0	0
1	0
2	5
3	8
4	12
5	5
Total	30

The probabilities are get as given below:

P(A) = (number of favourable elements for event A)/ (Total number of elements in the sample space)

Number of seatbelt-wearing drivers	Frequency	Probability
0	0	0
1	0	0
2	5	0.166667
3	8	0.266667
4	12	0.4
5	2	0.166667
Total	30	1

(d)

To determine

To find: actual probability model.

Explanation of Solution

Given:

n = 5, p = 0.75)

Calculation:

pmf of X is:

  $f\left(x\right)=\{\begin{array}{l}\left({}^5{C}_x\right){\left(0.75\right)}^x{\left(0.25\right)}^{5-x},x=0,1,\mathrm{......5}\\ 0,\text{otherwise}\text{.}\end{array}$

find the binomial probabilities.

X	$P\left(X=x\right)$	$P\left(x\right)$
0	$P\left(X=0\right)=\left({}^5{C}_0\right){\left(0.75\right)}^0{\left(0.25\right)}^{5-0}$	=0.00098
1	$P\left(X=1\right)=\left({}^5{C}_1\right){\left(0.75\right)}^1{\left(0.25\right)}^{5-1}$	=0.01465
2	$P\left(X=2\right)=\left({}^5{C}_2\right){\left(0.75\right)}^2{\left(0.25\right)}^{5-2}$	=0.08789
3	$P\left(X=3\right)=\left({}^5{C}_3\right){\left(0.75\right)}^3{\left(0.25\right)}^{5-3}$	=0.26367
4	$P\left(X=4\right)=\left({}^5{C}_4\right){\left(0.75\right)}^4{\left(0.25\right)}^{5-4}$	=0.39551
5	$P\left(X=5\right)=\left({}^5{C}_5\right){\left(0.75\right)}^5{\left(0.25\right)}^{5-5}$	=0.2373

(e)

To determine

To explain: the comparison of the distribution of results in the simulation to probability model.

Explanation of Solution

Compare the distribution of results to the probability model in the simulation:In the table below, the probabilities of the simulation phase and the probability model are given:

X	Probabilities of simulation result	Probabilities of probability model
0	0.00098	0
1	0.01465	0
2	0.08789	0.166667
3	0.26367	0.266667
4	0.39551	0.4
5	0.2373	0.166667

The probabilities corresponding to X=0 in both models are tiny. The odds of X=3 and X = 4 are high.