Concept explainers
(a)
To explain: among the five cars, that would simulate the amount of seatbelt drivers wearing.
(a)
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)
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 conduct: at least 30 trials.
(b)
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 calculate: the probabilities that there are not belted drivers, exactly one and two etc.
(c)
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
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 find: actual probability model.
(d)
Explanation of Solution
Given:
n = 5, p = 0.75)
Calculation:
pmf of X is:
find the binomial probabilities.
X | ||
0 | =0.00098 | |
1 | =0.01465 | |
2 | =0.08789 | |
3 | =0.26367 | |
4 | =0.39551 | |
5 | =0.2373 |
(e)
To explain: the comparison of the distribution of results in the simulation to probability model.
(e)
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.
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