Chapter 5, Problem 2CRE

a.

To determine

To describe:the design of simulation to estimate the probability.

Explanation of Solution

Given:

15% of the driver stopped for routine license checks are not wearing seat belts.

Calculation:

Choose a row table D.

Select the first two digit number from the selected row of table D.

If the number is between 00 and 14, then the person is NOT wearing a seat belt, else the person is wearing a seat belt.

Do the same for the following 9 two-digit numbers (Repetitions of the number is allowed).

Determine the number of people not wearing a seat belt.

Repeat the simeentaneously multiple times and determine the proportion of simulations that contain at least 2 consecutives drivers not wearing a seat belts.

b.

To determine

To carry out: the three repetition the simeentaneously.

Answer to Problem 2CRE

  $P\left(\text{At leeast two}\right)=33.33\%$

Explanation of Solution

Given:

15% of the driver stopped for routine license checks are not wearing seat belts.

The random digits are:

29077	14863	61683	47052	62224	51025
95052	90908	73592	75186	87136	95761
27102	56027	55892	33063	41842	81868
43367	49497	72719	96758	27611	91596

Calculation:

Repeat the simeentaneously multiple times and determine the proportion of simulations that contain at least 2 consecutives drivers not wearing a seat belts.

FIRST REPETITION

  $\begin{array}{l}29\Rightarrow \text{Wearing a seat belt}\\ 07\Rightarrow \text{Not wearing a seat belt}\\ 71\Rightarrow \text{Wearing a seat belt}\\ 48\Rightarrow \text{Wearing a seat belt}\\ 63\Rightarrow \text{Wearing a seat belt}\\ 31\Rightarrow \text{Wearing a seat belt}\\ 34\Rightarrow \text{Wearing a seat belt}\\ 70\Rightarrow \text{Wearing a seat belt}\\ 52\Rightarrow \text{Wearing a seat belt}\end{array}$

Thus, in first repetition 1 out of 10 is not wearing a seat belt.

SECOND REPETITION

  $\begin{array}{l}62\Rightarrow \text{Wearing a seat belt}\\ 22\Rightarrow \text{Wearing a seat belt}\\ 45\Rightarrow \text{Wearing a seat belt}\\ 10\Rightarrow \text{Not wearing a seat belt}\\ 25\Rightarrow \text{Wearing a seat belt}\\ 95\Rightarrow \text{Wearing a seat belt}\\ 05\Rightarrow \text{Not wearing a seat belt}\\ 29\Rightarrow \text{Wearing a seat belt}\\ 09\Rightarrow \text{Not wearing a seat belt}\\ 08\Rightarrow \text{Not wearing a seat belt}\end{array}$

Thus, in second repetition 4 out of 10 is not wearing a seat belt.

THIRD REPETITION

  $\begin{array}{l}73\Rightarrow \text{Wearing a seat belt}\\ 59\Rightarrow \text{Wearing a seat belt}\\ 27\Rightarrow \text{Wearing a seat belt}\\ 51\Rightarrow \text{Wearing a seat belt}\\ 86\Rightarrow \text{Wearing a seat belt}\\ 87\Rightarrow \text{Wearing a seat belt}\\ 13\Rightarrow \text{Not wearing a seat belt}\\ 69\Rightarrow \text{Wearing a seat belt}\\ 57\Rightarrow \text{Wearing a seat belt}\\ 61\Rightarrow \text{Wearing a seat belt}\end{array}$

Thus, in third repetition 1 out of 10 is not wearing a seat belt.

Then it is noted that 1 out of three repetitions contains at least two people not wearing repetitions.

Probability is the number of favourable outcomes divided by the number of possible outcomes.

  $P\left(\text{At leeast two}\right)=\frac{\text{favourable outcomes}}{\text{possible outcomes}}=\frac{1}{3}=0.3333=33.33\%$