Question

Simulation. Let X be the value of a randomly selected decimal digit. that is a whole number between 0 and 9 inclusive.

Part (a) Use simulation to estimate the mean of X Explain your reasoning.

Part (b) Obtain the exact mean of X by applying definition 5.9 on page 22.7. Compare your result with that in part (a)

Short answer

Part (a) 4.440

Part (b) 4.500 From part (a) are roughly equal to the estimated mean of X.

Step-by-step solution

Part (a) Step 1. Given information. 

It is assumed that X is the value of a randomly chosen decimal digit, i.e. a whole number between 0 and 9.

Part (a) Step 2. The simulation of X's mean

Let X be the value of a decimal digit chosen at random, i.e. a whole number between 0 and 9.

Now, using MINITAB, simulate 200 numbers to obtain the frequency and percent frequency distributions for the simulated values listed above:

Discrete Variables Tally: X

X	Count	Percent
0	19	9.50
1	22	11.00
2	18	9.00
3	25	12.50
4	17	8.50
5	19	9.50
6	24	12.00
7	21	10.50
8	14	7.00
9	21	10.50
N	200

Because variable X has a finite population, its probabilities are the same as its relative frequencies. As a result, the probabilities of the above-mentioned simulated values are as follows:

x	P(X = x)
0	0.095
1	0.110
2	0.090
3	0.125
4	0.085
5	0.095
6	0.120
7	0.105
8	0.070
9	0.105

According to definitions, the estimated mean value of X is as follows:

$$\begin{gathered} {\mu }_x=\sum _{}xP(X=x) \\ {\mu }_x=\left[\left(0\times 0.095\right)+\left(1\times 0110\right)+...+\left(9\times 0.105\right)\right] \\ {\mu }_x=\{0.000+0.110+0.180+0.375+0.340+0.475+0.720+0.735+0.560+ \\ 0.0945\} \\ {\mu }_x=4.440 \end{gathered}$$

Part (b) Step 1. The exact mean of X using definition and comparing it to part (a)

For a population of whole numbers ranging from 0 to 9, inclusive. The following is the probability distribution of X:

x	P(X = x)
0	0.100
1	0.100
2	0.100
3	0.100
4	0.100
5	0.100
6	0.100
7	0.100
8	0.100
9	0.100

The exact mean value of X is given by definition as follows:

$$\begin{gathered} {\mu }_x=\sum _{}xP(X=x) \\ {\mu }_x=\left[\left(0\times 0.100\right)+\left(1\times 0100\right)+...+\left(9\times 0.100\right)\right] \\ {\mu }_x=\{0.0+0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9\} \\ {\mu }_x=4.500 \end{gathered}$$

The results from part (a) are roughly equal to the estimated mean of X.