Q. 7.19 - Elementary Statistics

Question

Repeat parts (b)-(e) of Exercise 7.17 for samples of size 3. 

Step-by-Step Solution

Answer

Part b. Constructing the sample of size 3 for the given population is given below, 

S. No.	Sample	Wealth	Mean Wealth ($\overline{x}$)
1	G,B,E	72,59,41	$\frac{72+59+41}{3}=57.33$
2	G,B,C	72,59,36	$\frac{72+59+36}{3}=55.67$
3	G,B,D	72,59,36	$\frac{72+59+36}{3}=55.67$
4	G,B,W	72,59,35	$\frac{72+59+35}{3}=55.33$
5	G,E,C	72,41,36	$\frac{72+41+36}{3}=49.67$
6	G,E,D	72,41,36	$\frac{72+41+36}{3}=49.67$
7	G,E,W	72,41,35	$\frac{72+41+35}{3}=49.33$
8	G,C,D	72,36,36	$\frac{72+36+36}{3}=48$
9	G,C,W	72,36,35	$\frac{72+36+35}{3}=47.67$
10	G,D,W	72,36,35	$\frac{72+36+35}{3}=47.67$
11	B,E,C	59,41,36	$\frac{59+41+36}{3}=45.33$
12	B,E,D	59,41,36	$\frac{59+41+36}{3}=45.33$
13	B,E,W	59,41,35	$\frac{59+41+35}{3}=45$
14	B,C,D	59,36,36	$\frac{59+36+36}{3}=43.67$
15	B,C,W	59,36,35	$\frac{59+36+35}{3}=43.33$
16	B,D,W	59,36,35	$\frac{59+36+35}{3}=43.33$
17	E,C,D	41,36,36	$\frac{41+36+36}{3}=37.67$
18	E,C,W	41,36,35	$\frac{41+36+35}{3}=37.33$
19	E.D,W	41,36,35	$\frac{41+36+35}{3}=37.33$
20	C,D,W	36,36,35	$\frac{36+36+35}{3}=35.67$

Part c. The dot plot is given below, 

Part d. The chance that the sample mean is equal to the population mean is 0.

Part e. The probability that $\overline{x}$ is within $3$ billion of $\mu$ is $40\%$.

Part (b) Step 1. Given Information

We have been given these six people a population of interest.

Part (b) Step 2. Construct samples of size 3 of the given population.

The samples of size 3 and the corresponding means is given below, 

S. No.	Sample	Wealth	Mean Wealth ($\overline{x}$)
1	G,B,E	72,59,41	$\frac{72+59+41}{3}=57.33$
2	G,B,C	72,59,36	$\frac{72+59+36}{3}=55.67$
3	G,B,D	72,59,36	$\frac{72+59+36}{3}=55.67$
4	G,B,W	72,59,35	$\frac{72+59+35}{3}=55.33$
5	G,E,C	72,41,36	$\frac{72+41+36}{3}=49.67$
6	G,E,D	72,41,36	$\frac{72+41+36}{3}=49.67$
7	G,E,W	72,41,35	$\frac{72+41+35}{3}=49.33$
8	G,C,D	72,36,36	$\frac{72+36+36}{3}=48$
9	G,C,W	72,36,35	$\frac{72+36+35}{3}=47.67$
10	G,D,W	72,36,35	$\frac{72+36+35}{3}=47.67$
11	B,E,C	59,41,36	$\frac{59+41+36}{3}=45.33$
12	B,E,D	59,41,36	$\frac{59+41+36}{3}=45.33$
13	B,E,W	59,41,35	$\frac{59+41+35}{3}=45$
14	B,C,D	59,36,36	$\frac{59+36+36}{3}=43.67$
15	B,C,W	59,36,35	$\frac{59+36+35}{3}=43.33$
16	B,D,W	59,36,35	$\frac{59+36+35}{3}=43.33$
17	E,C,D	41,36,36	$\frac{41+36+36}{3}=37.67$
18	E,C,W	41,36,35	$\frac{41+36+35}{3}=37.33$
19	E.D,W	41,36,35	$\frac{41+36+35}{3}=37.33$
20	C,D,W	36,36,35	$\frac{36+36+35}{3}=35.67$

Here, Bill Gates is represented by G, Warren Buffett is represented by B, Larry Ellison is represented by E, Charles Koch is represented by C, David Koch is represented by D and Chris Walton is represented by W. 

Part (c) Step 1. Construct the dot plot.

On constructing the dot plot for the sampling distribution of the sample mean, 

Part (d) Step 1. Find the chance that the sample mean will equal the population mean.

The population mean wealth for six people is $\mu =46.5$ billion.

From the table in part (b), it is clear that none of the sample means is equal to the population mean. Also, the number of samples size 3 is 20.

Thus,

$$\begin{gathered} P(\overline{x}=\mu )=\frac{0}{20} \\ =0 \end{gathered}$$

So, there is zero chance that the sample mean is equal to the population mean.

Part (e) Step 1. Find the probability that x ¯ is within 3 billion of μ

We need to find $P(\mu -3\le \overline{x}\le \mu +3)$

Here, $\mu =46.5$.

So from the table constructed in part b, it can be seen that there are $8$ sample means in the range $\left[46.5-3,46.5+3\right]=\left[43.5,49.5\right]$.

Also, the number of samples size 3 is 20.

Thus,

$$\begin{gathered} P(\mu -3\le \overline{x}\le \mu +3)=\frac{8}{20} \\ =0.40 \end{gathered}$$

So, there is a probability of 40% that the mean wealth of the three people obtained will be within 3 billion of the population mean.