Do social robots walk or roll?Refer to the International Conference on Social Robotics(Vol. 6414, 2010) study onthe current trend in the design of social robots, Exercise 2.5(p. 72). Recall that in a random sample of social robotsobtained through a Web search, 28 were built with wheels.The number of wheels on each of the 28 robots is listed inthe accompanying table.

(4 4 3 3 3 6 4 2 2 2 1 3 3 3 3 4 4 3 2 8 2 2 3 4 3 3 4 2)

Source:Based on S. Chew et al., “Do Social Robots Walk or Roll?” International Conference on Social Robotics, Vol. 6414, 2010 (adapted from Figure 2).

a.Generate a histogram for the sample data set. Is the distribution of number of wheels mound-shaped and symmetric?

b.Find the mean and standard deviation for the sample data set.

c.Form the interval, xbar ±2s.

d.According to Chebychev’s Rule, what proportion of sample observations will fall within the interval, part c?

e.According to the Empirical Rule, what proportion of sample observations will fall within the interval, part c?

f.Determine the actual proportion of sample observations that fall within the interval, part c. Even though the histogram, part a, is not perfectly symmetric, does the Empirical Rule provide a good estimate of the proportion?

Step by step solution

01

Generating a histogram and determining its distribution shape

No, the distribution of number of wheels is not perfectly symmetric or mound-shaped.

02

Finding the mean and standard deviation

x	$(x-\overline{x})$	$(x-\overline{x}{)}^2$
4	0.79	0.6241
4	0.79	0.6241
3	-0.21	0.0441
3	-0.21	0.0441
3	-0.21	0.0441
6	2.79	7.7841
4	0.79	0.6241
2	-1.21	1.4641
2	-1.21	1.4641
2	-1.21	1.4641
1	-2.21	4.8841
3	-0.21	0.0441
3	-0.21	0.0441
3	-0.21	0.0441
3	-0.21	0.0441
4	0.79	0.6241
4	0.79	0.6241
3	-0.21	0.0441
2	-1.21	1.4641
8	4.79	22.9441
2	-1.21	1.4641
2	-1.21	1.4641
3	-0.21	0.0441
4	0.79	0.6241
3	-0.21	0.0441
3	-0.21	0.0441
4	0.79	0.6241
2	-1.21	1.4641
Sum = 90	0	50.7148

$$\begin{gathered} Mean=\frac{\sum x}{n} \\ =\frac{90}{28} \\ =3.21 \\ Variance=\frac{\sum {(x-\overline{x})}^2}{n-1} \\ =\frac{50.7148}{28-1} \\ =\frac{50.7148}{27} \\ =1.878 \\ StandardDeviation=\sqrt{Variance} \\ =\sqrt{1.878} \\ =1.37 \end{gathered}$$

Therefore, the mean and standard deviation are 3.21 and 1.37, respectively.

03

Creating the interval, x¯±2s

$\overline{x}\pm 2s$

From step 2, we have the values of mean and standard deviation, we will substitute them in the above equation to get the interval,

$$\begin{gathered} Upperrange=\overline{x}+2s \\ =3.21+2(1.37) \\ =3.21+2.74 \\ =5.95 \\ Lowerrange=\overline{x}-2s \\ =3.21-2(1.37) \\ =3.21-2.74 \\ =0.47 \end{gathered}$$

Therefore, the interval is 0.47 – 5.95.

04

Proportion of observations in the above interval according to the Chebyshev rule

According to the Chebyshev rule, ¾ proportion i.e., 75% of the observations should fall in the above interval.

05

Proportion of observations between 0.47 – 5.95 based on the Empirical rule

Based on the Empirical rule, 95% of the observations should fall in this interval.

06

Determining the actual proportion of sample observations that fall within 0.47 – 5.95

To find the actual proportion, we will first count the total number of observations that fall in this interval.

26 out of 28 observations fall in this interval. Therefore,

$$\begin{gathered} Actualproportion=\frac{26}{28} \\ =0.9285 \\ \%=0.9285\times 100 \\ =92.85\% \end{gathered}$$

92.85% of observations fall between 0.47 and 5.95.

Even though the histogram is not perfectly symmetric, the Empirical rule does provide a good estimate for the proportion.

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