2.4
a. Figure 1 illustrates the dot plot of the basketball players. There is a variety of heights presented on the plot.
Figure 1. The dot plot of the basketball players.
b. The tallest player is 86 inches, the shortest player is 72 inches.
c. The most common height is 74 inches, and 5 players share it.
d. The most common height is illustrated by horizontal clustering of the dots, or when the numerous dots are on the same heights line.
2.10
a. The frequency distribution is presented in Table 1.
Frequency Distribution of the Cars Speed
b. The class width is the difference between the higher and lower value of any class: 54-48 = 6. Therefore, 6 is the class width (Johnson & Kuby, 2012).
c. For the class 24-30, the lower class boundary is 24, the upper class boundary is 30; the class midpoint is 27 (Johnson & Kuby, 2012).
d. Figure 2 presents the frequency distribution of the car speed.
Figure 2. The frequency distribution of the car speed.
2.16
a. The mean value (Johnson & Kuby, 2012):
Mean=2+4+7+8+95=6.
b. The median for the data 2, 4, 7, 8, 9. The median is the middle value 7 (Johnson & Kuby, 2012).
c. The mode is the value that is the most frequent. Since each value in the sample is represented once, there is no mode for this set of data (Johnson & Kuby, 2012).
d. The midrange (Johnson & Kuby, 2012):
Midrange=Max+Min2=2+92=5.5.
2.22
The mean value is $150, the standard deviation is $125. The standard deviation is about 75% of the mean value. This indicates that the loss of customers varied significantly, or there are customers that lost more that $150, and there are many that lost much less than $150. The distribution of losses is not normal, and there are many extreme (maximum and minimum values). In this case, the mean value is not the best representation of the central tendency. Instead, median or mode should be used (Johnson & Kuby, 2012).
2.29
The mean lifetime is 30,000 miles, the standard deviation 2,500 miles.
a. What percentage of tires will last between 22,500 and 37,500 miles. Using the properties of the normal distribution:
- 68% of the values are within the range Mean ± Standard deviation;
- 95% of the values are within the range Mean ± 2Standard deviations;
- 99.7% of the values are witin the range Mean ± 3Standard deviations (Johnson & Kuby, 2012).
The tires mileage is in range 22,500 and 37,500. This is Mean ± 3Standard deviations, 30,000 ± 3 · 2,500. Therefore, 99.7% of the tires will last between 22,500 and 37,500 miles.
b. If we do not have any assumptions of the numerical data set, then we should apply the Chebyshev’s theorem. It states that for any data set:
- at least 75% of the data are within the range Mean ± 2Standard deviations;
- at least 8/9 of the data are within the range Mean ± 3Standard deviations;
- at least (1-1/k2) of the data are within the range Mean ± k·Standard deviations (Johnson & Kuby, 2012).
Therefore, for the problem the data are within Mean ± 3Standard deviations, and at least 8/9, or 88.9% of the data are within the given interval.
References
Johnson, R. R., & Kuby, P. (2012). Elementary statistics. Boston, MA: Brooks/Cole Cengage Learning.
