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The accompanying observations on stabilized viscosity (cP) for specimens of a certain grade of asphalt with \(18 \%\) rubber added are from the article "Viscosity Characteristics of Rubber-Modified Asphalts" (J. of Materials in Civil Engr., 1996: 153-156): \(\begin{array}{lllll}2781 & 2900 & 3013 & 2856 & 2888\end{array}\) a. What are the values of the sample mean and sample median? b. Calculate the sample variance using the computational formula.

Short Answer

Expert verified
The sample mean is 2887.6, the median is 2888, and the variance is 76915.8 cP².

Step by step solution

01

Calculating the Sample Mean

The sample mean is calculated using the formula:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]where \(n\) is the number of observations and \(x_i\) are the sample values.For the given data, we have:\[ \bar{x} = \frac{1}{5} (2781 + 2900 + 3013 + 2856 + 2888) \]\[ \bar{x} = \frac{1}{5} \times 14438 = 2887.6 \]
02

Calculating the Sample Median

To find the median, the data must be sorted in ascending order: 2781, 2856, 2888, 2900, 3013. The median is the middle value. Since there are 5 observations, the median is the third value. Thus, the median is 2888.
03

Calculating the Sample Variance - Computational Formula

The computational formula for variance is:\[ s^2 = \frac{1}{n-1} \left(\sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n}\right) \]First, calculate \(\sum_{i=1}^{n} x_i^2\):\[ \sum_{i=1}^{5} x_i^2 = 2781^2 + 2900^2 + 3013^2 + 2856^2 + 2888^2 = 40296341 \]Substitute the values in the formula:\[ s^2 = \frac{1}{4} \left( 40296341 - \frac{14438^2}{5} \right) \]\[ s^2 = \frac{1}{4} \left( 40296341 - 41608404.8 \right) \]\[ s^2 = \frac{1}{4} \times 307663.2 = 76915.8 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Asphalt Viscosity Analysis
Asphalt viscosity analysis is crucial in evaluating the performance characteristics of asphalt, especially when modified with additives such as rubber. Viscosity refers to a fluid's resistance to flow. A higher viscosity means the fluid is thicker and flows less easily. In pavement engineering, ensuring the right viscosity can help maintain the integrity and longevity of a road surface.
Adding rubber to asphalt can affect its viscosity, potentially enhancing the material's durability and flexibility. This is particularly important in varying weather conditions where the pavement might contract or expand. Understanding how additives alter viscosity allows engineers to select appropriate materials for specific environments.
In our specific analysis, viscosity observations for asphalt samples with 18% rubber were recorded. The values are as follows: 2781, 2900, 3013, 2856, 2888. These measurements are used to derive statistical insights such as mean, median, and variance, helping researchers understand how consistent the viscosity characteristics are across different asphalt samples.
Sample Median Calculation
Sample median calculation is a straightforward statistical method used to determine the middle value of an ordered data set. For the asphalt viscosity data provided, sorting is the first step: 2781, 2856, 2888, 2900, 3013.
Once ordered, locating the median is simple since there is an odd number of values in our data set. The median is the third value. This method is especially useful when the dataset might have outliers, as the median can provide a better central measure than the mean. In our example, despite variances in the values, the median gives us a clear indication of the central tendency, which is 2888 cP.
The practical significance of finding the median in asphalt viscosity analysis is that it shows a typical value around which other measurements cluster, giving a tangible understanding of expected performance.
Computational Formula for Variance
The computational formula for variance is essential to quantify the spread or dispersion of a data set. Variance helps us understand the variability of viscosity measurements in asphalt samples. A higher variance indicates that the data points are spread out over a wider range of values, which can be critical in quality control processes. The formula used here is\[ s^2 = \frac{1}{n-1} \left(\sum_{i=1}^{n} x_i^2 - \frac{(\sum_{i=1}^{n} x_i)^2}{n}\right)\] where \(s^2\) represents the sample variance, \(n\) is the number of observations, and \(x_i\) are the sample values. For asphalt viscosity values of 2781, 2900, 3013, 2856, and 2888, we calculated \(s^2\) to be 76915.8. This detailed breakdown helps in understanding how each observation compares with the mean, guiding quality assessments and aiding in decision-making for material usage.
Such analyses are pivotal when comparing multiple batches or determining the impact of various additives on material performance.

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Most popular questions from this chapter

A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of the paper "Effects of Short-Term Supplementation with Selenised Yeast on Milk Production and Composition of Lactating Cows" (Australian J. of Dairy Tech., 2004: 199-203) supplied the following data on milk selenium concentration \((\mathrm{mg} / \mathrm{L})\) for a sample of cows given a selenium supplement and a control sample given no supplement, both initially and after a 9-day period. $$ \begin{array}{rrrrr} \text { Obs } & \text { Init Se } & \text { Init } & & \text { Final } \\ 1 & 11.4 & \text { Cont } & \text { Final Se } & \text { Cont } \\ 2 & 9.6 & 8.7 & 104.0 & 8.8 \\ 3 & 10.1 & 9.7 & 96.4 & 8.8 \\ 4 & 8.5 & 10.8 & 89.0 & 10.1 \\ 5 & 10.3 & 10.9 & 88.0 & 9.6 \\ 6 & 10.6 & 10.6 & 103.8 & 8.6 \\ 7 & 11.8 & 10.1 & 147.3 & 10.4 \\ 8 & 9.8 & 12.3 & 97.1 & 12.4 \\ 9 & 10.9 & 8.8 & 172.6 & 9.3 \\ 10 & 10.3 & 10.4 & 146.3 & 9.5 \\ 11 & 10.2 & 10.9 & 99.0 & 8.4 \\ 12 & 11.4 & 10.4 & 122.3 & 8.7 \\ 13 & 9.2 & 11.6 & 103.0 & 12.5 \\ 14 & 10.6 & 10.9 & 117.8 & 9.1 \\ 15 & 10.8 & & 121.5 & \\ 16 & 8.2 & & 93.0 & \end{array} $$ a. Do the initial Se concentrations for the supplement and control samples appear to be similar? Use various techniques from this chapter to summarize the data and answer the question posed. b. Again use methods from this chapter to summarize the data and then describe how the final Se concentration values in the treatment group differ from those in the control group.

The article "Determination of Most Representative Subdivision" (J. of Energy Engr., 1993: 43-55) gave data on various characteristics of subdivisions that could be used in deciding whether to provide electrical power using overhead lines or underground lines. Here are the values of the variable \(x=\) total length of streets within a subdivision: $$ \begin{array}{rrrrrrr} 1280 & 5320 & 4390 & 2100 & 1240 & 3060 & 4770 \\ 1050 & 360 & 3330 & 3380 & 340 & 1000 & 960 \\ 1320 & 530 & 3350 & 540 & 3870 & 1250 & 2400 \\ 960 & 1120 & 2120 & 450 & 2250 & 2320 & 2400 \\ 3150 & 5700 & 5220 & 500 & 1850 & 2460 & 5850 \\ 2700 & 2730 & 1670 & 100 & 5770 & 3150 & 1890 \\ 510 & 240 & 396 & 1419 & 2109 & & \end{array} $$ a. Construct a stem-and-leaf display using the thousands digit as the stem and the hundreds digit as the leaf, and comment on the various features of the display. b. Construct a histogram using class boundaries 0,1000 , \(2000,3000,4000,5000\), and 6000 . What proportion of subdivisions have total length less than 2000 ? Between 2000 and 4000 ? How would you describe the shape of the histogram?

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