Mean and standard deviation

The method allows variables to be added or multiplied using basic statistical rules, and can be applied to dependent as well as independent variables. If input distributions can be represented by a mean, and standard deviation then the following rules are applicable for independent variables ... 

For a specified mean and standard deviation the number of degrees of freedom for a one-dimensional distribution (see sections on the least squares method and least squares minimization) of n data is (n — 1). This is because, given p and a, for n > 1 (say a half-dozen or more points), the first datum can have any value, the second datum can have any value, and so on, up to n — 1. When we come to find the... 

The F statistic, along with the z, t, and statistics, constitute the group that are thought of as fundamental statistics. Collectively they describe all the relationships that can exist between means and standard deviations. To perform an F test, we must first verify the randomness and independence of the errors. If erf = cr, then s ls2 will be distributed properly as the F statistic. If the calculated F is outside the confidence interval chosen for that statistic, then this is evidence that a F 2. 

Consider, for example, the data in Table 4.1 for the mass of a penny. Reporting only the mean is insufficient because it fails to indicate the uncertainty in measuring a penny s mass. Including the standard deviation, or other measure of spread, provides the necessary information about the uncertainty in measuring mass. Nevertheless, the central tendency and spread together do not provide a definitive statement about a penny s true mass. If you are not convinced that this is true, ask yourself how obtaining the mass of an additional penny will change the mean and standard deviation. 

How we report the result of an experiment is further complicated by the need to compare the results of different experiments. For example. Table 4.10 shows results for a second, independent experiment to determine the mass of a U.S. penny in circulation. Although the results shown in Tables 4.1 and 4.10 are similar, they are not identical thus, we are justified in asking whether the results are in agreement. Unfortunately, a definitive comparison between these two sets of data is not possible based solely on their respective means and standard deviations. 

Carbon has two common isotopes, and with relative isotopic abundances of, respectively, 98.89% and 1.11%. (a) What are the mean and standard deviation for the number of atoms in a molecule of cholesterol (b) What is the probability of finding a molecule of cholesterol (C27H44O) containing no atoms of... 

The mean and standard deviation for this sample are, respectively, 3.117 g and 0.051 g. Since the sample consists of seven measurements, there are six degrees... 

We begin by summarizing the mean and standard deviation for the data reported by each analyst. These values are... 

The mean and standard deviation for the differences are 2.25 and 5.63, respectively. The test statistic is... 

The following data were recorded during the preparation of a calibration curve, where S eas and s are the mean and standard deviation, respectively, for three replicate measurements of the signal. 

In this problem you will collect and analyze data in a simulation of the sampling process. Obtain a pack of M M s or other similar candy. Obtain a sample of five candies, and count the number that are red. Report the result of your analysis as % red. Return the candies to the bag, mix thoroughly, and repeat the analysis for a total of 20 determinations. Calculate the mean and standard deviation for your data. Remove all candies, and determine the true % red for the population. Sampling in this exercise should follow binomial statistics. Calculate the expected mean value and expected standard deviation, and compare to your experimental results. 

In Chaps. 5 and 6 we shall examine the distribution of molecular weights for condensation and addition polymerizations in some detail. For the present, our only concern is how such a distribution of molecular weights is described. The standard parameters used for this purpose are the mean and standard deviation of the distribution. Although these are well-known quantities, many students are familiar with them only as results provided by a calculator. Since statistical considerations play an important role in several aspects of polymer chemistry, it is appropriate to digress into a brief examination of the statistical way of describing a distribution. 

This result shows that the square root of the amount by which the ratio M /M exceeds unity equals the standard deviation of the distribution relative to the number average molecular weight. Thus if a distribution is characterized by M = 10,000 and a = 3000, then M /M = 1.09. Alternatively, if M / n then the standard deviation is 71% of the value of M. This shows that reporting the mean and standard deviation of a distribution or the values of and Mw/Mn gives equivalent information about the distribution. We shall see in a moment that the second alternative is more easily accomplished for samples of polymers. First, however, consider the following example in which we apply some of the equations of this section to some numerical data. 

The price of flexibility comes in the difficulty of mathematical manipulation of such distributions. For example, the 3-parameter Weibull distribution is intractable mathematically except by numerical estimation when used in probabilistic calculations. However, it is still regarded as a most valuable distribution (Bompas-Smith, 1973). If an improved estimate for the mean and standard deviation of a set of data is the goal, it has been cited that determining the Weibull parameters and then converting to Normal parameters using suitable transformation equations is recommended (Mischke, 1989). Similar estimates for the mean and standard deviation can be found from any initial distribution type by using the equations given in Appendix IX. 

The estimation of the mean and standard deviation using the moment equations as described in Appendix I gives little indication of the degree of fit of the distribution to the set of experimental data. We will next develop the concepts from which any continuous distribution can be modelled to a set of data. This ultimately provides the most suitable way of determining the distributional parameters. 

Figure 4.5 Cumulative frequency plot and determination of mean and standard deviation graphically...

Table 4.4 Mean and standard deviation of statistieally independent and eorrelated random variables v and y for some eommon funetions...

For example, the deterministic value for the yield strength, Sy, for SAE 1018 cold drawn steel for the size range tested is approximately 395 MPa (Green, 1992). Table 4.6 gives the mean and standard deviation as A(540,41) MPa. The lower bound value as used in deterministic design becomes ... 

We ean use a Monte Carlo simulation of the random variables in equation 4.83 to determine the likely mean and standard deviation of the loading stress, assuming that this will be a Normal distribution too. Exeept for the load, F, whieh is modelled by a 2-parameter Weibull distribution, the remaining variables are eharaeterized by the Normal distribution. The 3-parameter Weibull distribution ean be used to model... 

The mean and standard deviation of the hardness for the steel ean be determined from the regression eonstants TO and A as ... 

From equations 4.12 and 4.13, the mean and standard deviation for the ultimate tensile strength, Su, for steel ean be derived ... 

The above equations ean all be written in terms of the nominal dimensions, a, b and t, for the seetion. Solutions for the mean and standard deviation of eaeh property, for any seetion, ean be found using Monte Carlo simulation with knowledge of the likely dimensional variation for hot rolling of struetural steel seetions. The eoeffieient of variation for this proeess/material eombination is = 0.0083 (Haugen, 1980). 

In the above ealeulations of the mean, varianee and standard deviation, we make no prior assumption about the shape of the population distribution. Many of the data distributions eneountered in engineering have a bell-shaped form similar to that showed in Figure 1. In sueh eases, the Normal or Gaussian eontinuous distribution ean be used to model the data using the mean and standard deviation properties. 

Once the mean and standard deviation have been determined, the frequency distribution determined from the PDF can be compared to the original histogram, if one was constructed, by using a scaling factor in the PDF equation. For example, the expected frequency for the Normal distribution is given by ... 

If we plot a Normal distribution for an arbitrary mean and standard deviation, as shown in Figure 4, it ean be shown that at lcr about the mean value, the area under the frequeney eurve is approximately 68.27% of the total, and at 2cr, the area is 95.45% of the total under the eurve, and so on. This property of the Normal distribution then beeomes useful in estimating the proportion of individuals within preseribed limits. 

The following set of data represents the outeome of a tensile test experiment to determine the yield strength in MPa of a metal. There are 50 individual results and they are displayed in the order they were reeorded. It is required to find the mean and standard deviation when the data is represented by a histogram. It is also required to find the strength at —3cr from the mean for the metal and the proportion of individuals that eould be expeeted to have a strength greater than 500 MPa. 

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