Normal distributions, determination

Using Table 1.3 i.e. Table B for standard normal distribution, determine probabilities that correspond to the following Z intervals. 

For Problem 19.6 (assuming normal distribution), determine the probability that it will take a person between 5 to 8 minutes to assemble the phone. 

For these various reasons it is suggested that the way forward is to discard the concept of the match altogether, and to use in its place precise probabilities of getting the observed deviation between the two bands (scene-of-crime and suspect) under the two hypotheses. In the case of the hypothesis of guilt, this probability would be calculated from a normal distribution determined by the measurement error. In the case of innocence, the probability would be... 

Figure 4. Cumulated normal distribution determined with point, moment and graphic estimator.

Therefore an automatic method, which means an objective and reproducible process, is necessary to determine the threshold value. The results of this investigations show that the threshold value can be determined reproducible in the point of intersection of two normal distributed frequency approximations. 

The shape of a normal distribution is determined by two parameters, the first of which is the population s central, or true mean value, p, given as... 

In Section 4D.2 we introduced two probability distributions commonly encountered when studying populations. The construction of confidence intervals for a normally distributed population was the subject of Section 4D.3. We have yet to address, however, how we can identify the probability distribution for a given population. In Examples 4.11-4.14 we assumed that the amount of aspirin in analgesic tablets is normally distributed. We are justified in asking how this can be determined without analyzing every member of the population. When we cannot study the whole population, or when we cannot predict the mathematical form of a population s probability distribution, we must deduce the distribution from a limited sampling of its members. 

A second example is also informative. When samples are obtained from a normally distributed population, their values must be random. If results for several samples show a regular pattern or trend, then the samples cannot be normally distributed. This may reflect the fact that the underlying population is not normally distributed, or it may indicate the presence of a time-dependent determinate error. For example, if we randomly select 20 pennies and find that the mass of each penny exceeds that of the preceding penny, we might suspect that the balance on which the pennies are being weighed is drifting out of calibration. 

Vitha, M. F. Carr, P. W. A Laboratory Exercise in Statistical Analysis of Data, /. Chem. Educ. 1997, 74, 998-1000. Students determine the average weight of vitamin E pills using several different methods (one at a time, in sets of ten pills, and in sets of 100 pills). The data collected by the class are pooled together, plotted as histograms, and compared with results predicted by a normal distribution. The histograms and standard deviations for the pooled data also show the effect of sample size on the standard error of the mean. 

Interpreting Control Charts The purpose of a control chart is to determine if a system is in statistical control. This determination is made by examining the location of individual points in relation to the warning limits and the control limits, and the distribution of the points around the central line. If we assume that the data are normally distributed, then the probability of finding a point at any distance from the mean value can be determined from the normal distribution curve. The upper and lower control limits for a property control chart, for example, are set to +3S, which, if S is a good approximation for O, includes 99.74% of the data. The probability that a point will fall outside the UCL or LCL, therefore, is only 0.26%. The... 

Furthermore, when both np and nq are greater than 5, the binomial distribution is closely approximated by the normal distribution, and the probability tables in Appendix lA can be used to determine the location of the solute and its recovery. 

From Example A6.2 we know that after 100 steps of the countercurrent extraction, solute A is normally distributed about tube 90 with a standard deviation of 3. To determine the fraction of solute in tubes 85-99, we use the single-sided normal distribution in Appendix lA to determine the fraction of solute in tubes 0-84 and in tube 100. The fraction of solute A in tube 100 is determined by calculating the deviation z (see Chapter 4)... 

This shows that Schlieren optics provide a means for directly monitoring concentration gradients. The value of the diffusion coefficient which is consistent with the variation of dn/dx with x and t can be determined from the normal distribution function. Methods that avoid the difficulty associated with locating the inflection point have been developed, and it can be shown that the area under a Schlieren peak divided by its maximum height equals (47rDt). Since there are no unknown proportionality factors in this expression, D can be determined from Schlieren spectra measured at known times. 

To determine R(/) for the normal distribution, a standard normal variate must be calculated by the following formula ... 

Many distribution functions can be apphed to strength data of ceramics but the function that has been most widely apphed is the WeibuU function, which is based on the concept of failure at the weakest link in a body under simple tension. A normal distribution is inappropriate for ceramic strengths because extreme values of the flaw distribution, not the central tendency of the flaw distribution, determine the strength. One implication of WeibuU statistics is that large bodies are weaker than small bodies because the number of flaws a body contains is proportional to its volume. 

Determining the area under the normal cuiwe is a very tedious procedure. However, by standardizing a random variable that is normally distributed, it is possible to relate all normally distributed random variables to one table. The standardization is defined by the identity z = (x — l)/<7, where z is called the unit normal. Further, it is possible to standardize the sampling distribution of averages x by the identity = (x-[l)/ G/Vn). 

When experimental data is to be fit with a mathematical model, it is necessary to allow for the facd that the data has errors. The engineer is interested in finding the parameters in the model as well as the uncertainty in their determination. In the simplest case, the model is a hn-ear equation with only two parameters, and they are found by a least-squares minimization of the errors in fitting the data. Multiple regression is just hnear least squares applied with more terms. Nonlinear regression allows the parameters of the model to enter in a nonlinear fashion. The following description of maximum likehhood apphes to both linear and nonlinear least squares (Ref. 231). If each measurement point Uj has a measurement error Ayi that is independently random and distributed with a normal distribution about the true model y x) with standard deviation <7, then the probability of a data set is... 

The probabihty-density function for the normal distribution cui ve calculated from Eq. (9-95) by using the values of a, b, and c obtained in Example 10 is also compared with precise values in Table 9-10. In such symmetrical cases the best fit is to be expected when the median or 50 percentile Xm is used in conjunction with the lower quartile or 25 percentile Xl or with the upper quartile or 75 percentile X[j. These statistics are frequently quoted, and determination of values of a, b, and c by using Xm with Xl and with Xu is an indication of the symmetry of the cui ve. When the agreement is reasonable, the mean v ues of o so determined should be used to calculate the corresponding value of a. 

Finally, it is worth investigating how deterministic values of material strength are calculated as commonly found in engineering data books. Equation 4.14 states that the minimum material strength, as used in deterministic calculations, equals the mean value determined from test, minus three standard deviations, calculated for the Normal distribution (Cable and Virene, 1967) ... 

Theoretically, the effects of the manufacturing process on the material property distribution can be determined, shown here for the case when Normal distribution applies. For an additive case of a residual stress, it follows that from the algebra of random variables (Carter, 1997) ... 

The distributional parameters for Kt in the form of the Normal distribution can then be used as a random variable product with the loading stress to determine the final stress acting due to the stress concentration. Equations 4.23 and 4.24 show... 

Confidence levels on the reliability estimates from the SSI model can be determined and are useful when the PDFs for stress and strength are based on only small amounts of data or where critical reliability projects are undertaken. However, approaches to determine these confidence levels only strictly apply when stress and strength are characterized by the Normal distribution. Detailed examples can be found in Kececioglu (1972) and Sundararajan and Witt (1995). 

Before a probabilistic model can be developed, the variables involved must be determined. It is assumed that the variables all follow the Normal distribution and that they are statistically independent, i.e. not correlated in anyway. The scatter of the pre-load, F, using an air tool with a clutch is approximately 30% of the mean, which gives the coefficient of variation, = 0.1, assuming 3cr covers this range, therefore ... 

Assuming that all the variables follow a Normal distribution, a probabilistie model ean be ereated to determine the stress distribution for the first failure mode using the varianee equation and solving using the Finite Differenee Method (see Appendix XI). The funetion for the von Mises stress, L, on first assembly at the solenoid seetion is taken from equation 4.75 and is given by ... 

In the stress rupture ease, the interferenee of the stress, L, and strength, Sy, both following a Normal distribution ean be determined from the eoupling equation ... 

Equation 4.83 states that there are four variables involved. We have already determined the load variable, F, earlier. The load is applied at a mean distanee, /r, of 150 mm representing the eouple length, and is normally distributed about the width of the foot pad. The standard deviation of the eouple length, cr, ean be approximated by assuming that 6cr eovers the pad, therefore ... 

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... 

Both of the torque eapaeities ealeulated, the holding torque of the hub and the shaft torque at yield, are represented by the Normal distribution, therefore we ean use the eoupling equation to determine the probability of interferenee, where ... 

The material selected for the pin was 070M20 normalized mild steel. The pin was to be manufactured by machining from bar and was assumed to have non-critical dimensional variation in terms of the stress distribution, and therefore the overload stress could be represented by a unique value. The pin size would be determined based on the —3 standard deviation limit of the material s endurance strength in shear. This infers that the probability of failure of the con-rod system due to fatigue would be very low, around 1350 ppm assuming a Normal distribution for the endurance strength in shear. This relates to a reliability R a 0.999 which is adequate for the... 

There is no data available on the endurance strength in shear for the material chosen for the pin. An approximate method for determining the parameters of this material property for low carbon steels is given next. The pin steel for the approximate section size has the following Normal distribution parameters for the ultimate tensile strength, Su ... 

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 ... 

A value of Cp = 1.33 would indicate that the distribution of the product characteristics covers 75% of the tolerance. This would be sufficient to assume that the process is capable of producing an adequate proportion to specification. The numbers of failures falling out of specification for various values of Cp and Cp can be determined from Standard Normal Distribution (SND) theory (see an example later for how to determine the failure in parts-per-million or ppm). For example, at Cp = 1.33, the expected number of failures is 64 ppm in total. 

It is evident that an approximate — 1.5cr shift ean be determined from the data and so the Cpi value is more suitable as a model. Using the graph on Figure 6, whieh shows the relationship Cp, (at 1.5cr shift) and parts-per-million (ppm) failure at the nearest limit, the likely annual failure rate of the produet ean be ealeulated. The figure has been eonstrueted using the Standard Normal Distribution (SND) for various limits. The number of eomponents that would fall out of toleranee at the nearest limit, is potentially 30 000 ppm at = 0.62, that is, 750 eomponents of the 25 000 manufaetured per annum. Of eourse, aetion in the form of a proeess eap-ability study would prevent further out of toleranee eomponents from being produeed and avoid this failure rate in the future and a target Cp = 1.33 would be aimed for. 

From the Standard Normal Distribution (SND) it is possible to determine the probability of negative elearanee, P. 

Because the Normal distribution is difficult to work with, we can use a 3-parameter Weibull distribution as an approximating model. Given the mean, /i, and standard deviation, cr, for a Normal distribution (assuming [3 = 3.44), the parameters xo and 6 can be determined from ... 

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