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Contents 5 Normal Distribution

Example 2 In a Pet Tabs (pet vitamin tablets) production, the pharmaceutical manufacturer is using milling and micronizing machines to pulverize raw materials into fine particles. These finished particles are combined and processed further in mixing machines. The mixed ingredients are then pressed into tablets, dried, and sealed in packages. A normally distributed quality characteristic, moisture content, is monitored. Samples of n = 4 tablets are taken from the manufacturing process every hour. The data after 25 samples have been collected are shown in Table 5. [Pg.299]

Tables Va, b, and c attempt to answer these questions. In Table Va percent copper is ranked in increasing order along with the laboratory numbers and the methods. First, are any of the values outliers Should all the values be taken as representative of the copper content, or are some in error Assuming that the data are normally distributed (i.e., that they follow the Gaussian or bell-curve distribution) the V test can be applied to reject anomalous values this was done, and four values were rejected. Three come from one laboratory, and the fourth came from a laboratory which did not detect or take Zn into account on sample 3. Therefore, their calculation for copper was too high. Tables Va, b, and c attempt to answer these questions. In Table Va percent copper is ranked in increasing order along with the laboratory numbers and the methods. First, are any of the values outliers Should all the values be taken as representative of the copper content, or are some in error Assuming that the data are normally distributed (i.e., that they follow the Gaussian or bell-curve distribution) the V test can be applied to reject anomalous values this was done, and four values were rejected. Three come from one laboratory, and the fourth came from a laboratory which did not detect or take Zn into account on sample 3. Therefore, their calculation for copper was too high.
The standard deviation of tin content, when compared with the data, suggests that the tin percentages are normally distributed, the expected result for coins made by the same mint or by the same methods. Twenty-four of the thirty-six coins (approximately 67%) had tin percentages within one standard deviation of the mean, and thirty-four of the coins (approximately 95%) fell within two... [Pg.250]

Another piece of evidence to suggest that content uniformity data is not normally distributed can also be seen in the content uniformity data of Rohrs et al.4 They report the content uniformity assay results for 11 batches of tablets. In every case except one, the actual mean potency is below the target value. If the content uniformity data were normally distributed, as assumed by the Yalkowsky and Bolton approach,2 one might expect to see a more even distribution of mean values above and below the target. The one exception was for the lowest potency batch for which one tablet was assayed to be 292% of intented value, as mentioned above. This characteristic is... [Pg.60]

At this point let us return to the aluminum content data of Table 5.1. The skewed shape that is evident in all of Figs 5.2—5.5 makes a Gaussian distribution inappropriate as a theoretical model for (raw) aluminum content of such PET samples. But as is often the case with right skewed data, considering the logarithms of the original measurement creates a scale where a normal distribution is more plausible as a representation of the phenomenon under... [Pg.184]

Figure 4.3 Contours of constant density 17) for the bivariate normal distribution. Each contour encloses a probability content of 0.95. Figure 4.3 Contours of constant density 17) for the bivariate normal distribution. Each contour encloses a probability content of 0.95.
We recorded the water content of individual vials in the order of their moving out from the freeze-dryer and constructed the water content distribution over the vials in the freeze-drayer. Based on those data we obtained information about the operation state, i.e., the bias of heat flux in the shelf or the residence of water vapor in the chamber. The information was very helpful in the maintenance of the machine to achieve uniform conditions in the chamber. Furthermore, we determined the uniformity of the vials in each batch or the consistency among batches based on the information of the water content distribution by statistical analysis. An example of allocation of trays in a freeze-dryer and three-dimensional distribution of moisture levels in individual vials is shown in Figure 24. Frequency distributions of vial moisture and their normal distribution curves in different batches made with different freeze-dryers are shown in Figure 25. [Pg.458]

The F-distribution has great utility in a statistical test referred to as analysis of variance (ANOVA). ANOVA is a powerful tool for testing the equivalence of means from samples obtained from normally distributed, or approximately normally distributed, populations. As an example, suppose that the following are the content uniformity values on 20 tablets from each of four different lots lot A mean = 99.5%, standard deviation = 2.6% lot B mean = 100.2%, standard deviation = 2.8% lot C mean = 90.5%, standard deviation = 2.1% and lot D mean = 100.3%, standard deviation = 2.7%. [Pg.3492]

To study local concentrations of cytokines, it is necessary to define the (1) secreted concentration of the cytokine per cell per time unit, (2) number of secretory vesicles fusing with the plasma membrane per time unit, (3) distribution of fusing vesicles on cell surface (normally not evenly spaced), (4) size of secretory vesicles, (5) concentration of cytokines in secretory vesicles, (6) liberation of vesicle contents (normal versus propulsion), and (7) superficial diffusion (over ceU membranes) coefficient and time. ... [Pg.720]

In Figure 85, the area in the right-hand tail (above Zq) of the normal distribution curve represents the probability that the variable (an event) is Zq standard deviations above the mean. Applied to this problem, the area in the right-hand tail (above a Zq of 2.5) is the probability that the lead content in any drum exceeds 45 ppm. From the table, the area in the right-hand tail corresponding to a Zq of 2.5, is 0.006 or 6% of the total area under the curve. The probability that lead content in a drum exceeds 45 ppm is therefore 0.6%. [Pg.472]

Again from the normal distribution table, for zq = 0.98, the probability that the mean lead content in the receiving tank exceeds 20 ppm is 0.164 or 16.4%. [Pg.474]

The mean total anthocyanin content of the wines fermented with the ten yeast strains was 430 mg. The mean total adsorbate anthocyanin content was 74 mg/L (Table 1). The distribution of anthocyanins was very different between the wines and the cells walls. In general, in the wines (Fig. 6 (a)), anthocyanins glycosylated at position 3 (3G) were the most conunon (59 %), followed by acetyl derivatives (6Ac) (33 %), then ciimamoyl derivatives at S.4 %, and pyruvic derivatives at 2.9 % (Table 1). This is the normal distribution for Cabemet-Sauvignon wines. The glycosylated, aceQ l and pyruvic derivatives were slightly less common in the adsorbates than in the wines. However, the quantities of p-coumaryl and caffeyl derivatives were vety much greater (Fig. 6 (b)). [Pg.104]

EMULSION POLYMERIZATION Used for standard SBR. Monomer is emulsified in water with emulsifying agents. Polymerization is initiated by either decomposition of a peroxide or a peroxydisulfate. Hot SBR is initiated by free radicals generated by thermal decomposition of initiators at 50°C or higher. Cold SBR is initiated by oxidation-reduction reactions (redox) at temperatures as low as —40°C. Stjrrene content normally is 23%. Copolymer is randomly distributed. Structure of butadiene contents is about 18% ds-1,4, 65% frans-1,4, and 15-20% vinyl. [Pg.983]

Guttman 1. Constmction of y6-content tolerance regions at confidence level y for large samples from the k-variate normal distribution. Annals of Mathematical Statistics... [Pg.486]

The distribution of trans monoene isomers in intestinal contents of ruminants appears normally distributed when double bond position is arranged in ascending order. The trans- isomer is usually found in highest concentration. Proportions of the other isomers steadily decrease as the double bond is shifted in either direction from carbon 11. Because the trans- 1 monoene is converted to the anti-carcinogenic cis-9, trans-W CLA isomer by tissue desaturases (24), there is interest in enhancing trans- l Cl8 1 concentration in meat and milk products for human consumption. [Pg.204]


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