Read the pages and make a summary of them with your own words, please. It is what you understand. Calculating Standard Deviation (o)Repeating this process through the 10-bead row completesthe deviation column.These values are entered in Figure 15.15 to create Figure 15.18:To understand the process's variability, we must know itsstandard deviation. The formula for standard deviation isu - lo = 5.1 - 1.49 = 3,61Column 5 of Figure 15.17c is simply a list of thecolumn 4 deviation values squared. For example, in the0-bead row, column 4 shows d = -5.1. Column 5 lists thesquare of -5.1, or 26.01. The 1-bead row has d = -4.1.Column 5 lists its square, 16.81. This process is contin-ued through the 10-bead row to complete column 5 of thefigure.Column 6 of Figure 15.17c lists the results of thesquared deviations (column 5) multiplied by the numberof samples at the corresponding deviation value (column2). For the column 6 entry at the 0-bead row, we multi-ply 0 (from column 2) by 26.01 (from column 5); since,0 X 26.01 = 0, 0 is entered in column 6. For the 1-beadrow, we multiply 1 by 16.81; 16.81 is the second entry incolumn 6. At the 2-bead row, we multiply 3 by 9.61 andenter 28.83 in column 6. This process is repeated throughthe remaining rows of the figure.Next we add column 6's entries to obtain the sum of thesquared deviations, Ed. Ed for our bead process experi-u + lo = 5.1 + 1.49 = 6.59Ed/(n – 1).u - 20 = 5.1 - 2.99 = 2.11u + 20 = 5.1 + 2.99 = 8.09whereu - 30 = 5.1 - 4.47 = 0.63d = the deviation of any unit from the meanu + 30 = 5.1 + 4.47 = 9.57n = the number of units sampledSuppose we have a process that is operating like thecurve in Figure 15.18. We have specifications for the prod-uct output that require us to reject any part below 3.6 andabove 6.6. It turns out that these limits are approximately+lo. We know immediately that about one-third of theprocess output will be rejected. (Refer to the first bulletin the list on page 240.) If this is not acceptable, which ishighly probable, we will have to improve the process orchange to a completely different process. Even if morevariation could be tolerated in the product and we tookthe specification limits out to 2 and 8, about 5 of every 100pieces flowing out of the process would still be rejected.In a competitive world, this is poor performance indeed.Many companies no longer consider 2,700 parts per mil-lion defective (±30) to be good enough. A growing num-ber of organizations are seeking the Motorola version ofSix Sigma quality performance. These companies target adefect rate of 3.4 nonconformances per million opportu-nities (NPMO) for nonconformance. Technically speak-ing, 3.4 NPMO is not very close to the statistically pure6-sigma rate of 0.002 per million opportunities, or 1 non-conformance in 500 million. (We explain this difference inthe Six Sigma section of Chapter 19.) Although the popularSix Sigma does not match the true 6-sigma, 3.4 NPMO is aWe already have the value of n (100), but we have notcalculated the values of d, d, or Ed. We will perform thesecalculations and post the information in the remainingthree columns of Figure 15.17c. The values of the deviations(d) are determined by subtracting µ (5.1) from each of thered bead values (0 through 10) of column 1. The first entryin column 4 (deviation from u) is determined by subtract-ing u from the value in column 1, that is, 0 5.1 = -5.1Similarly, the second entry in column 4 is the value of col-umn 1 at the 1-bead row minus u, or 1 - 5.1 = -4.1.ment is 221.Now we have all the information we need to calculatethe standard deviation (o) for our process.o = VEd/(n – 1)24Multiply Col 1by Col 2Measured data fromDeviationDeviationsquared(Col 4)Sum ofdeviations squared(Col 2x Col 5)Figure 15.16from u(Col 1-o = V221+ 99# of RedBeads# ofSampleso = V2.23 (2.23 is called the mean squared deviation.)o = 1.49 (1.49 is called the root mean squared deviation.)X Value26.0116.819.614.41-5.14.116.81Note: Calculations are to two decimal places.Next calculate the positions of u ± lo, 20, and 30.3-3.1-2.128.8339.693271976-1.11.210.0122.990.3131155-0.1o = 1.49 20 = 2.99 30 = 4.47211260.90.8117.011177241.92.93.618.4139.7125.238.9.2183.915.2130.42104924.01n= 100EX= 510Ed= 221FIGURE 15.17c Completed Deviation Data Table. Or, you may use the formulafor example. One batch produces the left bell curve, andthe second batch the curve on the right. The two curvesmay be separated for a better view of what is going on bystratifying the data by batch. (See the "Stratification" sec-tion later in this chapter.)Histogram F suggests that someone is discarding thesamples below and above a set of limits. This typically happenswhen there is a 100% inspection and only data that are withinlimits are recorded. The strange Histogram G might have useddata from incoming inspection. The message here is that thevendor is screening the parts and someone else is getting thebest ones. A typical case might be electrical resistors that aregraded as 1,5, and 10% tolerance. The resistors that met 1 and3025k = VN20It is not necessary to be very precise with this. Thesemethods are used to get close and adjust one way or theother for a fit with your data.Suppose we are using steel balls in one of our prod-ucts and the weight of the ball is critical. The specificationis 5 ± 0.2 grams. The balls are purchased from a vendor,and because our tolerance is tighter than the vendor's, weweigh the balls and use only those that meet our specifi-cation. The vendor is trying to tighten its tolerance andhas asked for assistance in the form of data, Today 60 ballswere received and weighed. The data were plotted on ahistogram. To give the vendor the complete information,a histogram with intervals every 0.02 gram is established.Figure 15.20 does not look much like a bell curve be-cause we have tried to stretch a limited amount of data (60observations) too far. There are 23 active or skipped inter-vals. Our rule of thumb suggests 5 to 7 intervals for less than75 observations. If the same data were plotted into a histo-gram of 6 intervals (excluding the blank), it would look likeFigure 15.21. At least in this version, it looks like a histo-gram. With more data-say, 100 or more observations-onecould narrow the intervals and get more granularity. Don'ttry to stretch data too thin because the conversion to realinformation can become difficult and risky.5 15101.346810-20+20+305% criteria were screenedand sold at a higher price. YouRed Beads in Samplegot what wasHistogram H shows a normal distribution properlycentered between a set of upper and lower control limits.Histograms I and J illustrate what happens when the samenormal curve is allowed to shift left or right, respectively.There will be a significant loss of product as a control limitintersects the curve higher up its slope.Histograms K through P show a normal, centered curvethat went out of control and drifted. Remember that histo-left.FIGURE 15.18 Application of Standard Deviation Calculations to Red Bead Histogram.of histograms. Histogram A represents a normal distri-bution. So does B, except it is shallower. The differencebetween the process characteristics of these two histo-grams is that process A is much tighter, whereas the looserprocess B will have greater variances. Process A is usuallypreferred. Processes C and D are skewed left and right, re-spectively. Although the curves are normal, product will be careful about making judgments. If all the data that pro-lost because the processes are not centered. Process E is bi-modal. This can result from two batches of input material, all the data were combined to make a single histogram, youremarkable achievement. Whatever the situation, with thisstatistical sampling tool properly applied, there is no ques-tion about what can be achieved with any process becauseyou will be able to predict the results.Shapes of Histogramsgrams do not account for time and you must, therefore, beConsider the shape of some histograms and their positionrelative to specification limits. Figure 15.19 is a collectionduced Histograms K through P were averaged, or even ifcould be misled. You would not know that the process wasdrifting. Plotting a series of histograms over time, such as Kthrough P, clearly illustrates any drift right or left, shallowingof the bell, and the like.The number of samples or data points has a bearingon the accuracy of the histogram, just as with other tools.But with the histogram, there is another consideration: Howdoes one determine the proper number of intervals for thechart? (The intervals are, in effect, the data columns of thehistogram.) For example, Figure 15.15 is set up for 11 in-tervals: 0, 1, 2, and so on. The two outside intervals are notused, however, so the histogram plots data in nine intervals.The rule of thumb is as follows:A|H LCLUCLNumber of Observations (M)Number of Intervals (k)<755-775-3006-10>30010-20LCLUCLN108.6 -UCL4-LCL2Weight of Balls in GramsFIGURE 15.19 Histograms of Varying Shapes.FIGURE 15.20 Histogram with Limited Amount of Data StretchedNumber of Samples5.18-

0-bead row, column 4 shows d umn 5 lists the (n – 1). u- 20 = 5.1 square of -5.1, or 26.01. The 1-bead row has d = -4.1. Column 5 lists its square, 16.81. This process is contin- ued through the 10-bead row to complete column 5 of the figure. Column 6 of Figure 15.17c lists the results of the squared deviations (column 5) multiplied by the number of samples at the corresponding deviation value (column 2). For the column 6 entry at the 0-bead row, we multi- ply 0 (from column 2) by 26.01 (from column 5); since, 0X 26.01 = 0, 0 is entered in column 6. For the 1-bead row, we multiply 1 by 16.81; 16.81 is the second entry in column 6. At the 2-bead row, we multiply 3 by 9.61 and enter 28.83 in column 6. This process is repeated through the remaining rows of the figure. Next we add column 6's entries to obtain the sum of the squared deviations, Ed. Ed for our bead process experi- e + 20 5.1 %3D 30 = 5.1 ait from the mean u + 30 = 5.1 mpled of n (100), but we have not Ed. We will perform these rmation in the remaining The values of the deviations Suppose we have a proce curve in Figure 15.18. We hav uct output that require us to = above 6.6. It turns out that the ng µ (5.1) from each of the of column 1. The first entry is determined by subtract- 1, that is, 0 - 5.1 = -5.1 ±lo. We know immediately process output will be rejecte in the list on page 240.) If thi highly probable, we will hav change to a completely diffe variation could be tolerated olumn 4 is the value of col- ment is 221. the specification limits out to pieces flowing out of the pro In a competitive world, this Many companies no longer c lion defective (+30) to be go ber of organizations are seek Six Sigma quality performanc defect rate of 3.4 nonconforn us u, or 1 - 5.1 = -4.1. Now we have all the information we need to calculate the standard deviation (o) for our process. a = VEď(n – 1) a = V221 + 99 V2.23 (2.23 is called the mean squared deviation.) o = 1.49 (1.49 is called the root mean squared deviation.) 4 Deviation squared (Col 4 Sum of deviations squared (Col 2 X Col 5) viation am ol 1- 51 4.1 -3.1 -2.1 -1.1 01 0.9 26.01 1681 9.61 4.41 121 0.01 nities (NPMO) for nonconfo ing, 3.4 NPMO is not very c 6-sigma rate of 0.002 per mil conformance in 500 million. ( 16.81 28.83 39.69 Note: Calculations are to two decimal places. Next calculate the positions of u t lo, 20, and 30. 22.99 a = 1.49 20 0.31 2.99 30 = 4.47 081 17.01 19 3.61 39.71 the Six Sigma section of Chant