manufacturing technology (me461) lecture23

8/12/2019 Manufacturing Technology (ME461) Lecture23

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Manufacturing Technology

(ME461)

Instructor: Shantanu Bhattacharya


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Review of previous lecture

Type of control charts (variable control chart

and attribute control chart).

Benefits of control charts. Characteristics of data.

Alternate methods for finding the mean in

normal dataset as well as a frequency table.

Plotting of control charts.


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Checklist necessary for X and R chartsIt is helpful to visualize the decisions and calculations that must be made and the

actions that need to be taken for plotting a X and a R chart. They are the following:

1. Decisions preparatory to the control charts Some possible objectives of the charts.

Choice of variables.

Decisions on the basis of subgroups.

Decisions on size and frequency of subgroups.

Setting up the forms for recording the data.

Determining the method of measurement.2. Starting the Control Charts

Making and recording measurements and recording other relevant data.

Calculating the average X and range R of each subgroup.

Plotting the X and R charts.

3. Determining the trial control limits

Decision on required number of subgroups before control limits are calculated. Calculation of trial control limits.

Plotting the central lines and limits on the charts.

4. Drawing preliminary conclusions from the charts

Indication of control or lack of control.

Interpretation of processes in control.

Relationships between processes out of control and specification limits.


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Setting up the forms for recording the data

Layout of data should be as per convenience of calculation and analysis.

The forms should have a recording space for item of measurement, unit of measurement,

and operator remarks about tool change, operator change, machine change etc.


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Setting up the forms for recording the data

Layout of data should be as per convenience of calculation and analysis.

The forms should have a recording space for item of measurement, unit of measurement,

and operator remarks about tool change, operator change, machine change etc.


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Starting the Control charts

Making and recording the measurements and recording other relevant data:

The actual work of the control charts start with the first measurements.

Any method of measurement will have its own inherent variability; it is important that this isnot increased by mistakes in reading measurement instruments or errors in recording data.

Calculating the average and range for each subgroup:


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Plotting both the charts:Determining the trial control limits:

1. Decision is needed on the

required number of subgroups

before control limits are

calculated.

2. The fewer the subgroups used

the sooner the information thus

obtained will provide a basis foraction but the less the assurance

that the action will be sound.

Calculation of trial control limits:


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Drawing Preliminary Conclusions from the chart

Indication of control or lack of control:

Lack of control is indicated by points falling outside the control limits on either of the

charts.

When the points fall outside the control limits, we say that a process is out of control,this is equivalent to saying that assignable causes of variation are present and this is not a

constant cause system.

In contrast if none of the points fall outside the control limits then No assignable causes

are present.

Interpretation of processes in control:With the evidence from the control chart that a process is in control, we are in a

position to judge what is necessary to permit the manufacture of product that meets the

specifications for the quality characteristic charted. The control chart data gives us

estimate of :

1. The centering of the process.

2. The dispersion of the process.

Actions based on the relationship between the specifications and the centering and

dispersion of a controlled process depend somewhat on whether there are two

specification limits, a maximum or upper limit or a minimum or lower limit. This would

again depend heavily on the parameter that we choose to plot.


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Possible relationships of a process in control to upper

and lower specification limits

Case I: The spread of the process 6is

appreciably less than the difference

between the specification limits.

Case II: The spread is approximately

equal to the difference between the

specification limits.

Case III: The spread of the process is

appreciably greater than the difference

between the specification limits.


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Case I: When the process 6is appreciably less than difference between specification

limits

Frequency curves A,B,C, D and E indicate various

positions in which the process can be centered.

With any of the position A, B and C practically all

the products manufactured will meet thespecifications as long as the process stays in control.

In general when conditions A, B and C come it

represents the ideal manufacturing situation. When

the control chart shows that one of this control

chart exists, many different possible actions may be

considered depending on the relative economy.

For example it may be considered economically advisable to permit X to go out of control if it does not go

too far, i.e., the distributions may be allowed to move between positions B and C. This may avoid the cost

of frequent machine setup and of delays due to hunting of assignable causes of variation that will not be

responsible for unsatisfactory product.Or where acceptance has been based on 100% inspection, it may be economical to substitute acceptance

based on control charts.

Or where there is an economic advantage to be gained by tightening the specification limits, such action

may be considered.

With the process in the position D some points will fall above the upper specification limit. Similarly with

the process in position E some products will fall below the lower limit. In either case the obvious action is

to bring the centering of the process towards that of A.

C II Wh h 6 i l h diff b


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Case II: When the process 6is equal to the difference between

specification limitsIn this situation only the process

exactly centered between the

specification limits, as in position

A, will practically produce

everything conforming to the

specifications.

If the distribution shifts away

this exact centering as in B or C, it

is apparent that some of the

products will fall outside the

specification limits.

Here the obvious action is to take all steps possible to maintain the centering of

the process. This usually calls for continuous use of the control charts for X and R

with subgroups at frequent intervals and immediate attention to points out ofcontrol.

If fundamental changes can be made that reduce dispersion that eases the

pressure.

Consideration should also be given to changing of the tolerances.


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Case III: When the process 6is less than the difference between

specification limits

The third type of situation

arises when thespecification limits are so

tight that even with the

process in control and

perfectly centered some

non conforming parts still

get produced as in position

A.

This primarily calls for a review of tolerances.

It also calls for a fundamental change in the process that will reducethe process dispersion.

It is still very important to maintain the centering of the process; the

curves in position B and C show how a shift in process average will

increase the non conformity.


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Possible relationships of a process in control to a single specification

I: Low value of distribution ( ) isappreciably above LSL.

II: Low value of distribution is at LSL.

III: Low value of the distribution is

appreciably below LSL.


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Possible relationships of a process in control to a single

specification

The first situation is one in which there is a margin of safety.

The second is a one in which the specification is just barely met as long as the process

stays in control.

The third is one in which some non conforming products are produced unless there is a

fundamental change in the process spread or increase in process average.

All three distributions A,B and C have the same lower value. However, distribution B with a

greater dispersion must have a greater process average than A for the low points to be at

the same level. Similarly distribution C should have a greater process average than B. It is

evident that the greater the distribution the higher the average must be for the entire

distribution to fall above the lower specification limit.

The relationship between average and cost is vital in many instances. For example in the

filling of containers a reduction in dispersion may reduce cost by reducing the average

overfill.

On the other hand, the less the dispersion, the more important it is that the processaverage does not go out of control. This is illustrated by comparing distributions A and C in

Case III. In Case III both process averages have shifted an equal amount below their

position. However, the proportion of bad product, as indicated by the area of the

distribution below the specification limit, is much greater in A than in C.


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Use on control charts by a purchaser to help suppliers improve

their processes.

Facts of the case:

A manufacturer of electronic devices had trouble with the cracking of a certain small cross shaped

ceramic insulator used in the device.

The cracking generally took place after the manufacturing operations were nearly completed and did

so in a way that made it impossible to salvage the unit.

Hence the costs resulting from each cracked insulators during manufacturing operations suggested

that others might be likely to crack under service conditions.

In an effort to improve the situation, all incoming insulators of this type were given 100% inspection.This 100% inspection failed to decrease the percentage defective units.

A simple testing was then constructed to measure the actual strength in flexure by testing insulators

to destruction. From each incoming lot of insulators 25 were tested. As the insulators came from 2

suppliers, control charts (X and R) from both suppliers were separately maintained. The tests showed

that both suppliers had approximately the same % defectives but the explanations for defectives were

different.

Supplier A had higher average strength but complete lack of anything resembling statistical control.

Supplier B on the other hand had excellent statistical control but at a level such that an appreciable

part of frequency distribution was below the required minimum strength.

This diagnosis of the situation was brought to the attention of both suppliers and they were

encouraged to exchange information about production methods. Finally product control could be

achieved.


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Milling a slot in an aircraft terminal

block.Decisions preparatory to the control chart:

High percentages of rejections for many of the parts made in the machine shop of an

aircraft company indicated the need for examination of the reasons for trouble.

As most of the rejections were for failure to meet dimensional tolerances, it was decided

to try to find the causes of trouble by the use of X and R charts.

These charts which off-course required the measurements of dimensions, were to beused only for those dimensions that were causing numerous rejections.

Among the many dimensions, the ones selected for control charts were those having

high costs of spoilage and rework and those on which rejections were responsible for

delays in assembly operations.

This example deals with one such critical dimension, the width of a slot on the duralumin

forging used as a terminal block at the end of the rear tail or an airplane.The final matching of the slot width was a milling operation. The width of the slot was

specified as 0.8750 .

The design engineers had specified this dimension with an unilateral tolerance because

of the fit requirements of the terminal block; it was essential that the slot width be at

least .8750 in and desirable that it is very close to this value.

Milli l i d


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Milling example continuedDue to the small no. of available

inspection personnel it was decided

that for each chart the sample would

be inspected would be approximately5% of the total production of the parts

in question.

It was decided that all measurements

would be made at the place of

production and it was decided that a

part be measured every 20 partsproduced. The subgroup size was thus

fixed as 5. The figure on left shows all

observations.

The method to inspection to secure

data for each of the charts was to

measure two portions of the slot widthusing a micrometer screw gauge and

average these measurements.


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Milling example continuedStarting the control charts:

The actual measurements of the first 16 subgroups are shown in the table. These numbers of subgroups

correspond to a production order for 1600 of these terminal blocks.At the time of the 12thsubgroup, before the completion of this production order and before the calculation of the

central line or control limits, the quality control inspector noticed that the machine operator was occasionally

checking performance on a terminal block that had just come off the machine and was still hot.

After the 12thsubgroup the operator was asked to do measurements after some time of cooling is allowed.

Determining the trial control limits:

Calculations of the trial control limits was

made after the first 16 subgroups which

completed the production order. As shown in

the figure on the right these were done by

finding out the A2and D4factors for a

subgroup size 5.

Drawing preliminary conclusions from the

graphs:

Subgroup 1 is above the upper control limit

on the R chart.

Subgroup 10 is below the lower control limit

on the X chart. Moreover the last 10- of the 6points fall below the central line.


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Milling example continuedStarting the control charts:

The actual measurements of the first 16 subgroups are shown in the table. These numbers of subgroups

correspond to a production order for 1600 of these terminal blocks.At the time of the 12thsubgroup, before the completion of this production order and before the calculation of the

central line or control limits, the quality control inspector noticed that the machine operator was occasionally

checking performance on a terminal block that had just come off the machine and was still hot.

After the 12thsubgroup the operator was asked to do measurements after some time of cooling is allowed.

Determining the trial control limits:

Calculations of the trial control limits was

made after the first 16 subgroups which

completed the production order. As shown in

the figure on the right these were done by

finding out the A2and D4factors for a

subgroup size 5.

Drawing preliminary conclusions from the

graphs:

Subgroup 1 is above the upper control limit

on the R chart.

Subgroup 10 is below the lower control limit

on the X chart. Moreover the last 10- of the 6points fall below the central line.


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Milling Example ContinuedIt is obvious that the measurements made

are not the result of a constant system of

chance causes.

If subgroup 1 is eliminated fromconsideration, R for the remaining 15

subgroups is 536/ 15 = 36.

This gives us a revised upper control limit

D4 (R) = 2.11 (36) = 76.

Subgroup 5 falls exactly on the upper

control limit in the X chart. Hence a secondrevision of R with subgroup 5 eliminated

seems reasonable. Therefore R = 460/14 =

33. .

An estimate may be made of from the R

values. With a d2value of 2.362

corresponding to the subgroup size 5 we get

a = R/ d2 = 0.0014 in.

The value of the natural tolerance or spread

of the process is therefore 6= 6 (.0014) =

.0084 in.

This spread may be compared with

tolerance spread: U-L = .0050in

It is evident from the process that the natural tolerance of the

spread is substantially greater than the specified tolerance.

Therefore, unless the process dispersion is reduced the process

will keep producing non conforming components.


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Milling Example ContinuedIt is obvious that the measurements made

are not the result of a constant system of

chance causes.

If subgroup 1 is eliminated fromconsideration, R for the remaining 15

subgroups is 536/ 15 = 36.

This gives us a revised upper control limit

D4 (R) = 2.11 (36) = 76.

Subgroup 5 falls exactly on the upper

control limit in the X chart. Hence a secondrevision of R with subgroup 5 eliminated

seems reasonable. Therefore R = 460/14 =

33. .

An estimate may be made of from the R

values. With a d2value of 2.362

corresponding to the subgroup size 5 we get

a = R/ d2 = 0.0014 in.

The value of the natural tolerance or spread

of the process is therefore 6= 6 (.0014) =

.0084 in.

This spread may be compared with

tolerance spread: U-L = .0050in

It is evident from the process that the natural tolerance of the

spread is substantially greater than the specified tolerance.

Therefore, unless the process dispersion is reduced the process

will keep producing non conforming components.


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Revision of theory of ProbabilityDefinition:

Probability is concerned with the likelihood of an event occurring. The scale of

probability of an event varies from 0 to1.

If an event cannot occur on a trial, then the probability of its occurrence in 0.

If another event is certain to occur then its probability of occurrence is 1.

As an example suppose that a trial is drawing a piece at random from some production

line, and that the event in question is that the piece drawn is a defective or non

conforming one.Let us suppose that the probability of a defective is 0.05. This means that 5% of the

time when we draw a random piece from the line, it is defective.

The complementary event is that the piece drawn is a good one. Its probability is .95.

P(good or defective) = 1

Occurrence ratio

Suppose that we have a production process for which the probability of a piececontaining at least one minor defect is constantly 0.08.

We then say that the probability is 0.08 for a minor defective.

Now what happens to the observed proportion of minor defectives as we continue to

sample? This observed proportion of minor defectives is what we know as the

occurrence ratio.


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Samp e Tota s Occurrencei The proportion defective only tends to


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ratio

n D n d P = d/ n

10 0 10 0 .0000

10 1 20 1 .0050

10 1 30 2 .0667

10 1 40 3 .0750

10 0 50 3 .0600

50 5 100 8 .0800

50 4 150 12 .0800

50 6 200 18 .0900

50 4 250 22 .0880

50 8 300 30 .1000

50 1 350 31 .0886

50 3 400 34 .0850

50 5 450 39 .0867

50 3 500 42 .0840

50 5 550 47 .0855

50 5 600 52 .0867

50 5 650 57 .0877

50 4 700 61 .0871

50 3 750 64 .0853

50 6 800 70 .0875

50 4 850 74 .0871

50 7 900 81 .0900

50 4 950 85 .0895

50 4 1000 89 .0890

The proportion defective only tends to

approach p =0.08.

Sometimes it gets closer, sometimes it

backs away from p.

Before the total sample size was 100,

the occurrence ratio was below .08.

Between 100 and 150 it was 0.08 and

thereafter above.

Principle:

We can say that the p= d/n is an

estimate of the constant probability of

population p. How close the estimate

will be depends on sample size n, thevalue of p and also upon chance.


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Probability laws

Consider again the production line producing pieces with a constant probability .08 of

the piece being a minor defective, and such that each piece is independent of the other

produced.

This means that the probability of the next piece being a defective is 0.08 and the piece

being good is .92 irrespective of the preceding piece.

Now let us suppose that we draw a sample of two pieces and inspect them.

The outcomes are that the sample may contain 0, 1 or 2 defectives. Let us find the

probabilities of this outcome or events.

For the probability of the samples containing no defectives, we must have good pieces

on both draws.


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Probability Laws

P b bilit L


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Probability LawsIf 2 events A and B might occur on a trial or experiment, but the occurrence of either

one prevents the occurrence of the other, then events A and B are called mutually

exclusive.

For two such events P(A) + P (B) = P( A or B, mutually exclusive events)

We have seen this in the earlier example in the 1 good case P (1 Good) = .0736 +0.0736

If one of the two events A and A is certain to occur on a trial, but both cannot

simultaneously occur, then A and A are called complementary events.

For any such pair of events

We have seen this in the example of a single draw where p was given as 0.08.

L f P b bilit


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Laws of ProbabilityTwo events A and B are independent if the occurrence or non occurrence of A does not

affect the probability of B occuring.

Whenever a process produces defectives independently, or at random, so that theprobability of a defective on the next piece does not depend upon what the preceding

pieces were like, then we have the case of independent events. Such a process is said to

be in control, that is stable, even though some non conformity is produced.

Not all processes do behave in this manner.

For example: We consider the production of 3000 piston ring castings. The sample of 100

contain 25 defectives whereas the remaining 2900 only 4. This was because the

defectives occur in bunches from a certain defect producing condition.

Particularly in this case it was found that the castings were made from stacks of molds,

and if the iron is not hot enough when poured into a stack, many castings may be

defective. Under such conditions, whether a piece is good or defective does have an

influence on the probability of the next one being defective.

If two events A and B are independent, then we have :

P(Both A and B occurring) = P(A). P(B)

Example of Dependence and Equal


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Example of Dependence and Equal

LikelihoodAs a second example of probability, let us consider drawing without replacement from a lot

of N=6 speedometers, of which 1 is defective

Let N = no. of pieces in one lot and

D= no. of defective pieces in a lot

Now consider the very simple case in which we just draw a random sample of 1 from a lot

of 6. Random means that each of the six meters is equally likely to be chosen for thesample. Probability of each is 1/6.

There are only two kind of meters good and defective with P(good) = 5/6 and

P(defective) = 1/6.

Now next consider drawing a sample of n= 2 from the lot having N =6, of which D=1 isdefective. This may contain no. of defectives either d=0 or 1.

This is a case of two consecutive drawings which are not independent. Take first the case of

the sample yielding no defectives, that is, two good meters. We need

P(2 good) = P(good, good) = P(good on first draw). P(good on 2 nddraw given good on 1st

draw) = 5/6. 4/5 = 2/3

Counting samples (Permutations and


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combinations)In combinations we consider for example we have n objects which we can distinguish

between.Now how many distinct samples, each of one, can we draw from a lot of N=10?

Obviously the answer is 10. So, we call this a combination of N objects taken 1 at a time,

or in symbols :

C(N,1) = N.

Next consider samples of two, from say four good pieces (g1, g2, g3 and g4).

Then the number of distinct unordered samples may be found from the number of

distinct ordered samples. For example: for ordered samples

g1g2, g2g1, g1g3, g3g1, g1g4, g4g1, g2g3, g3g2, g2g4, g4g2, g3g4, g4g3

The no. of unordered samples are only half as much , that is, six, because, for example,

the one unordered pair g2g4 corresponds to two ordered pairs g2g4 and g4g2.

Now let us consider lots of 10 distinct pieces. The number of possible ordered samples,

each of two is 10.9, because there are 10 choices for the first piece and having made a

choice their remain 9 choices for the second piece. So we have 90 ordered samples and

exactly of this unordered. (45)

C ti l (P t ti d


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combinations)Now let us go to a sample of 3 from a lot of 10. The no. of distinct ordered samples is

10.9.8; 10 choices for the first, 9 choices for the second and 8 choices for the third. But

now six of this ordered samples correspond to just one unordered sample. For example:

g1g4g6, g1g6g4, g4g1g6, g4g6g1, g6g1g4, g6g4g1 all correspond to g1g4g6 unordered

sample.

Hence the number of distinct or unordered samples or combinations is 10.9.8/6 = 120

Ordered samples are also called permutations and they are calculated by the general

formulae

P(n,r) = n! / (n-r)! In the earlier case P(10,3) = 10!/ 7! = 10.9.8 = 720

We call the number of distinct unordered samples a combination and is given by thegenera formulae C(n,r) = n!/ r! (n-r)! In earlier case this would be = 10!/ 3!. 7! = 10.9.8/

3.2.1 = 120.

C t d d t D f t d d f ti


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Counted data: Defects and defectivesInspecting or testing n pieces , we may search for defects or non conformances in the

n pieces and record the total number of such defects. This is measuring quality by a

count of defects.

In inspecting or testing n pieces, we may consider whether each of n pieces does

contain any defects. Each piece having one or more defects is called a defective. This

measure is known as the count of defectives.

We shall consistently use the following symbols here and in later discussions:

n = number of units in a sample.

d = number of defective units in the sample of n units.

p = d/n= sample of fraction defective = proportion of defective units in the sample.

q= 1-p = sample fraction good

d= np = no. of defective units in the sample.

Binomial distribution for defectives:

Suppose we had a process with population fraction defective of p = .10. For a sample size

of 4 there could be

Either 0, 1,2, 3 or 4 defectives can exist.

Binomial distribution for defectives


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Binomial distribution for defectivesFirst find the probability of drawing a sample with all pieces good.

So for P(d=0) = P(4 good) = [P(good)]4 =(0.9)4 = .6561

Thus about 2/3 of the time, we draw a sample of n=4 pieces from the process, all four will

be good ones.

Now next we seek P (3 good, 1 defective) .

P (3 good, 1 defective) = P(g,g,g,d) + P(g,g,d,g) + P (g,d,g,g) + P (d,g,g,g) = (.9)(.9)(.9)(.1) + (.9)(.9)(.1)(.9) + (.9) (.1)(.9)(.9) + (.1) (.9)(.9)(.9) = 4(.9)^3 (.1) = .2916.

Next consider samples with d=2; they have two defectives and two good ones. How many

distinct orders are there for such samples? Six: ggdd. gdgd, gddg, dgdg, ddgg, dggd, the

probability for each one of these sample outcomes is (0.9)2(0.1)2.

P(d=2) = 6 (0.9)2

(0.1)2

= .0486

Similarly for d=3 there are only 4 orders of sampling results

P(d=3) = 4 (0.9)2(0.1)2 =.0036

Finally for d =4, all four must be defective

P(d=4) = .0001

Binomial distrib tion for Defecti es


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Binomial distribution for DefectivesThe sum of all the 4 outcomes are 1.

We can also represent the various coefficients viz, 4,6,4 of the products of the powers of .9

and .1 as

C(4,1) =4, C(4,2) = 6, C(4,3)= 4 respectively.

This reasoning enables us to write all the four probabilities in 1 formulae

P(d) = C(4,d) (0.9)4-d(0.1)d

This is also a representative of the dthterm of a Binomial distribution of n =4 and the p =0.9

and q= 0.1.

In general

P(d) = C(n,d) (p)n-d

(q)d

d P(d) P= d/n

0 .6561 .00

1 .2916 .25

2 .0486 .50

3 .0036 .75

4 .0001 1.00

Total 1 0000

manufacturing technology (me461) lecture23

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