SPC Methods

7/30/2019 SPC Methods

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Process control Concepts through statistics

Statistics: help to make a judgement of the properties of large quantities by evaluation of the results obtained from a function of the whole the so called representative random

sample.

Variable: whenever the quality characteristic can be measured and expressed as a quantityof length, weight, dia etc., the quality characteristic is said to be expressed as variable.

Attribute: whenever the quality characteristic can be expressed only as good or bad or conforming or not conforming, then it is said to be expressed as an attribute.

Frequency : is the number or times a particular value of a variable repeats itself in a givenset of data.

EX : 1,2,2,4,4,4,4,5,6,6,7,8,9,10

From the above data we see that 2 is appearing twice hence frequency is 24 is appearing four times hence frequency is 46 is appearing twice hence frequency is 2

Arithmetic mean: is the sum of values in a distribution divided by the number of values.It is one of the common measures of central tendency.

n

x

n

x x x x X

n

i

i

n

==

++++= 1321

EX : 5,6,7,8,8,9,10

AM , 57.77

10988765=

++++++= X

Median : another measure of central tendency. It divides an ordered set into two equalgroups, half of the numbers will have values that are less than the median and half willhave values greater than the median, for e.g. if we have 7 boys with marks in English28,31,27,45,43,52,38 then median of the marks obtained by those boys can be found byranking the marks & taking the 4 th fig. in order of the magnitude i.e., 27,28,31, 38 ,43,45,52median will be 38.

EX: 5 , 6, 7, 8, 8, 9, 10 Median is 8

Mode : the mode of a set of numbers is that value which occurs with the greatestfrequency. More precisely it is the value of the variable at which the concentration of theobservation is the densest. It is possible for the mode to be non-existent in a series of numbers or to have more than one value i.e., it could be unimodel, bimodel or multimodel.

EX : 5,6,7,8,8,9,10 Mode = 8


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Range : the range of a series of numbers is the difference between the largest & smallestvalues of observations.

R=X high X low

Ex : 5,6,7,8,8,9,10 Range = 10 5 = 5

Geometric mean : is the n th root of the product of all values of a series of numbers.

nn x x x xG = 321

n

n x x x xG = 321

Geometric mean is always smaller than the arithmetic mean.

Ex : 5,6,7,8,8,9,10

7 10*9*8*8*7*6*5=G

G = 7.39

Standard deviation : it is a measure of dispersion and is the one most commonly used.It is the square root of the arithmetic mean of the squares of deviations of all observationsin a lot from their arithmetic mean. This is also called lot standard deviation ( ).

( )n

x xn

i

i=

= 1

2

Ex : 5,6,7,8,8,9,10

X = 7.57

( ) ( ) ( ) ( ) ( ) ( ) ( )7

57.71057.7957.7857.7857.7757.7657.752222222 ++++++

=

= 1.59

Sample standard deviation ( S ) : the square root of the quotient obtained by dividing thesum of squares of deviations of the observations from their mean by one less than thenumber of observations in the sample.


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( )1

1

2

=

=

n

x xS

n

ii

Ex : 5,6,7,8,8,9,10 X = 7.57

( ) ( ) ( ) ( ) ( ) ( ) ( )17

57.71057.7957.7857.7857.7757.7657.752222222

++++++=S

S = 1.718

Variance : is the square of a standard deviation.

2

= (1.718)2

= 2.95

Defective item : an item, the quality of which does not meet specified requirements.

Defect : failure to meet the requirement imposed on an item with respect to a singlecharacteristic.

When the number of defects in an item exceeds a specified limit, the item turns out to bedefective.

Frequency distribution : frequency distribution may be defined as a tabulation or tally of the number of times a given quality characteristic measurement occurs within the sampleof product being checked.

a). Line diagram : the simplest method of representing a discrete series graphically is theline or bar diagram.The magnitude of the item is plotted on the horizontal axis and the frequency is plotted onthe vertical axis.

25-

20-

15-

10-

5 -

5 10 15 20 25 30 35

F

R EQUENCY

SIZE OF ITEMS


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b). Histograms : this is another graphical method of portraying frequency distribution of both continuous and discrete types. In this case, the class boundaries are plotted along thehorizontal axis and the frequencies on the vertical axis. The upper ends of the vertical linesare joined.

c. Frequency polygon : if instead of setting up a separate rectangle for each class weconnect the ends of the vertical lines, representing the frequency in each class, set atmidpoints or class intervals, and connect the ends of vertical lines, the curve soconstructed would better indicate the nature of variations in character or size of the

variable under consideration. The graph so constructed is known as frequency polygon.

d) Frequency curves : if instead of joining up the vertical lines by straight lines a freehand curve is drawn to fit into the histogram in a manner that it passes through the tops of rectangular blocks of the histograms excluding approximately as much area of therectangular blocks as the area outside the histograms has been included therein, thediagram so constructed would be known as the frequency curve.

Probability concept:

-A probability generally refers to totality of possible occurrences.- Probability is the relative frequency.

- The probability of an event 'A' is the relative frequency of occurrence of the event 'A'in all possible occurrences of events.

( )cases possibleof noTotal

cases successful of Non s

A P .

.==

Events: there are two types of statistical events.

No.of

Em

p

loyees

Wage Group (Rs.)

200-

150-

100-

50-

10 15 20 25 30 35 40

Frequency Curve

Frequency Polygon


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a. Simple event : this type of event cannot be broken down into further sub events.

b. Compound event : this event can be further broken down into sub events.

E.g., Consider two 6-sided dices- dice A & dice B. In this case, getting say 1, in dice A or B is a simple event, which, getting a combined total, say 6 on the two dices is a singlecompound event.

i) Two events are said to be mutually exclusive if the occurrence of one precludes theoccurrence of the other for e.g. tossing a coin results in 2 mutually exclusive events(heads or tails) getting a head implies not getting a tail.

ii). Two events are said to be independent of each other when occurrence of first does notin any way affect the occurrence of the second for e.g. if a coin is tossed twice insuccession getting a head for the first time does not affect the second throw & its still

possible to get a head or a tail.

LAWS OF PROBABILITY:

a) Addition law: If the events are mutually exclusive that is, only one event can occur ata time, and if the probability of occurrence of one event is P 1 & the second event is P 2then the probability of occurrence P of first or second event is P 1+P 2i.e., P = P 1+P 2

EX 1 : If a bag contains 50 red balls then probability of picking a red ball is

5.0100

501 == P

And probability picking a black ball

5.0100

502 == P

Therefore probability occurrence of first or second events is P = 0.5+0.5 = 1 i.e. 100%

EX 2: The probability of obtaining one defective in a sample of 5 is 0.08 and the probability of obtaining 2 defectives is 0.03

The probability of obtaining one or 2 defectives in a sample of 5 is as follows :

Assume that the probability of obtaining one defective in an event A, i,e P(A) = 0.08

The probability of obtaining 2 defectives in an event B , i,e P(B) = 0.03

The probability of occurrence of either A or B is given by

P(A+B) = P(A) + P(B) = 0.03 + 0.08 = 0.11


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b) Multiplication law: when the events are mutually independent the probability of theevents occurring together or in a particular order is given by the product of individual

probabilities of the events.

EX 1: With 4 dice the probability that all dice will show odd no.In this case in each dice there are 3 odd nos. (1,3,5) & 3 even nos. (2,4,6)

Therefore probability of occurrence of odd no. in each

5.063

== Dice

And probability that all-4 dice will show odd no. is 0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Ex 2 : During 500 hrs of operation of an amplifier the probability of the transistor failingis 5% and that of diode is 9% what is the probability of the failure of both during 500 hrsof operation.

Assume that the probability of the failure of the transistor is an event A i,e P(A) = 0.05

The probability of the failure of the diode in an event B i,e P(B) = 0.09

The probability of failure of both of them is given by

P(A.B) = P(A) * P(B) = 0.05 * 0.09 = 0.0045

Probability distribution function : is a mathematical formula that relates the values of the characteristic with their probability of occurrence in the population.

The collection of these probabilities is called a probability distribution.

Distributions are of 2 typesa). Continuous (for variables, data)

b). Discrete (for attributes data)

i). Continuous data (variable) : a basic characteristic of continuous data is that it isinfinite. When the characteristics being measured can take on any value (subject to thefineness of the measuring process), its probability distribution is called a continuous

probability distribution.

Experience has shown that most continuous characteristics follow one of several common probability distributions i.e.,a. the normal distribution

b. the exponential distributionc. the weibull distribution

ii). Discrete data (attributes) : basic property of discrete data is that it is finite. When thecharacteristics being measured can take on only certain specific values (e.g., integers0,1,2,3, etc.) its probability distribution is called a discrete probability distribution.The common discrete distributions are:


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a. Poissons distribution b. Binomial distribution

NORMAL DISTRIBUTION

The normal distribution curve is called the normal curve and is of the familiar bell shape.Most of the populations observed in real life do approximate the normal population curve& 3 limits have been accepted to contain 99.73% of the population.If x is a normal random variable, then the probability distribution of x is

( )


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The area from - to any variable x can be got by calculating Z & referring to the table.

=

x Z

x = Value of random variable = Mean of the distribution = Std. deviation of the distributionZ = No. of std. deviation from x to the mean of distribution

EX 1: From past experience, a manufacturer concludes that the burn out time of a particular bulb follows a normal distribution. A sample of 50 bulbs has been tested & theaverage life is found to be 60 days with a std. deviation of 20 days. How many bulbs in theentire population of light bulbs can be expected to be still working after 100 days of life?

Solution:

The area beyond 100 days is the probability that the bulb will last more than 100 days.

=

x Z

220

60100=

= Z

From Table - Area under the normal curve for Z=2, probability = 1- 0.9773

Z x-

60100


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Thus, the probability that the bulb will last more than 100 days= 0.0227 or 2.27 %

i.e., 2.27 % of the bulb in the population will still be working after 100 days.

EX 2 : The width of a board lock is normally distributed with mean 10cms and = 0.03cma) What %age of the board locks will have the width exceeding 10.075

b) Below what value of width will 15% of the board locks fall.

a).% age of Specimen falling below 10.075 is

5.203.0

10075.10=

=

=

x Z

From the Table Area under the normal curve for Z = 2.5 the probability is 1- 0.9938

The %age of specimen above 10.075 = 0.0062 * 100 = 0.62 %

b).03.0

10=

=

x x Z

( ) 1003.004.1 = x

x = -0.0312 +10 = 9.96 cm.

Note : In this case since the probability of board locks falling below a certain value isgiven we can find the Z value. For Probability of 0.1515 value of Z is 1.04.

EX 3 : The mean weight of stamped contact reel is 9.07 kgs with a Standard deviation of 0.4 kg.If the distribution is approximately normal. Determine the %age of reels with

a) Less than 8.3 kgs b) Greater than 10 kgs c) between 8 and 10

a) % age of reels less than 8.3 kgs is

92.14.0

07.93.8=

=

=

x Z

= 10 10.075 = 0.03

= 10

= 0.03

15 %

8.3 = 9.07

= 0.40


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From the normal tables the probability isFrom the Table Area under the normal curve for Z= -1.92 ,probability is 0.0274i,e is 2.74 % of the reels will have weight less than 8.3 kgs

b) Greater than 10 Kg

% age of reels falling below 10 kgs is

32.24.0

07.910=

=

=

x Z

From the Table Area under the normal curve for Z = 2.32the probability is 1-0.9898 = 0.0102

The %age of reels above 10 kgs = 0.0102 *100 = 1.02 %

c) Between 8 and 10 kgs

% age of reels falling below 8 kgs is

67.24.0

07.98=

=

=

x Z

From the table Area under the normal curve the probability is 1- 0.0038 = 0.9962 i,e99.62 %

% of reels below 10 kgs is

32.24.0

07.910=

=

=

x Z

From table Area under the normal curve for Z=2.32 the probability is 0.0102 i,e 1.02 %

The %age of reels falling between 8 and 10 kgs is= 99.62 % - 1.02%= 98.60 %

BINOMIAL DISTRIBUTION

Binomial distribution can be used; if following conditions are fulfilled.

= 9.07 10 = 0.40

8 = 9.07 10

= 0.40


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1. There are only 2 possible outcomes for each trial, arbitrarily called success (e.g.,obtaining a defective) and failure, (e.g., not obtaining a defective) without inferringthat a success is necessarily a desirable outcome.

2. The probability of success (p) the same is constant from trial to trial. It will be denoted by the letter p & hence probability of failure (1-p)=q

3. There are n different trials

4. The n trials are independent

The probability of r success in n trials

( )( )

r nr q pr nr

nr y

= !!!

Where p = probability of successq = probability of failurer = no. of success desiredn = no. of trials undertaken

EX 1 : Samples of 20 pieces each are taken at random from very large lots of

components. Determine the probability that such a sample will contain

p =0.1 p=0.3 p=0.5

r


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a) 2 or more defectives b) Exactly 2 defectives c) More than 3 defectives d) Less than

or equal to 3 defectives e) Less than 2 defectives f) Utmost 2 defectives g) Atleast 2

defectives . The lot is 15 % defective

n = 20, p = 0.15, q = 1 0.15 = 0.85

( )( )

r nr q pr nr

nr y

= !!!

a) 2 or more defectives = 1 y(0) - y(1)

( )( )

0200 85.015.0!020!0

!200

= y

( ) 0387.00 = y

( )( )

1201 85.015.0!120!1

!201

= y

( ) 1367.01 = y

2 or more defectives = 1 y(0) - y(1)

=1 y(0) y(1) = 1 0.0387 - 0.1367 = 0.8246

b) Exactly 2 defectives

( )( )

2202 85.015.0!220!2

!202

= y

( ) 229.02 = y

c) More than 3 defectives = 1 y(0) y(1) y(2) y(3)

( )( )

3203 85.015.0!320!3

!203

= y

( ) 2428.03 = y


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More than 3 defectives = 1 y(0) y(1) y(2) y(3)

= 1 0.0387 - 0.1367 0.229 - 0.2428 = 0.3528

d) Less than or equal to 3 defectives = y(0) + y(1) + y(2) + y(3)

= 0.0387 +0.1367 + 0.229 + 0.2428 = 0.6472

e) Less than 2 defectives = y(0) + y(1) = 0.0387 + 0.1367 = 0.1754

f) Utmost 2 defectives = y(0) + y(1) + y(2)

= 0.0387+ 0.1367 + 0.229 = 0.4044

g) Atleast 2 defectives = 1 y(0) y(1) = 1- 0.0387 0.1367 = 0.8246

POISSONS DISTRIBUTION:

Same as binomial but particularly applicable when there are many opportunities for occurrence of an event, but a low probability (less than 0.1) on each trial.

If the probability of occurrence pof an event is constant on each of n independent trialsof the event, the probability of r occurrence in n trials is

( ) ( )( )

!r enp

r ynpr

=

wheren = no. of trials

p = probability occurrencer = no. of occurrence

n is large & p is small for Poisson distribution.

p=0.3

p=0.5

r

p=0.1


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EX 1: The probability that a casting produced by a certain foundry has blow holes is0.002. Find the probability that less than 5 of the next 2000 produced by the foundry hasthe blow holes.

p = 0.002 , n = 2000 , np = 4

The probability of less than 5 blow holes = y(0) + y(1) + y(2) + y(3) + y(4)

( ) ( )( )

!r enpr ynpr

=

( ) ( ) 0184.0!0

40

40

=

=

e y

( ) ( ) 0732.0!1

41

41

==e

y

( ) ( ) 1465.0!2

42

42

=

=

e y

( )( )

1953.0!34

3

43

=

=

e y

( ) ( ) 1953.0!4

44

44

=

=

e y

The probability of less than 5 blow holes = y(0) + y(1) + y(2) + y(3) + y(4)

= 0.018+ 0.0732 + 0.1465 + 0.195 + 0.1953

= 0.6287

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