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Statistical methods

1. Pareto diagrams and dot diagrams:

It is essential to collect data to provide the vital information

necessary to solve engineering problems. Graphicalillustration is an effective tool to describe the information.Pareto diagrams and dot diagrams fall into this category.

Example 1:

A computer-controlled lathe whose performance wasbelow par, workers recorded the following causes andtheir frequencies.

Power fluctuations 6

Controller not stable 22Operator error 13

Worn tool not replaced 2

Other 5

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The data is presented asa special case of barchart called a Paretodiagram shown in Fig.1.This diagram showsParetos empirical lawthat any assortment of

events consists of a fewmajor and many minorelements. Typically, twoor three elements will

account for more thanhalf of the totalfrequency.

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The figure visually emphasizes the importance of reducing

the frequency of controller misbehavior.

As a second step toward improvement of the process, data

were collected on the deviations of cutting speed from the

target value set by the controller. The seven observed

values of (cutting speed)-(target)3 6 -2 4 7 4 3

are plotted as a dot diagram shown in Fig.2. The dot diagram

visually summarizes the information that the lathe is,

generally, running fast.

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It helps to develop efficient experimental designs and

methods for identifying primary causal factors that

contribute to the variability in a response such as cuttingspeed.

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2. Frequency distributions:

Properties of frequency distributions relating to their steps are

best exhibited by means of graphs. Most common form is

the histogram. The histogram of a frequency distribution isconstructed of adjacent rectangles, the height of rectangles

represent external between successive boundaries.

Example 2: A histogram reveals the solution to a grinding

operation problem.

A metallurgical engineer was experiencing trouble with a

grinding operation. The grinding action was produced by

pellets. After some thought he collected the sample of

pellets used for grinding, took them home, spread them outof his kitchen table, and measured their diameters. His

histogram displayed in Fig.3. What does the histogram

reveal?

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Solution:

The histogram exhibitstwo distinct peaks, onefor a group of pelletswhose diameters arecentered near 25 andthe other concerned

near 40.By getting his supplier todo a better sort, so allthe pellets would be

essentially from the firstgroup, the engineercompletely solved hisproblem.

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3. Descriptive Measures:

(a) Mean:

Mean is the average defined by

(1)

To emphasize that it is based on a set of observation, it is

often referred as Sample mean , .

(b) Median:

It is a descriptive measure of the centre, or location of aset of data. It is used to eliminate the effect of extreme

(very large or very small) values. The median of n

observations can be defined loosely as the middle most

value when the data is arranged to size. (i.e.)

1

n

i

i

x

xn

==

x

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if nis even

average of observation if nis odd.

Example 3: Calculation of the sample median with even

sample size

An engineering group receives email requests for

technical information from sales and service persons. The

daily numbers for six days were 11,9,17,19,4, and 15. Find

the mean and the median.

Solution: The mean is

requests

1

2

n +

12 2n n ++

11 9 17 19 4 1512.5

6x

+ + + + += =

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and, ordering the data from the smallest to largest

the median, the mean of the third and fourth largest values,is 13 requests.

(c)Sample Variance ( ): It is the average of the squared

deviations from the mean .

(2)

There are (n-1) independent deviations .

4 9 11 15 17 19

2s

x

( )2

2 1

1

n

i

i

x x

sn

=

=

( )ix x

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Example 4: Calculation of sample variance

The delay times (handling, setting, and positioning the

tools) for cutting 6 parts on an engine lathe are 0.6, 1.2,0.9, 1.0, 0.6, and 0.8 minutes. Calculate .

Solution: First we calculate the mean:

Then we set up the work required to find in

the following table:

2s

0.6 1.2 0.9 1.0 0.6 0.80.85

6x

+ + + + += =

( )2

ix x

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We divide 0.2750 by (6 1) = 5 to obtain

2 20.27500.055 (min )

5

s ute= =

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(d) Standard deviation: In the above example, is inwrong limits (minutes)2. Hence, standard deviation of nobservations is defined as the square root

of their variance.

(3)

Example 5: Calculation of sample standard deviation

With reference to the previous example, calculate s.

Solution: From the previous example, . Take thesquare root and get

minutes

2s

1 2

, ,....,n

x x x

( )2

1

1

n

i

i

x x

sn

=

=

2

0.055s =

0.055 0.23s = =

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The standard deviation and the variance are measures of

absolute variation, that is, they measure the actual

amount of variation in a set of data, and they depend on thescale of measurement. To compare the variation in several

sets of data, it is generally desirable to use a measure of

relative variation, for instance, the coefficient of

variation, which gives the standard deviation as apercentage of the mean.

(e)Coefficient of variation:

(4).100s

Vx

=

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Example 6: The coefficient of variation for comparing

relative preciseness

Measurements made with one micrometer of the diameter ofa ball bearing have a mean of 3.92 mm and a standard

deviation of 0.015 mm, whereas measurements made with

another micrometer of the unstretched length of a spring

have a mean of 1.54 inches and a standard deviation of0.008 inch. Which of these two measuring instruments is

relatively more precise?

Solution: For the first micrometer the coefficient of variationis 0.015

.100 0.38%3.92

V = =

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And for the second micrometer the coefficient of variation is

Thus, the measurements made with the first micrometer arerelatively more precise.

4. Permutations and Combinations:

(a) Permutations: The number permutations of r objectsselected from a set of n distinct objects is

(5)

0.008

.100 0.52%1.54V = =

( 1)( 2).......( 1)( )! !

( )! ( )!n r

n n n n r n r nP

n r n r

+ = =

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Example 7: Evaluating a permutation

In how many different ways can one make a first, second,third, and fourth choice among 12 firms leasingconstruction equipment?

Solution: For n= 12 and r= 4, the first formula yields

Example 8: The number of ways to assemble chips in acontroller

An electronic controlling mechanism requires 5 identicalmemory chips. In how many ways can this mechanism beassembled by placing the 5 chips in the 5 positions withinthe controller?

12 4

12! 12! 12.11.10.9.8!11,880

(12 4)! 8! 8!

P = = = =

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Solution: For n= 5 and r= 5, the first formula yields

And the second formula yields

5 5

5.4.3.2.1 120P = =

( )5 55! 5! 5! 120

5 5 ! 0!P = = = =

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(b)Combinations:

Combination of n objects taken r at a time and denoted by

(6a)

(6b)

n r

nC or

r

( 1)( 2).....( 1)

!

n n n n n r

r r

+=

!

!( )!

n nor

r r n r

=

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Example 9: Evaluating a combination

In how many ways different ways can 3 of 20 laboratory

assistants be chosen to assist with an experiment?

Solution: For n= 20 and r = 3, the first formula for

yields

Example 10: Selection of machines for an experiment

A calibration study needs to be conducted to see if the

readings on 15 test machines are giving similar results. In

how many ways can 3 of the 15 be selected for the initialinvestigation?

Solution:

n

r

20 20.19.181,140

3 3!

= =

15 15.14.13455

3 3.2.1

ways

= =

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Note that selecting which 3 machines to use is the same as

selecting which 12 not to include. That is, according to the

second formula

Example 11: The number of choices of new researchers

In how many different ways can the director of a research

laboratory choose 2 chemists from among 7 applicants and

3 physicists from among 9 applicants?

Solution: The 2 chemists can be chosen in ways

and the 3 physicists can be chosen in ways. Bythe multiplication rule, the whole selection can be made in

21 * 84 = 1,764 ways

15 1515! 15!

12 312!3! 3!12!

= = =

721

2

=

9

843

=

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5. Probability:

If there are nequally likely possibilities, of which one must

occur and sare regarded as favorable, or as a success,then the probability of a success is given by .

Example 12: Well-shuffled cards are equally likely to be

selected.What is the probability of drawing an ace from a well-

shuffled deck of 52 playing cards?

Solution: There are s= 4 aces among the n= 52 cards, so

we get

s

n

4 1

52 13

s

n= =

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Example 13: Random selection results in the equallylikely case

If 3 of 20 tires in storage are defective and 4 of them are

randomly chosen for inspection (that is, each tire has thesame chance of being selected), what is the probability thatonly one of the defective tires will be included?

Solution: There are equally likely ways of choosing4 of 20 tires, so n = 4,845. The number of favorableoutcomes is the number of ways in which one of thedefective tires and three of the nondefective tires can be

selected, or

It follows that the probability is

or approximately 0.42

20

4,8454

=

3 173* 680 2,040.

1 3s

= = =

2, 040 8

4,845 19

s

n= =

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(a)The frequency interpretation of probability

The probability of an event (or outcome) is the proportion

of times the event would occur in a long run of repeatedexperiments.

Example 14: Long-run relative frequency approximation

of probability

If records show that 294 of 300 ceramic insulators testedwere able to withstand a certain thermal shock, what is the

probability that any one untested insulator will be able to

withstand the thermal shock?

Solution: Among the insulators tested, were able to

withstand the thermal shock, and we use this figure as an

estimate of the probability.

294 0.98300

=

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(b) Axioms of Probability:

Axiom 1 for each event A in S. (7a)

Axiom 2 (7b)Axiom 3 If A and Bare mutually exclusive events in S, then

(7c)

The first axioms states that probabilities are real numbers

on the interval from 0 to 1. The second axiom states that

the sample space as a whole is assigned a probability of 1

and this expresses the idea that the probability of a certainevent, an event which must happen, is equal to 1. The

third axiom states that probability functions must be

additive.

0 ( ) 1P A

( ) ( ) ( ).P A B P A P B = +

( ) 1P S =

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(c) General addition rule of probability

Theorem: If A and Bare any events in S, then

(8)

Example 15: Using the general addition rule for

probability

With reference to the lawn-mower-rating example, find theprobability that a lawn mower will be rated easy to operate

or having high average cost of repairs, namely

Solution: Given,

we substitute into the formula of above theorem and get

( ) ( )( ) ( )P A B P A P B P A B = +

( )1 1 .P E C

( ) ( ) ( )1 1 1 10.40, 0.30 0.12P E P C and P E C = = =

( )1 1 0.40 0.30 0.12 0.58P E C = + =

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Example 16: The probability of requiring repair underwarranty.

If the probabilities are 0.87, 0.36, and 0.29 that, while

under warranty, a new car will require repairs on theengine, drive train, or both, what is the probability that acar will require one or the other or both kinds of repairsunder the warranty?

Solution: Substituting these given values into the formula ofabove theorem, we get

(d) Conditional Probability:

If A and B are any events in S and , the

conditional probability of A given Bis

(9)

0.87 0.36 0.29 0.94+ =

( ) 0P B

( )( )

( )

P A BP A B

P B

=

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Example 17: Calculating a conditional probability

If the probability that a communication system will have highfidelity is 0.81 and the probability that it will have highfidelity and high selectivity is 0.18, what is the probabilitythat a system with high fidelity will also have highselectivity?

Solution: If A is the event that a communication system has

high selectivity and B is the event that it has high fidelity,we have

and substitution into the formula yields

( ) ( )0.81 0.18,P B and P A B= =

( )0.18 2

0.81 9P A B = =

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Example 18: The conditional probability that a lawn

mover is easy to operate given that repairs are costly

Referring to the lawn-mower-rating example, for which theprobabilities of the individual outcomes are given in fig 1,

use the results obtained to find

Solution: Since we had

substitution into the formula for a conditional probability

yields

1 1( )P E C

( ) ( )1 1 10.12 0.30,P E C and P C = =

( )1 10.12

0.400.30

P E C = =

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(e) General multiplication rule of probability

Theorem: If A and Bare any events in S, then

(10)

Example 19: Using the general multiplication rule of

probability

The supervisor of a group of 20 construction workers wants

to get the opinion of 2 of them ( to be selected at random)

about certain new safety regulations. If 12 workers favor

the new regulations and the other 8 are against them,

what is the probability that both of the workers chosen by

the supervisor will be against the new safety regulations?

( ) ( ). ( ) ( ) 0( ). ( ) ( ) 0

P A B P A P B A if P A

P B P A B if P B =

=

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Solution: Assuming equal probabilities for each selection

(which is what we mean by the selections being random),

the probability that the first worker selected will be againstthe new safety regulations is , and the probability that

the second worker selected will be against the new safety

regulations given that the first one is against them is

Thus, the desired probability is

(f) Special product rule of probability

Two events A and Bare independent events if and only if

(11)

8

20

719

8 7 14. .20 19 95

=

( ) ( ). ( )P A B P A P B =

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Example 20: The outcome to unrelated parts of an

experiment can be treated as independent

What is the probability of getting two heads in two flips ofa balanced coin?

Solution: Since the probability of heads is for each flip

and the two flips are independent the probability is

Example 21: Independence and selection with and

without replacement

Two cards are drawn at random from an ordinary deck

of 52 playing cards. What is the probability of getting two

aces if

1

2

1 1 1. .2 2 4

=

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(a) the first card is replaced before the second card isdrawn;

(b) The first card is not replaced before the second card is

drawn?Solution:

(a) Since there are four aces among the 52 cards, we get

(b) Since there are only three aces among the 51 cardsthat remain after one ace has been removed from thedeck, we get

Note that so independence is violated when the

sampling is without replacement.

4 4 1. .

52 52 169

=

4 3 1. .

52 51 221=

1 4 4. ,

221 52 52

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Example 22: Checking if two events are independent

under an assigned probability

If are the eventsC and D independent?

Solution: Since and not 0.24,

the two events are not independent.

Example 23: Assigning probability by the special

product rule

Let A be the event that raw material is available when

needed and B be the event that the machining time is less

than 1 hour. If assign

probability to the event

( ) ( ) ( )0.65, 0.40P C P D and P C D= =

( ) ( ) ( )( ). 0.65 0.40 0.26P C P D = =

( ) ( )0.8 0.7,P A and P B= =A B

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Solution: Since the events A and B concern unrelated stepsin the manufacturing process, we invoke independenceand make the assignment

Example 24: The extended special product rule ofprobability

What is the probability of not rolling any 6s in four rolls ofa balanced die?

Solution: The probability is

( ) ( ) ( ) ( ) ( ). 0.8 . 0.7 0.56P A B P A P B = = =

5 5 5 5 6 2. . . .6 6 6 6 1, 2 9

=

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6. Bayes Theorem:

(a)Rule of elimination:

Theorem: If are mutually exclusive eventsof which one must occur, then

(12)

(b)Bayes Theorem: If are mutually exclusive

events of which one must occur, then

(13)

for

1 2, , ...,

nB B B

( )1

( ). ( )n

i i

i

P A P B P A B=

=

1 2, , ..., nB B B

( )

( ) ( )

1

.

( ). ( )

r r

r n

i i

i

P B P A B

P B AP B P A B

=

=

1, 2,......, .r n=

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Example 25: Using Bayes Theorem

Four technicians regularly make repairs when breakdowns

occur on an automated production line. Jagat, who

services 20% of the breakdowns, makes an incomplete

repair 1 time in 20; Tyagarajan, who services 60% of the

breakdowns, makes an incomplete repair 1 time in 10;

Guna, who services 15% of the breakdowns, makes an

incomplete repair 1 time in 10; and Pandyan, who services

5% of the breakdowns, makes an incomplete repair 1 time

in 20. For the next problem with the production line

diagnosed as being due to an initial repair that wasincomplete, what is the probability that this initial repair

was made by Jagat?

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Solution: Substituting the various probabilities into theformula of Bayes Theorem, we get

and it is of interest to note that although Jagat makes anincomplete repair only 1 out of 20 times, namely, 5% of thebreakdowns, more than 11% of the incomplete repairs arehis responsibility.

7. Mathematical expectation of Decision making.

If the probabilities of obtaining the amounts

are , then themathematical expectation is

(14)

( )1 (0.20)(0.05) 0.114(0.20)(0.05) (0.60)(0.10) (0.15)(0.10) (0.05)(0.05)

P B A = =+ + +

1 2, ,......., ka a or a 1 2, ,......., kp p and p

1 1 2 2 ..... k kE a p a p a p= + + +

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Example 26: Mathematical expectation of net profit

A distributor makes a profit of Rs 200 on an item if it is

shipped from the factory in perfect condition and arriveson time, but it is reduced by Rs 20 if it does not arrive on

time, and by Rs 120 regardless of whether it arrives on

time if it is not shipped from the factory in perfect

condition. If 70% of such items are shipped in perfect

condition and arrive on time, 10% are shipped in perfect

condition but do not arrive on time and 20 % are not

shipped in perfect condition, what is the distributors

expected profit per item?

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Solution: We can argue that the distributor will make the Rs

200 profit 70 % of the time, the Rs 180 profit 10 % of the

time, and the Rs 80 profit 20 % of the time, so that theaverage profit per item ( the expected profit) is

The result is the sum of the products obtained by

multiplying each amount by the corresponding proportion

or probability.

200(0.70) 180(0.10) 80(0.20) 174Rs+ + =

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8.Probability Distributions:

Random variables: They are functions defined over the

elements of a sample space. They are denoted with

capital letters X, Y and so on to distinguish them from their

possible values given in lower case.

The function f(x) = P(X=x) which assigns probability to

each possible outcome x that is called the probabilitydistribution.

Random variables are usually classified according to the

number of values which they can assume. Discrete

random variables can take only a finite number, or acountable infinity of values.

for1( )6

f x = 1,2,3,4,5,6x =

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Gives the probability distribution for the number of pointswe roll with a balanced die.

Since the values of probability distributions are

probabilities and one value of a random variable mustalways occur (i.e)

for all x

and

Example 27: Checking for nonnegativity and totalprobability equals one

Check whether the following can serve as probabilitydistributions:

(a) for x= 1,2,3,4

(b) for x= 0,1,2,3,4

( ) 0f x ( ) 1

all x

f x =

2( )2

xf x =

2

( )

25

xh x =

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Solution: (a) This function cannot serve as a probabilitydistribution because f(1)is negative

(b) The function cannot serve as a probability distribution

because the sum of the five probabilities is andnot 1.

9.Binomial Distribution:

Many statistical problems deal with the situationsreferred to as repeated trials. For example we may wantto know

(i) The probability that one of five rivets will rupture in a

tensile test.

(ii) The probability that 9 of 10 VCRs will run at least1000hours

65

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In determining the probability the following assumptionsare made

1. There are only two possible outcomes for each trial.

2. The probability of a success is the same for each trial.

3. There are ntrials, where nis a constant.

4. The ntrials are independent.

Trials satisfying these assumptions are referred to asBernoulli trials.

Let X: be the random variable that equals thenumber of successes in n trials.

p: Probability of success on any one trial.

(1-p): Probability of failure on any one trial.

The probability of getting x success and (n x) failures in

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The probability of getting x success and (n-x) failures in

some specific order is The number of ways in

which x trials can be selected is , the number of

combinations of xobjects selected from a set of nobjects,and we arrive at

for x= 0,1,2,,n

This probability distributions is called the binomial

distributionbecause for x = 0,1,2,.., and n, the values of

the probabilities are the successive terms of binomial

expansion of ; for the same reason, the

combinatorial quantities are referred to as binomialcoefficients. Actually, the preceding equation defines a

family of probability distributions with each member

characterized by a given value of theparameterp and the

number of trials n.

(1 )x n x

p p

( )nx

( )( ; , ) (1 )x n xnb x n p p px =

[ ](1 )n

p p+

( )nx

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Example 28: Evaluating binomial probabilities

It has been claimed that in 60% of all solar heat installations

the utility bill is reduced by at least one-third.Accordingly, what are the probabilities that the utility bill

will be reduced by at least one-third in

(a) four of five installations;

(b) at least four of five installations?

Solution: (a) Substituting x=4, n=5, and p=0.60 into the

formula for the binomial distribution, we get

( )( )4 5 45(4;5, 0.60) 0.60 (1 0.60) 0.2594b = =

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(b) Substituting x=5, n=5, and p=0.60into the formula for thebinomial distribution, we get

and the answer is

( )( )5 5 55(5;5, 0.60) 0.60 (1 0.60) 0.078

5b = =

(4;5, 0.60) (4;5, 0.60) 0.259 0.078 0.337b b+ = + =