DETERMINATIONOFARITHMETICMEAN,MEDIAN, AND STANDARD DEVIATION

The standard viation is a measure of the dispersion, or scatter, of the values of the random variable; it is de-approximately equal to the amount by which a given value of the variable m

Project time, days

FIGURE 19 Workforce requirements based on project schedule.

Statistics, Probability, and Their Applications

Basic Statistics

If the value assumed by a variable on a given occasion cannot be predicted because it is

influenced by chance, the variable is known as a random, or stochastic, variable The number of times the variable assumes a given value is called the frequency of that value.

A number that may be considered representative of all values of the variable is called anaverage There are several types of averages, such as arithmetic mean, geometric mean,

harmonic mean, median, mode, etc Each is used for a specific purpose The standard viation is a measure of the dispersion, or scatter, of the values of the random variable; it is

de-approximately equal to the amount by which a given value of the variable may be

expect-ed to differ from the arithmetic mean, in either direction The variance is the square of thestandard deviation Where no mention of frequency appears, it is understood that all fre-quencies are 1

Notational System

Here X= random variable; X1 = ith value assumed by X\ft - frequency of X 1 ; n = sum of frequencies; X = arithmeticmean of values of X\ Xmed = median of these values; d t = devi- ation of Xb from X = Xi -X\ A = an assumed arithmetic mean; d A>i = deviation of X 1 from A=X 1 -A', s = standard deviation; s 2 = variance In the following material, the subscript /

will be omitted

DETERMINATIONOFARITHMETICMEAN,

MEDIAN, AND STANDARD DEVIATION

Column 2 of Table 22 presents the number of units of a commodity that were sold

month-ly by a firm for 7 consecutive months Find the arithmetic mean, median, and standard viation of the number of units sold monthly

TABLE 22

(2)(1) Number of (3) (4) (5) (6)

Calculation Procedure:

1 Compute the arithmetic mean

Let X= number of units_sold monthly Find the sum of the values of X9 which is 301 Then setX = QX)In, or X = 301/7 = 43.

2 Find the median

Consider that all values of X are arranged in ascending order of magnitude If n is odd, the value that occupies the central position in this array is called the median If n is even, the

median is taken as the arithmetic mean of the two values that occupy the central positions

In either case, the total frequency of values below the median equals the total frequency

of values above the median The median is useful as an average because the arithmeticmean can be strongly influenced by an extreme value at one end of the array and therebyoffer a misleading view of the data

In the present instance, the array is 32, 37, 41, 44, 47, 49, 51 The fourth value in the

array is 44; then Xmed = 44.

3 Compute the standard deviation

Compute the deviations of the X values from X, and record the results in column 3 of

Table 22 The sum of the deviations must be O Now square the deviations, and record theresults in column 4 Find the sum of the squared deviations, which is 278 Set the variance

s 2 = &d 2 )/n = 278/7 Then set the standard deviation s = V278/7 = 6.30.

4 Compute the arithmetic mean by using an assumed

arithmetic mean

SQtA = 40 Compute the deviations of the Jf values from A 9 recordjthe results in column 5

of Table 22, and find the sum of the deviations, which is 21 Set X = A + (2dA )/n, or X =

Related Calculations: Note that the equation applied in step 5 does not contain

the true mean X This equation serves to emphasize that the standard deviation is purely a

measure of dispersion and thus is independent of the arithmetic mean For example, if all

values of Xincrease by a constant /z, then X increases by h, but s remains constant Where

X has a nonintegral value, the use of an assumed arithmetic mean A of integral value can result in a faster and more accurate calculation of s.

DETERMINATION OFARITHMETIC

MEAN AND STANDARD DEVIATION

OF GROUPED DATA

In testing a new industrial process, a firm assigned a standard operation to 24 employees

in its factory and recorded the time required by each employee to complete the operation.The results are presented in columns 1 and 2 of Table 23 Find the arithmetic mean andstandard deviation of the time of completion

Calculation Procedure:

1 Record the class midpoints

Where the number of values assumed by a variable is very large, a comprehensive listing

of these values becomes too cumbersome Therefore, the data are presented by grouping

the values in classes and showing the frequency of each class The range of values of a

given class is its class interval, and the end values of the interval are the class limits Thedifference between the upper and lower limits is the class width, or class size Thus, inTable 23, all classes have a width of 4 min The arithmetic mean of the class limits is themidpoint, or mark In analyzing grouped data, all values that fall within a given class arereplaced with the class midpont The midpoints are recorded in column 3 of Table 23, and

they are denoted by X.

2 Compute the arithmetic mean

Set ^ = (S/9/W, or X = (3 x 22 + 9 x 26 + 7 x 30 + 5 x 34)724 = 680/24 = 28.33 min.

3 Compute the standard deviation

Set s 2 J=GfJ 2 Vn, ors 2 = [3(-6.33)2 + 9(-2.33)2 + 7(1.67)2 + 5(5.67)2]/24 = 14.5556 Then

j = Vl4.5556 = 3.82min

4 Compute the arithmetic mean by the coding method

This method simplifies the analysis of grouped data where all classes are of uniformwidth, as in the present case Arbitrarily selecting the third class, assign the integer O to

TABLE 23

(1) (2)

Time of Number of (3) (4)

completion, min employees Midpoint Code

(class interval) (frequency/) X c

this class, and then assign integers to the remaining classes in consecutive and ascending

order, as shown in column 4 of Table 23 These integers are the class codes Let c = class code, w = class width, and A = midpoint of class having the code O Compute X/c, or 2/c

= 3(-2) + 9(-l) + 7(0) + 5(1) - -10 Now set X = A + w@fc)/n> or X = 30 + 4(-10)/24 =28.33 min

5 Compute the standard deviation by the coding method

Using the codes previously assigned, compute 2/c2, or 2/c2 = 3(-2)2 + 9(-l)2 + 7(O)2 +5(1)2 = 26 Now set s2 = w*{(2fi?yn - [&fc)/n] 2 } Then s 2 = 16[26/24 - (-10/24)2] =

14.5556, and s = V14.5556 = 3.82 min.

Permutations and Combinations

An arrangement of objects or individuals in which the order or rank is significant is called

a permutation A grouping of objects or individuals in which the order or rank is not nificant, or in which it is predetermined, is called a combination Assume that n objects are available and that r of these objects are selected to form a permutation or combination.

sig-If interest centers on only the identity of the r objects selected, a combination is formed; if interest centers on both the identity and the order or rank of the r objects, a permutation is formed In the following material, the r objects all differ from one another.

Where necessary, the number of permutations or combinations that can be formed iscomputed by applying the following law, known as the multiplication law: If one task can

be performed in ml different ways and another task can be performed in m2 different

ways, the set of tasks can be performed in W1W2 different ways

Notational System

Here n\ (read "w factorial" or "factorial n") = product of first n integers, and the integers

are usually written in reverse order Thus, 5! = 5 x 4 * 3 x 2 x 1 = 120 For mathematicalconsistency, O! is taken as 1

Also, Pn ^ = number of permutations that can be formed of n objects taken r at a time; and Cn r = number of combinations that can be formed of n objects taken r at a time.

NUMBER OF WAYS OFASSIGNING WORK

A firm has three machines, A, B, and C, and each machine can be operated by only oneindividual at a time The number of employees who are qualified to operate a machine is:machine A, five; machine B, three; machine C, seven In addition to these 15 employees,Smith is qualified to operate all three machines In how many ways can operators be as-signed to the machines?

2 Compute the number of possible assignments

if Smith is selected

If Smith is assigned to machine A, the number of possible assignments to B and C = 3 x 7

= 21 If Smith is assigned to machine B, the number of possible assignments to A and C =

5 x 7 = 35 if Smith is assigned to machine C, the number of possible assignments to Aand B = 5x3 = 15 Thus, the number of possible assignments with Smith selected = 21 +35'+15 = 71

3 Compute the total number of possible assignments

By summation, the number of ways in which operators can be assigned to the three

Calculation Procedure:

1 Compute the number of permutations in the absence

of any restriction

Use the relation Pn , = n\l(n - r)!, or P 1 ^ = 7!/3! - 7 x 6 x 5 x 4 - 840.

2 Compute the number of permutations that violate

the imposed restriction

Form permutations that violate the restriction Start by placing d in the first position ter c can be placed in any of the three subsequent positions Two positions now remainunoccupied, and five letters are available; these positions can be filled in 5 x 4 = 20 ways.Thus, the number of permutations in which d occupies the first position and c some sub-sequent position is 3 x 20 = 60 Similarly, the number of permutations in which d occu-pies the second position and c occupies the third or fourth position is 2 x 20 = 40, and thenumber of permutations in which d occupies the third position and c occupies the fourthposition is 1 x 20 = 20

Let-By summation, the number of unacceptable permutations = 60 + 40 + 20 = 120

3 Compute the number of permutations that satisfy

the requirement

By subtraction, the number of acceptable permutations = 840 - 120 = 720

FORMATION OF COMBINATIONS SUBJECT

2 Compute the number of possible committees that violate

the imposed restriction

Assign McCarthy to the committee, but exclude Polanski Five members remain to be lected, and 13 individuals are available The number of such committees = C135 =13!/(5!8!) = (13x 12 x U x 10 x 9)/(5 x 4 x 3 x 2) = 1287

se-3 Compute the number of possible committees that satisfy

the requirement

By subtraction, the number of ways in which the committee can be formed = 5005 - 1287-3718

Probability

If the outcome of a process cannot be predicted because it is influenced by chance, the

process is called a trial, or experiment The outcome of a trial or set of trials is an event Two events are mutually exclusive if the occurrence of one excludes the occurrence of the other Two events are independent of each other if the occurrence of one has no effect on

the likelihood that the other will occur

Assume that a box contains 17 objects, 12 of which are spheres If an object is to bedrawn at random and all objects have equal likelihood of being drawn, then the probabili-

ty that a sphere will be drawn is 12/17 Thus, the probability of a given event can rangefrom O to 1 The lower limit corresponds to an impossible event, and the upper limit cor-responds to an event that is certain to occur If two events are mutually exclusive, theprobability that either will occur is the sum of their respective probabilities If two events

are independent of each other, the probability that both will occur is the product of their

respective probabilities

Assume that a random variable is discrete and the number of values it can assume is

fi-nite A listing of these values and their respective probabilities is called the probability distribution of the variable Where the number of possible values is infinite, the probabil-

ity distribution is expressed by stating the functional relationship between a value of thevariable and the corresponding probability Where the random variable is continuous, themethod of expressing its probability distribution is illustrated in the calculation procedurebelow pertaining to the normal distribution

PROBABILITY OF A SEQUENCE OF EVENTS

A box contains 12 bolts Of these, 8 have square heads and 4 have hexagonal heads

Sev-en bolts will be removed from the box, individually and at random What is the

probabili-ty that the second and third bolts drawn will have square heads and the sixth bolt willhave a hexagonal head?

Calculation Procedure:

1 Compute the total number of ways in which the bolts

can be drawn

The sequence in which the bolts are drawn represents a permutation of 12 bolts taken 7 at

a time, and each bolt is unique The total number of permutations = P12 J — 121/5!.

2 Compute the number of ways in which the bolts can be drawn

in the manner specified

If the bolts are drawn in the manner specified, the second and third positions in the mutation are occupied by square-head bolts and the sixth position is occupied by a hexag-onal-head bolt Construct such a permutation, in these steps: Place a square-head bolt inthe second position; the number of bolts available is 8 Now place a square-head bolt inthe third position; the number of bolts available is 7 Now place a hexagonal-head bolt inthe sixth position; the number of bolts available is 4 Finally, fill the four remaining posi-tions in any manner whatever; the number of bolts available is 9

per-The second position can be filled in 8 ways, the third position in 7 ways, the sixth

po-sition in 4 ways, and the remaining popo-sitions in P94 ways By the multiplication law, the

number of acceptable permutations is 8 x 7 * 4 x P94 = 224(9!/5!)

3 Compute the probability of drawing the bolts

in the manner specified

Since all permutations have an equal likelihood of becoming the true permutation, theprobability equals the ratio of the number of acceptable permutations to the total number

of permutations Thus, probability = 224(9!/5!)/(12!/5!) = 224(9!)/!2! = 224(12 x U x10) = 224/1320-0.1697

4 Compute the probability by an alternative approach

As the preceding calculations show, the exact positions specified (second, third, andsixth) do not affect the result For simplicity, assume that the first and second bolts are to

be square-headed and the third bolt hexagonal-headed The probabilities are: first boltsquare-headed, 8/12; second bolt square-headed, 7/11; third bolt hexagonal-headed, 4/10.The probability that all three events will occur is the product of their respective probabili-ties Thus, the probability that bolts will be drawn in the manner specified =(8/12)(7/l 1)(4/10) = 224/1320 = 0.1697 Note also that the precise number of bolts drawnfrom the box (7) does not affect the result

PROBABILITY ASSOCIATED

WITH A SERIES OF TRIALS

During its manufacture, a product passes through five departments, A, B, C, D, and E.The probability that the product will be delayed in a department is: A, 0.06; B, 0.15; C,0.03; D 0.07; E, 0.13 These values are independent of one another in the sense that the

time for which the product is held in one department has no effect on the time it spends inany subsequent department What is the probability that there will be a delay in the manu-facture of this product?

2 Compute the probability of a delay in manufacture

Probability of delay in manufacture = 1 probability of no delay in manufacture = 1 0.6271-0.3729

-Related Calculations: This method of calculation can be applied to any situation

where a series of trials occurs, either simultaneously or in sequence, and any trial cancause the given event Thus, assume that several projectiles are fired simultaneously andthe probability of landing in a target area is known for each projectile The above methodcan be used to find the probability that at least one projectile will land in the target area

BINOMIAL PROBABILITYDISTRIBUTION

A case contains 14 units, 9 of which are of type A Five units will be drawn at random

from the case; and as a unit is drawn, it will be replaced with one of identical type IfX denotes the number of type A units drawn, find the probability distribution of X and the average value of X in the long run.

case, since each unit drawn is replaced with one of identical type, each drawing is

inde-pendent of all preceding drawings; therefore, X has a binomial probability distribution The event E consists of drawing a type A unit.

With respect to every drawing, probability of drawing a type A unit = 9/14, and ability of drawing a unit of some other type = 5/14 Arbitrarily set Jf = 3, and assume thatthe units are drawn thus: A-A-A-N-N, where N denotes a type other than A The proba-bility of drawing the units in this sequence = (9/14)(9/14)(9/14)(5/14)(5/14) =(9/14)3(5/14)2 Clearly this is also the probability of drawing 3 type A units in any othersequence Since the type A units can occupy any 3 of the 5 positions in the set of draw-ings and the exact positions do not matter, the number of sets of drawings that contain 3

prob-type A units = C53 Summing the probabilities, we find P(3) = C53(9/14)3(5/14)2 =[5!/(3!2!)](9/14)3(5/14)2 = 0.3389.

2 Write the equation of binomial probability distribution

Generalize from step 1 to obtain P(X) = Cn ^(I - P) n ~ x where P = probability event E will occur on a single trial Here P = 9/14.

3 Apply the foregoing equation to find the probability

distribution of X

The results are P(O) = 1(9/14)°(5/14)5 = 0.0058 Similarly, P(Y) = 0.0523; P(2) = 0.1883;

P(3) = 0.3389 from step 1; P(4) = 0.3050; P(S) = 0.109C

4 Verify the values of probability

Since it is certain that X will assume some value from O to 5, inclusive, the foregoing

probabilities must total 1 There sum is found to be 1.0001, and the results are thus firmed

con-5 Compute the average value of X in the long run

Consider that there are an infinite number of cases of the type described and that 5 unitswill be drawn from each case in the manner described, thereby generating an infinite set

of values of X Since the chance of obtaining a type A unit on a single drawing is 9/14, the

average number of type A units that will be obtained in 5 drawings is 5(9/14) = 45/14 =

3.21 Thus, the arithmetic mean of this infinite set of values of X is 3.21.

Alternatively, find the average value of X by multiplying all X values by their

respec-tive probabilities, to get 0.0523 + 2(0.1883) + 3(0.3389) + 4(0.3050) + 5(0.1098) = 3.21

The arithmetic mean of an infinite set of Jf values is also called the expected value of X.

PASCAL PROBABILITY DISTRIBUTION

Objects are ejected randomly from a rotating mechanism, and the probability that an ject will enter a stationary receptacle after leaving the mechanism is 0.35 The process of

ob-ejecting objects will continue until four objects have entered the receptacle Let X denote the number of objects that must be ejected Find (a) the probability corresponding to every X value from 4 to 10, inclusive; (b) the probability that more than 10 objects must

be ejected; (c) the average value of Xin the long run.

Calculation Procedure:

1 Compute the probability corresponding to a particular

value of X

Consider that a trial is performed repeatedly, each trial being independent of all preceding

trials, until a given event E has occurred for the Mi time Let X denote the number of als required The variable X is said to have a Pascal probability distribution (In the spe- cial case where k = 1, the probability distribution is called geometric.) In the present situ- ation, the given event is entrance of the object into the receptacle, and k = 4.

tri-Use this code: A signifies the object has entered; B signifies it has not Arbitrarily set

X = 9, and consider this sequence of events: A-B-B-A-B-B-B-A-A, which contains four

A's and five B's The probability of this sequence, and of every sequence containing fourA's and five B's, is (0.35)4(0.65)5 Other arrangements corresponding to X= 9 can be ob-

tained by holding the fourth A in the ninth position and rearranging the preceding letters,which consist of three A's and five B's Since the A's can be assigned to any 3 of the 8

positions, the number of arrangements that can be formed is C83 Thus, P(9) =

C83(0.35)4(0.65)5 = 56(0.35)4(0.65)5 = 0.0975

2 Write the equation of Pascal probability distribution

Generalize from step 1 to obtain P(X) = C^^P^l - Pf-*, where P = probability that event E will occur on a single trial Here P = 0.35.

3 Apply the foregoing equation to find the probabilities

corresponding to the given X values

The results are P(4) = 1(0.35)4(0.65)° = 0.0150; P(5) = 4(0.35^(0.6S)1 = 0.0390; P(G) =

10(0.35)4(0.65)2 = 0.0634 Similarly, P(I) = 0.0824; P(S) = 0.0938; P(9) = 0.0975 from

step 1; P(IO) = 0.0951

Thus, as X increases, P(X) increases until X= 9, and then it decreases The variable X

can assume an infinite number of values in theory, and the corresponding probabilitiesform a converging series having a sum of 1

4 Compute the probability that IO or fewer ejections

will be required

Sum the values in step 3; P(X < 10) = 0.4862.

5 Compute the probability that more than 10 ejections

will be required

Since it is certain that X will assume a value of 10 or less or a value of more than 10, P(X> 1O)-I- 0.4862 = 0.5138.

6 Compute the average number of ejections required

in the long run

Consider that the process of placing a set of four objects in the receptacle is continued

in-definitely, thereby generating an infinite set of values of X Since there is a 35 percent

chance that a specific object will enter the receptacle after being ejected from the nism, it will require an average of 1/0.35 = 2.86 ejections to place one object in the recep-tacle and an average of 4(1/0.35) = 11.43 ejections to place four objects in the receptacle

mecha-Thus, the infinite set of values of X has an arithmetic mean of 11.43.

POISSON PROBABILITY DISTRIBUTION

A radioactive substance emits particles at an average rate of 0.08 particles per second suming that the number of particles emitted during a given time interval has a Poissondistribution, find the probability that the substance will emit more than three particles in a20-s interval

As-Calculation Procedure:

1 Compute the average number of particles emitted in 20 s

Let T denote an interval of time, in suitable units Consider that an event E occurs domly in time but the average number of occurrences of E in time T9 as measured over a relatively long period, remains constant Let m = average (or expected) number of occur- rences of E in T9 and X = true number of occurrences of E in T The variable X is said to

ran-have a Poisson probability distribution

In the present case, X= number of particles emitted in 20 S9 and m = 20(0.08) = 1.6.

2 Compute the probability that X < 3

Use the relation P(X) = n^l\em (^\ where e = base of natural logarithms = 2.71828 Thus, em = e 1 - 6 = 4.95303 Then P(O) = (1.6)°/(4.95303 x 1) = 0.2019; P(I) = (1.6)3/(4.95303 x i) = 0.3230; P(2) = (1.6)2/(4.95303 x 2) = 0.2584; P(3) = (1.6)3/(4.95303 x 6)

= 0.1378 Sum these results to obtain P(X < 3) = 0.9211.

3 Compute the probability that X > 3

P(X > 3) =1-0.9211=0.0789.

Related Calculations: The probabilities in step 2 also can be found by referring

to a table of Poisson probability The foregoing discussion pertains to an event that occurs

in time, but analogous comments apply to an event that occurs in space For example,

as-sume that a firm manufactures long rolls of tape Defects in the tape occur randomly, butthe average number of defects in a 300-m length, as measured across long distances, isconstant The number of defects in a given length of tape has a Poisson distribution ThePoisson distribution is an extreme case of the binomial distribution As the probability

that event E will occur on a single trial becomes infinitesimally small and the number of

trials becomes infinitely large, the binomial distribution approaches the Poisson tion as a limit

distribu-COMPOSITE EVENT WITH

POISSON DISTRIBUTION

With reference to the preceding calculation procedure, a counting device is installed todetermine the number of particles emitted The probability that the device will actuallycount an emission is 0.90 Find the probability that the number of emissions counted in a20-s interval will be 3

Calculation Procedure:

1 Compute the average number of emissions counted in 20 s

In the present case, event E is that an emission is counted This event is a composite of two basic events: a particle is emitted, and the device functions properly Thus, m =

20(0.08)(0.90) =1.44

2 Compute the probability thatXs 3

Use the equation given in the preceding calculation procedure, or P(3) = (1.44)3/[eL44(3!)] = 0.1179

NORMAL DISTRIBUTION

A continuous random variable X has a normal probability distribution with an arithmetic

mean of 14 and a standard deviation of 2.5 Find the probability that on a given occasion

X will assume a value that (a) lies between 14 and 17; (b) lies between 12 and 16.2; (c) is

less than 10

Calculation Procedure:

1 Compute the values of z corresponding to the specified

boundary values of X

If a random variable X is continuous, the probability that X will assume a value between

Xj and X k is represented graphically by constructing a probability diagram in this manner:

Plot values o f X o n the horizontal axis; then construct a curve such that P(Xj <X<X k ) = area bounded by the curve, the horizontal axis, and vertical lines at Xj and X k The ordi- nate of this curve is denoted by f(X) and is called the probability density function The to-

tal area under the curve is 1, the probability of certainty

A continuous random variable has a normal or gaussian probability distribution if the

range of its possible values is infinite and its probability curve has an equation of this

form: f (X) = (\lb^/2^^) Q -< x -^ 2 ' 2b2 where a and b are constants and e = base of natural

log-arithms Figure 20 is the probability diagram; the curve is bell-shaped and symmetricabout a vertical line through the summit

Consider that the trial that yields a value of X is repeated indefinitely, generating an finite set of values of X Let JJL and a = arithmetic mean and standard deviation, respec- tively, of this set of values The summit of the probability curve lies at X= p By symme- try, the area under the curve to the left and to the right of X =IJL is 0.5.

in-The deviation of X, from JJL is expressed in standard units in this form: Z 1 = (X 1 - IJL)/cr Thus, for JT= 14, z = O; for X= 17, z - (17 - 14)72.5 = 1.20; for ,T= 12, z = (12 - 14)72.5

= -0.80; etc Record the z values in Table 24

10 -1.60 0.44520

FIGURE 20 Curve of normal probability distribution.

2 Find the values ofA(z)

Let A(ZJ) = area under probability curve from centerline (where X= JJL) to X=X 1 ; this area

= P(IJL <X<Xi) Obtain the values of A(Z) from the table of areas under the normal ability curve Refer to Table 25, which is an excerpt from this table Thus, if z = 1.60, A(Z)

prob-= 0.44520; if z prob-= 1.78, A(z) prob-= 0.46246 Note that A(-z t ) = A(Z) by symmetry Record the values of A(Z) in Table 24.

3 Compute the required probabilities

Refer to Fig 21 Apply the areas in Table 24 to obtain these results: P(14 < X< 17) = 0.38493; P(12 <X< 16.2) = 0.28814 + 0.31057 = 0.59871; P(X< 10) = 0.5 - 0.44520 -

0.05480

Related Calculations' Many random variables that occur in nature have a

nor-mal probability distribution For example, the height, weight, and intelligence of bers of a species have normal distributions Although in theory this distribution applies

mem-solely where the range of X values is infinite, in practice the distribution is applied as a

valid approximation where the range of lvalues is finite

APPLICATION OF NORMAL DISTRIBUTION

The time required to perform a manual operation is assumed to have a normal tion Studies of past performance disclose that the average time required is 5.80 h and thestandard deviation is 0.50 h Find the probability (to three decimal places) that the opera-tion will be performed within 5.25 h

FIGURE 21

2 Find the value of A(z)

Refer to the table of areas under the normal probability curve, and take the absolute value

of z If z =1.10, ,4(z) = 0.364

3 Compute the required probability

The area under consideration lies to the left of a vertical line at X= 5.25, and the area found in step 2 lies between this line and the centerline Then P(X< 5.25) = 0.5 - 0.364 =

0.136

NEGATIVE-EXPONENTIAL DISTRIBUTION

The mean life span of an electronic device that operates continuously is 2 months If thelife span of the device has a negative-exponential distribution, what is the probability thatthe life span will exceed 3 months?

Calculation Procedure:

1 Write the equation of cumulative probability

Refer to the calculation procedure on the normal distribution for definitions pertaining to

a continuous random variable A variable X is said to have a negative-exponential (or ply exponential) probability distribution if its probability density function is of this form: f(X) = O if X< O and/^Y) = ae'^if X > O Eq a, where a = positive constant and e = base

sim-of natural logarithms Figure 22 shows the probability diagram The arithmetic mean sim-of X

is jji = I/a, Eq b.

Let X = life span of device, months, and let K denote any positive number Integrate

Eq a between the limits of O and K, giving P(X < K)=I- e~aK , Eq c Then P(X >K) = e-° K , Eq d.

2 Compute the required probability

Compute a by Eq b, giving a = I/JJL = 1/2 = 0.5 Set K = 3 months By Eq d, P(X> 3) =

^r1-5 = !/*?1-5 = 0.2231

Statistical Inference

Consider that there exists a set of objects, which is called the population, or universe.

Also consider that interest centers on some property of these objects, such as length, lecular weight, etc., and that this property assumes many values Thus, associated with the

mo-population is a set of numbers This set of numbers has various characteristics, such as

arithmetic mean and standard deviation A characteristic of this set of numbers is called a

parameter For example, assume that the population consists of five spheres and that they

have the following diameters: 10, 13, 14, 19, and 21 cm The diameters have an metic mean of 15.4 cm and standard deviation of 4.03 cm, and these values are parame-ters of the given population

arith-Now consider that a subset of these objects is drawn This subset is called a sample, and a characteristic of the sample is called a statistic Thus, using the previous illustration,

assume that the sample consists of the spheres having diameters of 14, 19, and 21 cm.These diameters have an arithmetic mean of 18 cm and standard deviation of 2.94 cm,and these values are statistics of the sample drawn The number of objects in the sample is

the sample size.

FIGURE 22 Negative-exponential probability distribution.

In many instances, it is impossible to evaluate a parameter precisely, for two reasons:The population may be so large as to preclude measurement of every object, and meas-urement may entail destruction of the object, as in finding the breaking strength of a ca-ble In these cases it is necessary to estimate the parameter by drawing a representativesample and evaluating the corresponding statistic The process of estimating a parameter

by means of a statistic is known as statistical inference.

Since a statistic is a function of the manner in which the sample is drawn and thus isinfluenced by chance, the statistic is a random variable The probability distribution of a

statistic is called the sampling distribution of that statistic Consider that all possible ples of a given size have been drawn and the statistic S corresponding to each sample has been calculated A characteristic of this set of values of S, such as the arithmetic mean, is referred to as a characteristic of the sampling distribution of S In the subsequent material, the term mean refers exclusively to the arithmetic mean.

sam-Notational System

Table 26 is presented for ease of reference Here N = number of objects in the population; IJL and a = arithmetic mean and standard deviation, respectively, of the population; n = number of objects in the sample; X and s = arithmetic mean and standard deviation, re- spectively, of the sample; fjis and <rs = arithmetic mean and standard deviation, respec- tively, of the sampling distribution of the statistic S.

The quantity JJL^ is an index of the diversity of the sample means As the sample size

in-creases, the samples become less diverse

Since a sample represents a combination of TV objects taken w at a time, the number of

samples that may be drawn is C N>n = N\/[n\(N- «)!].

SAMPLING DISTRIBUTION OF THE MEAN

The population consists of 5 objects having the numerical values 15, 18, 27, 36, and 54;the sample size is 3 Find the mean and standard deviation of the sampling distribution ofthe mean

Calculation Procedure:

1 Compute the mean and variance of the population

Mean IUL = (15 + 18 + 27 + 36 + 54)/5 = 30; variance o 2 = [(15 - 3O)2 + (18 - 3O)2 + (27 3O)2 + (36 - 3O)2 + (54 - 30)2]/5 = 198

-2 Compute the properties of the sampling distribution

Apply Eq 14 to find the mean of the sampling distribution of the mean, or /^ = 30 ply Eq 15 to find the variance of the sampling distribution, or crj = 198(5 - 3)/(3 x 4) =

Ap-33 Then cry V33 = 5.74

3 Compute the required properties without recourse to any set equations

If the population is finite, the number of possible samples is finite Since all samples have

an equal likelihood of becoming the true sample, the sampling distribution of a statisticcan be found by forming all possible samples and computing the statistic under consider-ation for each

Record all possible samples in the first column of Table 27; the number of these

sam-ples is C 5 3 = 10 Now compute the mean X of each sample, record the results in the

sec-TABLE 27 Properties of Sampling Distribution of

ond column, and total them This column contains full information concerning the

sam-pling distribution of the mean Thus, since no duplications occur^P(X = 20) = 1/10; P(X

= 23) = 1/10; etc Compute the mean of the possible values of X 9 or ^ x = 300/10 = 30.

Record the deviations from 30 in Table 27, square the deviations, and total the results

Compute the variance of the possible values of X, or (T x = 330/10 = 33 Then (TX V33 =

5.74 These results are consistent with those in step 2

ESTIMATION OF POPULATION MEAN

ON BASIS OF SAMPLE MEAN

A firm produces rods, and their lengths vary slightly because of unavoidable differences

in manufacture Assume that the lengths are normally distributed One hundred rods wereselected at random, and they were found to have a mean length of 1.856 m and a standarddeviation of 0.074 m Estimate the mean length of all rods manufactured by this firm, us-ing a 95 percent confidence level

Calculation Procedure:

1 Compute the z value corresponding to the given

confidence level

Let X= length of a rod The population consists of all rods manufactured by the firm, and

it may be considered infinite There are two types of estimates: a point estimate, which

as-signs a specific value to X 9 and an interval estimate, which states that X lies within a given

interval This interval is called a confidence interval, its boundaries are called the dence limits, and the probability that the estimate is correct is called the confidence level,

confi-or confidence coefficient Statistical inference can supply only interval estimates

The central-limit theorem states: (a) If the population is extremely large and the ability distribution of X is normal, then the sampling distribution of the sample mean X is also normal; (b) if the population is extremely large but the probability distribution of X is not normal, the sampling distribution of X is approximately normal if the sample size is

prob-30 or more Thus, in the present case, the sampling distribution of X is considered to be

normal _

Figure 23 is the sampling distribution diagram of the sample mean X Let M= area der curve from B to^ C If a sample is drawn at random, there is a probability M that the true sample mean Xj lies within the interval BC, or P(JJLX ~ 2 Px < ^j < №x + zi°"jr) = M This equation can be transformed to P(Xj - z t ^ x < ^ x < Xj + Z 1 Cr x ) = M 9 Eq a.

un-In the present case, Xj = 1.856 m, s = 0.074 m, n = 100, and M= 0.95 Then area under curve from A to C= (0.50)(0.95) = 0.475 From the table of areas under the normal curve,

if A(Z 1 ) = 0.475, z t =1.96

2 Set up expressions for the mean and standard deviation

of the sampling distribution of the mean

By Eq 14, fjb x = fa where JJL is the population mean to be estimated Use the standard viation s of the sample as an estimate of the standard deviation cr of the population Then o- = 0.074 m, and by Eq 1 Sa 9 a x = 0.074/VKJO = 0.0074 m.

de-3 Estimate the mean length of the rods

Refer to Eq a, and compute zpz = (1.96)(0.0074) = 0.015 m Compute the confidence

limits in Eq a, or 1.856 - 0.015 = 1.841 m and 1.856 + 0.015 = 1.871 m Equation a

FIGURE 23 Sampling distribution of the mean.

becomes P(1.841 < /x < 1.871) = 0.95 Thus, there is a 95 percent probability that themean length of all rods lies between 1.841 and 1.871 m

Related Calculations: Note that the confidence interval is a function of the

de-gree of probability that is demanded, and the two quantities vary in the same direction.For example, if the confidence level were 90 percent, the confidence interval would be1.844 to 1.868m

DECISION MAKING ON STATISTICAL BASIS

Units of a commodity are produced individually, and studies have shown that the time quired to produce a unit has a mean value of 3.50 h and a standard deviation of 0.64 h Anindustrial engineer claims that a modification of the production process will substantiallyreduce production time The proposed method was tested on 40 units, and it was foundthat the mean production time was 3.37 h per unit Management has decided that it willmake the proposed modification only if there is a probability of 95 percent or more thatthe engineer's claim is valid What is your recommendation?

re-Calculation Procedure:

1 Formulate the null and alternative hypotheses

The population consists of all units that will be produced under the modified method if it

is adopted, and the sample consists of the 40 units actually produced under this method

An assumption based on conjecture is termed a hypothesis A hypothesis that is

formulat-ed merely to provide a basis for investigation is a null hypothesis, and any hypothesis that contradicts the null hypothesis is an alternative hypothesis However, interest centers on

the particular alternative hypothesis that is significant in the given case The null and ternative hypotheses are denoted by Tf0 and HI , respectively.

al-Let X= time required to produce 1 unit, h Place the burden of proof on the industrial

engineer by assuming that production time under the modified method is identical with

Sample mean XCross-hatched area = MStandard deviation = O"x

that under the present method Thus, the hypotheses are //0: JJL = 3.50 h and a = 0.64 h;

H 1 IfJL <3.50h.

2 Compute the properties of the sampling distribution

of the mean as based on the null hypothesis

Apply Eqs 14 and 15«, giving JJL^ = 3.50 h and a^ = 0.64/40 = 0.101 h.

3 Compute the critical value of X

By the central-limit theorem given in the preceding calculation procedure, the sampling

distribution of the sample mean X may be considered normal, and the sampling tion diagram is shown in Fig 24a Management has imposed a requirement of 95 percent

distribu-probability Therefore, the null hypothesis is disproved and the alternative hypothesis idated if the true sample mean has a value less than that corresponding to 95 percent of all

val-possible samples In Fig 24«, locate B such that the area to the right of B = 0.95; then area from A to B = 0.95 - 0.50 = 0.45 The null hypothesis is to be accepted or rejected accord- ing to whether the true value of X lies to the right or left of B, respectively, and the re-

Area=0.05

FIGURE 24 Sampling distribution of mean production time corresponding to

three distinct values of the population mean

gions are labeled as shown The values of X and z at the boundary of the acceptance and rejection regions are called the critical values.

At B, A(z) = 0.45 From the table of areas under the normal curve, z — -1.645 Thus, at

B, X = 3.50 + (-1.645)(0.101) = 3.334 h.

4 Make a recommendation

Since the true sample mean of 3.37 h falls to the right of B, the null hypothesis stands.

Therefore, we must recommend that the modified method of production be disapprovedand the present method retained

This recommendation does not necessarily imply that the industrial engineer's claim isinvalid The decision must be based on probability rather than certainty, and the test re-sults have failed to demonstrate a 95 percent probability that the modified method is supe-rior to the present method The difference between the assumed population mean of 3.50

h and the sample mean of 3.37 h can be ascribed to chance

Related Calculations' If the null hypothesis is rejected when, in fact, it is true, a

Type I error has been committed The probability of committing this error is denoted by

a, and the acceptable value of a is termed the level of significance In this case, if the null hypothesis is correct, the sampling distribution of X is as shown in Fig 24a The hypoth- esis will be rejected if X assumes a value to the left of B, and the probability of this event

is 1 - 0.95 = 0.05 Thus, the level of significance is 0.05

Whether a null hypothesis is accepted or rejected depends largely on the level of nificance imposed Therefore, selecting an appropriate level of significance is one of thecrucial problems that arise in statistical decision making The selection must be based onthe amount of the loss that would result from a false decision

sig-PROBABILITY OFACCEPTING A FALSE

NULL HYPOTHESIS

With reference to the preceding calculation procedure, the time required to produce 1 unitunder the modified method has these characteristics: The standard deviation remains 0.64

h, but the arithmetic mean is (a) 3.40 h; (b) 3.20 h Determine the probability that the

in-dustrial engineer's proposal will be vetoed despite its merit

Calculation Procedure:

1 Compute the critical value of z in each case

Since cr remains 0.64 h and the sample size is still 40, CT^ remains 0.101 h Refer to Fig 24b and c, which gives the sampling distributions of X when LL = 3.40 h and LL = 3.20 h, respectively The null hypothesis will be accepted if X > 3.334, and it is necessary to cal-

culate the probability of this event

In Fig 246, at B', z = (3.334 3.40)70.101 = 0.653 In Fig 24c, at B", z = (3.334

-3.20)70101-1.327

2 Compute the required probabilities

Refer to the table of areas under the normal curve When z = -0.653, A(z) = 0.243 In Fig.

246, area to right of B' = 0.243 + 0.5 = 0.743 Thus, when LL = 3.40 h, there is a ity of 74.3 percent that the proposal will be vetoed Similarly, when z = 1.327, A(z) = 0.408 In Fig 24c, area to right of B" = 0.5 - 0.408 = 0.092 Thus, when LL = 3.20 h, there

probabil-is a probability of 9.2 percent that the proposal will be vetoed

Related Calculations: If the null hypothesis is accepted when, in fact, it is false,

a Type II error has been committed, and the probability of committing this error is

denot-ed by /3 Thus, this calculation procdenot-edure involves the determination of'/3 It follows that 1

- /3 is the probability that a false null hypothesis will be rejected The process of drawing

and analyzing a sample represents a test of the null hypothesis, and the quantities ft and 1

- ft are called the operating characteristic and power, respectively, of the test Thus, the

power of a test is its ability to detect that the null hypothesis is false if such is truly thecase

Consider that a diagram is constructed in which assumed values of the parameter are

plotted on the horizontal axis and the corresponding values of ft resulting from the null hypothesis are plotted on the vertical axis The curve thus obtained is called an operating- characteristic curve Similarly, the curve obtained by plotting values of 1 - ft against as- sumed values of the parameter is called a power curve.

DECISION BASED ON PROPORTION

OFSAMPLE

A firm receives a large shipment of small machine parts, and it must determine whetherthe number of defectives in a shipment is tolerable Its policy is as follows: A shipment isaccepted only if the estimated incidence of defectives is 3 percent or less, the decision isbased on an inspection of 250 parts selected at random, and a shipment is rejected only ifthere is a probability of 90 percent or more that the incidence of defectives exceeds 3 per-cent What is the highest incidence of defectives in the sample if the shipment is to beconsidered acceptable?

Calculation Procedure:

1 Formulate the null hypothesis

Consider that a set of objects consists of type A and type B objects The ratio of the

num-ber of type A objects to the total numnum-ber of objects is called the proportion of type A jects Let P and p = proportion of type A objects in the population and sample, respective-

ob-ly

In this case, the population consists of all machine parts in the shipment, the sampleconsists of the 250 parts that are inspected, and interest centers on the proportion of de-fective parts To provide a basis for investigation, assume that the proportion of defec-

tives in the shipment is precisely 3 percent Thus, the null hypothesis is H0 : P = 0.03.

2 Compute the properties of the sampling distribution of the proportion as based on the null hypothesis

Consider that all possible samples of a given size are drawn and their respective values of

p determined, thus obtaining the sampling distribution of p As before, let N and n = ber of objects in the population and sample, respectively The sampling distribution of p has these values: the mean fip = P, Eq a; the variance (Tp=P(I- P)(N- n)/[n(N- I)], Eq.

num-b Where the population is infinite, (Tp=P(I -P)/«, Eq c In the present case, P = 0.03, N may be considered infinite, and n = 250 By Eq a, /JLP = 0.03; by Eq c, a* = (0.03)(0.97)/250 - 0.0001164 Then standard deviation o-p = 0.0108.

3 Compute the critical value of p

For simplicity, treat the number of defective parts as a continuous rather than a discrete

variable; then P and p are also continuous Since the sample is very large, the sampling distribution ofp is approximately normal, and it is shown in Fig 25a The null hypothesis

is to be rejected if.p assumes a value greater than that corresponding to 90 percent of all possible samples In Fig 25<z, locate B such that area to the left of B = 0.90; then area from A to B = 0.90 - 0.50 = 0.40 At B9 A(z) = 0.40 From the table of areas under the nor- mal curve, z = 1,282 Thus, at B9 p = 0.03 + (1.282)(0.0108) = 0.044.

4 State the decision rule

If the proportion of defectives in the sample is 4.4 percent or less, accept the shipment; ifthe proportion is greater, reject the shipment

By setting the limiting proportion of defectives in the sample at 4.4 percent as pared with the limiting proportion of 3 percent in the population, allowance is being madefor the random variability of sample results

com-PROBABILITY OF ACCEPTING AN

UNSATISFACTORY SHIPMENT

With reference to the preceding calculation procedure, what is the probability that a ment in which the incidence of defectives is 5 percent will nevertheless be accepted?

ship-Acceptance region Rejection region

FIGURE 25 Sampling distribution of proportion of defective units.

2 Compute the probability that the shipment will be accepted

The shipment will be accepted if/? < 0.044, and it is necessary to calculate the probability

of this event In Fig 256, at B', z = (0.044 - 0.05)70.0138 = -0.435 From the table of eas under the normal curve, A(Z) = 0.168 Then the area to the left of B' = 0.5 - 0.168 =

ar-0.332 Thus, there is a probability of 33.2 percent that the shipment will be accepted

Refer to Fig 26, which is the assumed life-span curve of a device The diagram is

con-structed so that P(J1 < T ^ t2 ) = area under curve from t l to t2 Thus, a life-span curve is the probability curve of the continuous variable T Left/(f) = ordinate of life-span curve = probability-density function From the definition of reliability, it follows that R(t) = area under curve to right of t, or