Quantitative Methods for Business chapter 12 pptx

Accidents and incidence – discrete probability distributionsand simulation 12 Chapter objectives This chapter will help you to: ■ work out the probabilities for a basic discrete probabil

Accidents and incidence – discrete probability distributions

and simulation

12

Chapter objectives

This chapter will help you to:

■ work out the probabilities for a basic discrete probabilitydistribution

■ calculate the mean and standard deviation of a discrete ability distribution

prob-■ model business processes with the binomial distribution

■ model business processes with the Poisson distribution

■ simulate simple random business processes with randomnumbers

■ use the technology: discrete probability distributions inEXCEL, MINITAB and SPSS, random number generation inEXCEL

■ become acquainted with business uses of discrete probabilitydistributions and simulation

In this chapter we will bring together two key concepts from earlierchapters The first of these is the idea of a frequency distribution, whichshows the frequency or regularity with which the values of a variableoccur, in other words how they are distributed across their range Thesecond key concept is that of probability, which we considered in

Chapter 9 Here we will be looking at probability distributions which

por-tray not the frequency with which values of a distribution actually occurbut the probability with which we predict they will occur

Probability distributions are very important tools for modelling orrepresenting processes that occur at random, such as customers visit-

ing a website or accidents on a building site These are examples of

dis-crete random variables as they vary in a random fashion and can have only

certain values, whole numbers in both cases; we cannot conceive of half

a customer visiting a website or 0.3 of an accident happening We use

discrete probability distributions to model these sorts of variables.

In studying probability distributions we will look at how they can be

derived and how we can model or represent the chances of different

combinations of outcomes using the same sort of approach as we use toarrange data into frequency distributions Following that we will exam-ine two standard discrete probability distributions; the binomial andthe Poisson Lastly we will look at how random numbers and discreteprobability distributions can be used to simulate the operation of ran-dom business processes

12.1 Simple probability distributions

In section 4.4.2 of Chapter 4 we looked at how we could present data

in the form of a frequency distribution This involved defining egories of values that occurred in the set of data and finding out howmany observed values fell into each category, in other words, the fre-quency of each category of values in the set of data The results of thisprocess enabled us to see how the observations were distributed overthe range of the data, hence the term frequency distribution

cat-A probability distribution is very similar to a frequency distribution.

Like a frequency distribution, a probability distribution has a series ofcategories, but instead of categories of values it has categories of types

of outcomes The other difference is that each category has a ability instead of a frequency

prob-In the same way as a frequency distribution tells us how frequentlyeach type of value occurs, a probability distribution tells us how probableeach type of outcome is

In section 5.2.1 of Chapter 5 we saw how a histogram could be used

to show a frequency distribution We can use a similar type of diagram

to portray a probability distribution

In Chapter 6 we used summary measures, including the mean and standard deviation, to summarize distributions of data We can usethe mean and standard deviation to summarize distributions ofprobabilities

Just as we needed a set of data to construct a frequency distribution,

we need to identify a set of compound outcomes in order to create aprobability distribution We also need the probabilities of the simpleoutcomes that make up the combinations of outcomes

Example 12.1

Imported Loobov condoms are sold in packets of three Following customer complaintsthe importer commissioned product testing which showed that due to a randomly occur-ring manufacturing fault 10% of the condoms tear in use What are the chances that apacket of three includes zero, one, two and three defective condoms?

The probability that a condom is defective (D) is 0.1 and the probability it is good (G) is 0.9.

The probability that a packet of three contains no defectives is the probability that asequence of three good condoms were put in the packet

P(GGG) 0.9 * 0.9 * 0.9  0.729The probability that a packet contains one defective is a little more complicatedbecause we have to take into account the fact that the defective one could be the first

or the second or the third condom to be put in the packet

so P(1 Defective)  P(DGG or GDG or GGD)

Because these three sequences are mutually exclusive, according to the addition rule

of probability (you may like to refer back to section 10.3.1 of Chapter 10):

P(1 Defective)  P(DGG)  P(GDG)  P(GGD)

The probability that the first of these sequences occurs is:

P(DGG) 0.1 * 0.9 * 0.9  0.081The probability of the second:

P(GDG) 0.9 * 0.1 * 0.9  0.081

It is no accident that the probabilities of these sequences are the same Although theexact sequence is different the elements that make up both are the same To work outthe compound probabilities that they occur we use the same simple probabilities but in

In Example 12.1 the probability distribution presents the number of

defectives as a variable, X, whose values are represented as x The able X is a discrete random variable It is discrete because it can only take

vari-a limited number of vvari-alues – zero, one, two or three It is rvari-andombecause the values occur as the result of a random process

The symbol ‘P(x)’ represents the probability that the variable X takes

a particular value, x For instance, we can represent the probability that

the number of females is one, as

P(X 1)  0.243

a different order, and the order does not affect the result when we multiply them

together If you work out P(GGD) you should find that it also is 0.081.

The probability of getting one defective is therefore:

Figure 12.1 shows the probability distribution we compiled inExample 12.1 in graphical form.

We can find summary measures to represent this distribution, in thesame way as we could use summary measures to represent distributions

of data However, as we don’t have a set of data to use to get our

sum-mary measures we use the probabilities to ‘weight’ the values of X, just

as we would use frequencies to obtain the mean from a frequency tribution You work out the mean of a probability distribution by multi-

dis-plying each value of x by its probability and then adding up the

products:

␮  ∑x P(x)

Notice that we use the Greek symbol ␮ to represent the mean of the

distribution The mean of a probability distribution is a population meanbecause we are dealing with a distribution that represents the prob-abilities of all possible values of the variable

Once we have found the mean we can proceed to find the varianceand standard deviation We can obtain the variance, ␴2, by squaring each

x value, multiplying the square of it by its probability and adding the

products From this sum we subtract the square of the mean

␴2 ∑x2P(x)  ␮2The standard deviation, ␴, is simply the square root of the variance.

Again you can see that we use a Greek letter in representing the ance and the standard deviation because they are population measures

vari-0 0.0

0.1 0.2 0.3

0.5 0.6 0.7 0.8

The mean of a probability distribution is sometimes referred to as the

expected value of the distribution Unlike the mean of a set of data,

which is based on what the observed values of a variable actually were,the mean of a probability distribution tells us what the values of the

variable are likely, or expected to be.

We may need to know the probability that a discrete random variable

takes a particular value or a lower value This is known as a cumulative probability because in order to get it we have to add up or accumulate

other probabilities You can calculate cumulative probabilities directlyfrom a probability distribution

Example 12.2

Calculate the mean and the standard deviation for the probability distribution inExample 12.1

The mean, ␮, is 0.300, the total of the x P(x) column.

The variance, ␴2, is 0.360, the total of the x2P(x) column minus the square of the mean:

␴2 0.360  0.3002 0.360  0.090  0.270The standard deviation: ␴ √␴2√0.270 0.520

We can use the symbol ‘’ to represent ‘less than or equal to’, so we are looking for

P(X 2) (It may help you to recognize this symbol if you remember that the small end of the ‘ ’ part is pointing at the X and the large end at the 2, implying that X

is smaller than 2.)

The cumulative probabilities like those we worked out in Example12.3 are perfectly adequate if we want the probability that a variabletakes a particular value or a lower one, but what if we need to know theprobability that a variable is higher than a particular value?

We can use the same cumulative probabilities if we manipulate themusing our knowledge of the addition rule of probability If, for instance,

we want to know the probability that a variable is more than two, we canfind it by taking the probability that it is two or less away from one

these cumulative probabilities in the right-hand column of the following table:

The cumulative probability that X is zero or less, P(X 0), is the probability that X

is zero, 0.729, plus the probability that X is less than zero, but since it is impossible for

X to be less than zero we do not have to add anything to 0.729.

The second cumulative probability, the probability that X is one or less, P(X 1), is

the probability that X is one, 0.243, plus the probability that X is less than one, in other

words, that it is zero, 0.729 Adding these two probabilities together gives us 0.972

The third cumulative probability is the probability that X is two or less, P(X 2) We

obtain this by adding the probability that X is 2, 0.027, to the probability that X is less

than 2, in other words that it is one or less This is the previous cumulative probability,0.972 If we add this to the 0.027 we get 0.999

The fourth and final cumulative probability is the probability that X is three or less Since we know that X cannot be more than three (there are only three condoms in a

packet), it is certain to be three or less, so the cumulative probability is one We would get

the same result arithmetically if we add the probability that X is three, 0.001, to the lative probability that X is less than three, in other words that it is two or less, 0.999.

Although the situation described in Example 12.1, considering packets

of just three condoms, was quite simple, the approach we used to obtainthe probability distribution was rather laborious Imagine that you had

to use the same approach to produce a probability distribution if therewere five or six condoms in a packet instead of just three We had to becareful enough in identifying the three different ways in which therecould be two defectives in a packet of three If the packets containedfive condoms, identifying the different ways that there could be say twodefectives in a packet would be far more tedious

Fortunately there are methods of analysing such situations that donot involve strenuous mental gymnastics These involve using a type of

probability distribution known as the binomial distribution.

At this point you may find it useful to try Review Questions 12.1 and

12.2at the end of the chapter

12.2 The binomial distribution

The binomial distribution is the first of a series of ‘model’ statisticaldistributions that you will meet in this chapter and the two that follow

it The distribution was first derived theoretically but is widely used indealing with practical situations It is particularly useful because itenables you not only to answer a specific question but also to explorethe consequences of altering the situation without actually doing it.You can use the binomial distribution to solve problems that have

what is called a binomial structure These types of problems arise in

situ-ations where a series of finite, or limited number of ‘experiments’, or

‘trials’ take place repeatedly Each trial has the same two mutually

exclu-sive and collectively exhaustive outcomes, as the bi in the word binomial

might suggest By convention one of these outcomes is referred to as

‘success’; the other as ‘failure’

To analyse a problem using the binomial distribution you have to knowthe probability of each outcome and it must be the same for every trial Inother words, the results of the trials must be independent of each other.Words like ‘experiment’ and ‘trial’ are used to describe binomial situations because of the origins and widespread use of the binomial

distribution in science Although the distribution has become widelyused in many other fields, these scientific terms have stuck.

The process in Example 12.1 has a binomial structure Putting threecondoms in a packet is in effect conducting a series of three trials Ineach trial, that is, each time a condom is put in a packet, there can beonly one of two outcomes: either it is defective or it is good

In practice, we would use tables such as Table 2 in Appendix 1 onpage 618 to apply the binomial distribution These are produced using

an equation, called the binomial equation, which you will see below.You won’t need to remember it, and you shouldn’t need to use it Wewill look at it here to illustrate how it works

We will use the symbol X to represent the number of ‘successes’ in a certain number of trials, n X is what is called a binomial random variable The probability of success in any one trial is represented by the letter p The probability that there are x successes in n trials is:

You will see that an exclamation mark is used several times in the

equa-tion It represents a factorial, which is a number multiplied by one less

than itself then multiplied by two less itself and so on until we get to one.For instance four factorial, 4!, is four times three times two times one,

Example 12.4

Use the binomial equation to find the first two probabilities in the probability distributionfor Example 12.1

We will begin by identifying the number of trials to insert in the binomial equation

Putting three condoms in a packet involves conducting three ‘trials’, so n 3

The variable X is the number of defectives in a packet of three We need to find the probabilities that X is 0, 1, 2 and 3, so these will be the x values.

Suppose we define ‘success’ as a defective, then p, the probability of success in any

one trial, is 0.1

We can now put these numbers into the equation We will start by working out the

probability that there are no defectives in a packet of three, that is X 0

This expression can be simplified considerably Any number raised to the power zero

is one, so 0.10 1 Conveniently zero factorial, 0!, is one as well We can also carry outthe subtractions

Finding binomial probabilities using printed tables means we don’thave to undertake laborious calculations to obtain the figures we arelooking for We can use them to help us analyse far more complexproblems than Example 12.1, such as the problem in Example 12.5.

If you look back at Example 12.1, you will find that this is the same as the first figure

in the probability distribution The figure below it, 0.243, is the probability that there is

one defective in a packet of three, that is X 1 Using the binomial equation:

Look carefully at this expression You can see that the first part of it, which involvesthe factorials, is there to reflect the number of ways there are of getting a packet with

one defective, 3(DGG, GDG and GGD) In the earlier expression, for P(X 0), the firstpart of the expression came to one, since there is only one way of getting a packet with

If 10% of the population are vegetarians, what is the probability that on a fullybooked flight there will be at least one vegetarian passenger for whom a meat-freeSnack Pack will not be available?

This problem has a binomial structure We will define the variable X as the number

of vegetarians on a fully booked flight Each passenger is a ‘trial’ that can be a ‘success’,

We can show the binomial distribution we used in Example 12.5 graphically.

In Figure 12.2 the block above 0 represents the probability that X 0,

P(0), 0.349 The other blocks combined represent the probability that

X is larger than 0, P(X

a vegetarian, or a ‘failure’, a non-vegetarian The probability of ‘success’, in this case theprobability that a passenger is a vegetarian, is 0.1 There are ten passengers on a fully

booked flight, so the number of trials, n, is 10.

The appropriate probability distribution for this problem is the binomial distribution

with n  10 and p  0.1 Table 2 on page 618 contains the following information about

the distribution:

For 10 trials (n  10); p  0.1

The column headed P(x) provides the probabilities that a specific number of cesses’, x, occurs, e.g the probability of three ‘successes’ in ten trials, P(3), is 0.057 The column headed P(X x) provides the probability that x or fewer ‘successes’ occur, e.g the probability that there are 3 or fewer ‘successes’, P(X 3), is 0.987

‘suc-If there is only one vegetarian passenger they can be given the single vegetarian SnackPack available on the plane It is only when there is more than one vegetarian passengerthat at least one of them will object to their Snack Pack So we need the probability that

there is more than one vegetarian passenger, which is the probability that X is greater than one, P(X

except the first and second ones, the probability that X is zero, P(0), and the probability that X is one, P(1) However, it is easier to take the probability of one or fewer, P(X 1),away from one:

It is quite easy to find the mean and variance of a binomial tion The mean, ␮, is simply the number of trials multiplied by the

In Example 12.5 the number of trials, n, was 10 and the probability of success, p, was

0.1, so the mean number of vegetarians on fully booked flights is:

␮  n * p  10 * 0.1  1.0

The variance is: ␴2 n * p(1  p)  10 * 0.1(1  0.1)  1.0 * 0.9  0.9

The standard deviation is: ␴ √␴2√0.9 0.949

The binomial distribution is called a discrete probability distribution

because it describes the behaviour of certain types of discrete randomvariables, binomial variables These variables concern the number oftimes certain outcomes occur in the course of a finite number of trials

But what if we need to analyse how many things happen over aperiod of time? For this sort of situation we use another type of discrete

probability distribution known as the Poisson distribution.

At this point you may find it useful to try Review Questions 12.3 to

12.8at the end of the chapter

12.3 The Poisson distribution

Some types of business problem involve the analysis of incidents that areunpredictable Usually they are things that can happen over a period

of time, such as the number of telephone calls coming through to anoffice worker However, it could be a number of things over an area,such as the number of stains in a carpet

The Poisson distribution describes the behaviour of variables like thenumber of calls per hour or the number of stains per square metre Itenables us to find the probability that a specific number of incidentshappen over a particular period The distribution is named after theFrench mathematician Simeon Poisson, who outlined the idea in 1837,but the credit for demonstrating its usefulness belongs to the Russianstatistician Vladislav Bortkiewicz, who applied it to a variety of situationsincluding famously the incidence of deaths by horse kicks amongst soldiers of the Prussian army

Using the Poisson distribution is quite straightforward In fact you mayfind it easier than using the binomial distribution because we need toknow fewer things about the situation To identify which binomial dis-tribution to use we had to specify the number of trials and the probabil-

ity of success; these were the two defining characteristics, or parameters

of the binomial distribution In contrast, the Poisson distribution is asingle parameter distribution, the one parameter being the mean

If we have the mean of the variable we are investigating we can obtainthe probabilities of the Poisson distribution using Table 3 on page 619

in Appendix 1

Example 12.7

The medical tent at the Paroda music festival has the capacity to deal with up to threepeople requiring treatment in any one hour The mean number of people requiringtreatment is 2 per hour What is the probability that they will not be able to deal with allthe people requiring treatment in an hour?

If we had to produce the Poisson probabilities in Example 12.7 out the aid of tables we could calculate them using the formula for thedistribution You won’t have to remember it, and probably won’t need

with-to use it, but it may help your understanding if you know where thefigures come from

The probability that the number of incidents, X, takes a particular value, x, is:

The variable, X, in this case is the number of people per hour that require treatment.

We can use the Poisson distribution to investigate the problem because it involves a

dis-crete number of occurrences, or incidents over a period of time The mean of X is 2.

The medical facility can deal with three people an hour, so the probability that there

are more people requiring treatment than they can handle is the probability that X is more than 3, P(X

The appropriate distribution is the Poisson distribution with a mean of 2 Table 3 onpage 619 contains the following information about the distribution:

The letter e is the mathematical constant known as Euler’s number Thevalue of this, to 4 places of decimals, is 2.7183, so we can put this in theformula:

The symbol ␮ represents the mean of the distribution and x is the

value of X whose probability we want to know In Example 12.7 the

mean is 2, so the probability that there are no people requiring

treat-ment, in other words the probability that X is zero, is:

To work this out you need to know that if you raise any number (inthis case ␮) to the power zero the answer is one, so:

The first part of the expression, 2.7183 2, is 1/ 2.71832since any numberraised to a negative power is a reciprocal, so:

If you are unsure of the arithmetic we have used here you may find ithelpful to refer back to section 1.3.3 of Chapter 1

0 0.0 0.1 0.2 0.3