1

Unit5 SPECIAL PROBABILITY DISTRIBUTIONS

SESSION 1-5: DISCRETE PROBABILITY DISTRIBUTIONS

1-5.1 Bernoulli Process

A single trial of an experiment may result in one of the two mutually exclusive

outcomes, such as, defective or non-defective in testing of an item from production

process, dead or alive as a person is rescued from disaster, yes or no in a consumer

preference for a product, male or female in a child birth, head or tail in a toss of a coin,

etc. Such a trial is called Bernoulli trial and a sequence of these trials form a Bernoulli

Process, satisfying the following conditions:

Each trial results in one of the two possible mutually exclusive outcomes,

success and failure.

The probability of a success p remains constant, from trial to trial. The

probability of failure, denoted 1 ,q p remains the same.

The trials are independent. That is, the outcome of any particular trial is not

affected by the outcome of any other trial.

The random variable of this experiment is a binary variable which assumes the

values, 0 and 1.

(a) Definition 5.1:

A random variable, x is said to have a Bernoulli distribution if it assumes the

values, 0 and 1 for the two outcomes. The probability distribution for the

success in the trial, x is defined by

,1010,)1()( 1 pandorxppxp xx

where the mean and variance of the distribution are as follows:

( ) ,E x p and 2 ( ) (1 )Var x p p

1-5.2 The Binomial Distribution

The Binomial distribution is a discrete probability distribution used to model

experiments consisting of sequence of observations of identical and independent trials,

each of which results in one of the two outcomes. Such experiments which are actually

2

generalization of Bernoulli trials are called Binomial Experiment. A Binomial

Experiment exhibits the following properties:

The experiment consists of n independent and identical trials.

Each trial results in one of the two outcomes called success and failure.

The probability of success in a single trial is p and remains the same from trial

to trial. The probability of a failure, also in a single trial is 1 .q p

The random variable of interest, x is the number of successes observed during

the n trials.

There are several situations which may result in a random variable that may be satisfied

by the conditions of the Binomial Experiment.

Some examples of these situations are a random selection of items from a

manufacturing process for inspection is either defective or non-defective, an opinion

poll during electioneering campaign where each of n persons interviewed will either

vote or not vote for a particular candidate, interviewing a random sample of n students

to determine whether a policy being introduced by the university authorities is favoured

or not favoured, a number of patients undergoing through a medical treatment may

either come out successfully or not successfully, a sequence of n shots at a target may

result in a number of hits or misses, tossing a coin n times and observing the number of

successes, heads or tails.

Definition 5.2:

A random variable, x is said to have a Binomial distribution based on n trials with

probability of a success, p if and only if

nxppx

nxp xnx ,...,2,1,0,)1()(

n where 0 1p and the mean and variance are respectively,

( )E x np and 2 ( ) (1 )Var x np p

The random variable, x with Binomial distribution is simply denoted as ( , )x B n p .

The following diagrams (Figure 5.1 and Figure 5, 2) illustrate graphically the

probability histograms of Binomial distribution.

3

( )p x

0.4

0.3

0.2

0.1

0.0 0 1 2 3 4 x

Figure 5.1: Binomial distribution for 10n and 0.10p

( )p x

0.25

0.20

0.15

0.10

0.05

0.0 0 1 2 3 4 5 6 7 8 9 x Figure 5.2: Binomial distribution for 20n and 0.30p

0.20 ( )p x

0.15

0.10

0.05

0 4 5 6 7 8 9 10 11 12 13 14 x

Figure 5.3: Binomial distribution for 50n , 0.50p

4

The term binomial experiment is derived from the fact that the probabilities, p(x) at x =

0, 1, 2. . . n are terms of the Binomial expansion,

0

nn x n x

x

np q p q

x

where 0 0

( ) 1n n

nx n x

x x

np x p q p q

x

The generalization of the Binomial is distribution is Multinomial distribution which

arises when each trial of the experiment has more than two possible outcomes of the

probabilities of the respective outcomes are the same for each trial. These two

probability distributions have many applications because the binomial and multinomial

experiments occurs in sampling for defectives in industrial quality control, sampling of

consumer preference for products, voting intentions in an opinion polls and in many

other physical situations.

Example 5.2:

5.2(a) Suppose that a consignment of 300 electrical fuses contains 5% defectives. If a

random sample of ten fuses is selected and tested, find the probability of

observing at least three defectives.

5.2(b) A large retailer purchases a certain kind of electrical device from a

manufacturer. The manufacturer indicates that the defective rate of the device

is 3% in a shipment. The inspector of the retailer randomly picks 20 items from

a shipment. (i) What is probability that there will be at most two (2) defective

items?

(ii) Suppose that the retailer receives 10 shipments in a month and the

inspector randomly tests 20 devices per shipment. What is the

probability that there will be at least 3 shipments containing at least one

defective device?

Solution:

(a) The number of detective fuses, ( 10, 0.05)x B n p and so the probability,

210

0

( 3) 1 ( 2)

101 (0.05) (0.95)x x

x

P x P x

x

5

0 10 1 9

2 8

10 100.05 0.95 0.05 0.95

0 11

100.05 0.95

2

10 9 2 81 (0.95) 10(0.05)(0.95) 45(0.05) (0.95)

1 0.9884996441

0.011503558

(b) Let the number of defective devices among the 20 randomly selected from a

shipment be y and so we have the following:

(i) ( 20, 0.03).y B n p The required probability becomes

220

0

0 20 1 19 2 18

20( 2) 0.03 0.97

20 20 200.03 0.97 0.03 0.97 0.03 0.97

0 1 2

y y

y

P yy

20 19 18

0.97 20 0.03 0.97 190 0.0009 0.97

978991643

(ii) The probability of having at least one defective device in a shipment,

0

20

( 1) 1 ( 0)

201 0.03

0

1 0.97 0.456205657 0.4562,

P y P y

which can be considered as a result in a Bernoulli trial with 0.4562.p

Assuming this outcome is independent from a shipment to shipment and we de

denote x as the number of shipments containing at least one defective device.

Then x has another Binomial distribution, ( 10, 0.4562),x B n p and the

required probability is

210

0

0 10 1 9 2 8

( 3) 1 ( 2)

101 0.4562 0.5438

10 10 101 0.4562 0.5438 0.4562 0.5438 0.4562 0.5438

0 1 2

x x

x

P x P x

x

6

10 9 2 81 0.5438 10 0.4562 0.5438 45 0.4562 0.5438

1 0.09285365 0.90714635

5.2(c) An oil exploration firm is formed with enough capital to finance 5 explorations.

The equipment of particular exploration being successful is 0.15. Assume that

the explorations are independent and conform to the properties of Binomial

experiment.

(i) Find the expected number of unsuccessful explorations and its variance.

(ii) Suppose the firm has a fixed cost of $10,000 and that it costs $25,000 to

make a successful exploration. Find the expected total cost in the 5

explorations.

(iii) What is the probability that no or one successful exploration is made?

5.2(d) A multiple-choice test consists of 15 questions each with five possible answers

of which only one is correct. Suppose one of the students taking the test answers

the questions by guessing. What is the probability that he answers at most 3

questions correctly?

Solution:

(c) Let y be the number of successful explorations. Then given 5n and

probability of a successful exploration, 0.15,p we have

(i) The expected number of unsuccessful explorations, x is

( ) (1 )

5(0.85) 4.25 4

E x n p

and the variance,

( ) (1 )

5(0.85)(0.15) 0.6375

Var x n p p

(ii) The total cost for making y explorations is given by

10,000 25,000 ,C y (in dollars), where the expected cost is

( ) (10,000 25,000 )

10,000 25,000 ( )

10,000 25,000(5)(0.15)

$28,750

E C E y

E y

(iii) The probability that 0 or 1 successful exploration is made,

7

0 5 1 4

5 4

( 0 1) ( 0) ( 1)

5 50.15 0.85 0.15 0.85

0 1

0.85 5 0.15 0.85 0.8352

P x or x P x P x

= (0.85)5 + 5(0.15)(0.85)4

= 0.4437 + 0.3915 = 0.8352

(d) The number of correct answers,1

15,5

x B

. Then probability of answering at

most 3 questions,

153

0

0 15 1 14 2 13 3 12

15 14

15 1 4( 3)

5 5

15 15 15 151 4 1 4 1 4 1 4

0 1 2 35 5 5 5 5 5 5 5

4 1 4 115 105

5 5 5 5

x x

x

P xx

2 13 3 124 1 4

4555 5 5

0.03518 0.13194 0.23090 0.25014 0.64816

5.2(e) An early warning detection system for aircraft consists of five identical radar

units operating independently of one another. Suppose that each has a

probability of 0.95 of detecting an intruding aircraft. When an intruding aircraft

enters the scene, the interest is to note the number of radar units that do not

detect the aircraft. Is this a binomial experiment?

Solution:

To decide whether this is a binomial experiment, we must determine whether each of

the five requirements is met.

(i) The experiment consists of five trials where each trial determines whether a

particular radar unit detects the aircraft.

(ii) Each trial in one of two outcomes, the success being “not detecting aircraft” and

the failure, “detecting aircraft”

(iii) The probability of success (a radar unit not detecting aircraft) is 0.05.

(iv) The trials are independent because the radar units operate independently.

(v) The random variable, y is the number of successes in the operation of five radar

units. Hence 5, 0.05y B n p .

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1-5.3 The Multinomial Distribution

Suppose that an experiment consists of n independent trials such that each trial results

in any one of 2k possible outcomes. This can be seen in many real-life situations

like persons being interviewed in an opinion poll to indicate whether they are for,

against or undecided for a candidate, rating of manufactured product as excellent, very

good, good, average or inferior and the recording of accidents at an intersection

according to the day of the week. Such situations conform to the multinomial

experiment, a generation of the binomial experiment where 2k ). The multinomial

experiment has the following properties.

The experiment consists of n independent and identical trials.

A trial of an experiment results in any one of the k mutually exclusive possible

outcomes with respective probabilities 1 2 3, , . . ., kp p p p such that1

1k

i

i

p

The probability of an outcome i in a single trial, ( 1,2,3,. . ., )ip i k remains the

same from a trial to trial.

The random variables are 1 2 3, , . . ., ky y y y the number of successes in each class

of outcomes where 1 2 3

1

. . .k

i k

i

y y y y y n

Definition 5.3

The random variables, 1 2 3, , . . ., ky y y y with probabilities 1 2 3, , , . . ., kp p p p in a

multinomial experiment have a joint probability distribution known as multinomial

distribution given by, 31 2

1 2 3 1 2 3

1 2 3

!( , , , . . . ) . . . . . .

! ! ! . . . !ky yy y

k k

k

np y y y y p p p p

y y y y

Where 0, 1, 2, 3, . . .,iy n ,1

k

i

i

y n

, 1

1k

i

i

p

, ( )i iE y np ( )i iVar y np and

( , )i j i jCov y y n p p , where i j

Example 5.3

5.3(a) A certain city has three television (TV) stations, namely channels 12, 10 and 3.

The percentages of viewing audience at channels 12, 10 and 3 during prime

time on Saturday nights are 50, 30 and 20 respectively.

If a random sample of 8 TV viewers in the city is chosen on Saturday night,

then the probability that 5, 2 and 1 will be viewing channels 12, 10 and 3

respectively is

9

5 2 1

1 2 3

8!( 5, 2, 1) 0.5 0.3 0.2

5! 2!1!

0.0945

p y y y

5.3(b) Suppose that a die is rolled 9 times. The probability that one(1) appears 3 times,

two(2) and three(3) twice each, four(4) and five(5) once each and six(6) not at

all is given by

1 2 3 4 5 6

3 2 2 1 1 0

( 3, 2, 2, 1, 1, 0)

9! 1 1 1 1 1 1

3! 2! 2! 1!1! 0! 6 6 6 6 6 6

0.001500342

P y y y y y y

5.3(c) Items in a large lot under inspection are subject to two types of defects. It is

judged that 70% of the items are to be defect free whereas 20% have a type A

defect and 10% have a type B defect. If 6 of these items are randomly selected

from the lot find the probability that 3 have no defects, 1 has a type A defect and

2 have type B defect.

Solution

Assume the outcomes are independent from a selection of an item to item.

Letting 1 2,y y and 3y be number of defect free, type A and type B defects with

probabilities 0.70, 0.20, 0.10 respectively, then

3 1 2

1 2 3

6!( 3, 1, 2) 0.7 0.3 0.1

3!1! 2!

0.06174

P y y y

where the expected values and the variances are as follows:

1 1 1 1 1( ) 6 0.7 4.2; ( ) (1 ) 6(0.7)(0.3) 1.26E y n p Var y n p p

2 2 2 12 2( ) 6 0.1 0.6; ( ) (1 ) 6(0.1)(0.9) 0.54E y n p Var y n p p

3 3 3 3 3( ) 6 0.2 1.2; ( ) (1 ) 6(0.2)(0.8) 0.96E y n p Var y n p p

1-5.4 The Hypergeometric Distribution

Sampling from a finite population can be done in one of two ways, sampling with

replacement (where items are selected, examined and returned to the lot/population for

possible reselection) or sampling without replacement (where items are selected,

examined and kept, thus preventing its reselection in the subsequent draws). Sampling

with replacement guarantees that the draws are independent while in sampling without

replacement the draws not independent. If we are interested in computing probabilities

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for the number of observations or items that fall into a particular category, then

sampling of the observations/items would be done without replacement and this ceases

to be a binomial random variable which rather it follows a probability known as

Hypergeometric distribution. A hypergeometric random variable is often found in

many fields with heavy uses in acceptance sampling, electronic testing and quality

assurance where testing is done at the expense of the items being tested (i.e. the items

are destroyed and hence cannot be replaced in the sample).

In general, suppose a random sample of n items is drawn from a finite population of

N items consisting of k items of a particular type, say A (call them successes) and the

rest, ( )N k of particular type B (called failures). If the number if items of type A in

the sample of n items, x is observed then a hypergeometric experiment results which

possesses the following properties.

The experiment consists of randomly drawing a sample of n elements without

replacement from a lot of N elements.

k Of the N items are classified as successes and the rest, ( )N k are failures.

The random variable, x is the number of successes in the draw of n elements.

Definition 5.4:

The probability distribution of the hypergeometric random variable x, the number of

successes in a random sample of size n selected from N items of which k successes

and ( )N k failures called Hypergeometric distribution is

( ) , 0, 1, 2, . . .,min( , ),

k N k

x n xp x x k n

N

n

where ,N n and k are positive integers with mean and variance given as follows:

( ) ,nk

E xN

2 ( ) 1 11

N n nk k nk kVar x

N N N N N

The Hypergeometric random variable as already noted arises from a situation quite

similar to the Binomial random variable. The main distinction between the two is that

the trials of Hypergeometric are not independent (sampling without replacement) while

that of the Binomial are independent (sampling with replacement).

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Example 5.4

5.4(a) Suppose a radio contains six transistors, two of which are defective. Three

transistors are selected at random, removed from the radio and inspected.

(i) Determine the probability distribution of ,x the number of defectives

observed.

(ii) Find the mean and the variance of .x

(iii) Compute the likelihood of having at least one defective transistor in that

random sample of three transistors.

5.4(b) A town council consists of members six men and four women. A committee of

three is selected. Let x be the number of women selected on the committee.

(i) Compute the mean and standard deviation of x .

(ii) What is the probability of getting fewer than two women on the

committee?

Solution.

(a) (i) The probability distribution of x

2 4 2 4

3 3( ) , 0, 1, 2

6 20

3

x x x xp x x

2 4

0 3 4 1(0)

20 20 5p

2 4

1 2 12 3(1)

20 20 5p

2 4

2 1 4 1(2)

20 20 5p

(ii) )()(2

0

xpxxEx

or N

nk

= 5

10 + 5

31 + 5

12 or 6

)2(3

= 0 + 5

3 + 5

2 = 1

)(xVar =

11

N

nN

N

k

N

nk

=

16

36

6

21

6

)2(3

12

= 5

25

36

4

(iii) 3 1 4

( 1) (1) (2)5 5 5

P x p p

(b) Given that N = 10, n = 3 and k = 4

(i) )(xE = N

nk

= 10

)4(3 = 1.2

)(xVar =

11

N

nN

N

k

N

nk

= (1.2)

110

310

10

41

= (1.2)

9

7

10

6

= 0.26133 = (0.5112)2

Hence the standard deviation, = 0.5112

(ii) 1

0

4 6 10( 2)

3 3x

P xx x

= 120

2

6

1

4

120

3

6

0

4

= 120

60

120

20 =

3

2

120

80

1-5.5 The Negative Binomial and Geometric Distributions

The binomial random variable is a count of number of successes in a series of n

Bernoulli trials. The number of trials is fixed (predetermined) and the number of

successes varies randomly from experiment to experiment. The negative binomial

random variable is a count of the number of trials required to obtain k successes. The

number of successes here is fixed and the number of trials varies randomly. In this

sense, the negative binomial random variable is considered as the opposite or negative

of the binomial random variable. In particular, the negative binomial random variable

arises in situations characterized by these properties:

The experiment consists of a series of independent and identical Bernoulli trials,

each with probability of successes, p .

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The trials are observed until the thk success is obtained ( k is fixed by the

experimenter)

The random variable y is defined as the number of identical and independent

trials required to obtain k successes.

Typical of such situations are the number of job applicants interviewed until the

thk suitable applicant is found, the number of oil will needed to be drilled until the thk

successful oil well is hit, the number of shots fired before the first, second or third

target was hit, the number of pregnancies required before the fifth girl-child is born.

Definition 5.5

(i) The random variable y (the number of identical and independent trials

required to obtain k successes) has probability distribution called the negative

binomial distribution is defined by:

1

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

y kky

p y p p y k k kk

; 0 1p

where mean and variance given as follows:

( )k

E yp

and 2

(1 )( )

k pVar y

p

(ii) A special type of Negative Binomial distribution is Geometric distribution

where the random variable y is defined as the number of identical and

independent trials the experiment is performed until the first success occurs

( 1).k

The probability distribution is defined by

1

( ) 1 , 1, 2, 3, . . .y

p y p p y

; 0 1p

where mean and variance become,

1( )E y

p and

2

1( )

pVar y

p

Example 5.5

5.5. (a) A geological study indicates that an exploratory oil well drilled in a certain part

of a state strikes oil with probability of 0.20.

(i) Find the probability that the first strike of oil comes on the third well

drilled.

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(ii) Find the probability that the third strike of oil comes on the sixth well

drilled.

(iii) Find the mean and variance of the member of wells that must be drilled

if the company wants to set up three producing wells.

5.5(b) Ten percent of engines manufactured on an assembly line are defective.

If engines are randomly and independently selected one at a time and tested,

what is the probability that

(i) The first non-defective engine is found on the third trial;

(ii) The third non-defective engine is found on or before the fifth trial.

Solution

(a) The probability of striking an oil well, 0.2p

(i) The probability of striking first oil on third well drilled,

1( 3) (1 )yp y p p

3 10.2(0.8) 20.2(0.8) = 0.128

(ii) The probability of striking third oil on the sixth well drilled

3 3

3 3

1( 6) 1 , 3

1

6 10.2 0.8

3 1

10 0.2 0.8 0.04096

y kky

P y p p where kk

(b) (i) The use of Geometric distribution,

3 1( 3) 0.9(0.1) 0.009p x

(ii) The use of Negative Binomial distribution,

3 36 1

( 6) . 0.9 0.1 0.007293 1

P x

1-5.6 The Poisson Distribution

1-5.6.1 Introduction

The Poisson probability distribution provides a good model for a discrete random

variable which results from an experiment called Poisson Process. The Poisson Process

is characterised by the following assumptions or properties.

The experiment consists of counting the number of times a particular event

occurs during a given unit of time, area or volume.

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The occurrence or non-occurrence of the event in any interval of time space or

volume is random and independent of the occurrence or non-occurrence of the

event in any other interval.

An infinite number of occurrences of event must be possible in the interval.

Also in any infinitesimally small portion of the interval the probability of more

than one occurrence of the event is negligible.

The probability of the occurrence of an event in a given interval is proportional

to the length of the interval.

The mean number of events in an interval, denoted or , is equal to its

variance.

Typical examples of some experiments which may result a random variable that can be

modelled by the Poisson process are the number of industrial accidents during a given

period of time, number of flaws or defects on a square metre piece of material, number

of radioactive or particles that decayed or emitted in a particular period of time, number

of errors a typist makes in typing a page of a text, number of insurance claims received

by an insurance company during a period of time, number of repairs on a kilometre of a

road and number of telephone calls received by a switchboard in a unit time interval.

In each of the above examples the random variable represents the number of events

occurring in a unit period of time or space during which the mean number of events,

or is expected to occur.

Definition 5.6:

The Poisson distribution for the random variable, x, representing the number of

occurrence of an event in a given interval of time, space or volume is defined by

,0,...,2,1,0,!

)(

andxx

exp

x

where the mean and variance are the same. That is,

( ) ( )E x Var x

The distribution of x may simply be denoted as ( )x P .

16

1-5.6.2 Poisson Approximation to Binomial

When the number of trials, n in a Binomial process is large, the computations of the

binomial probabilities may be too tedious. The Poisson distribution can be used, as an

alternative, to approximate the Binomial distribution. This is based on the convergence

of the Binomial distribution as n becomes large ( )n as illustrated by the

following theorem.

Theorem 5.1:

Let be a fixed number and n an arbitrary positive integer. For each non-negative

integer x, lim (1 ) ,!

xx n x

n

n ep p where p

x x n

,

Proof: of Theorem

lim (1 )

lim 1

!lim 1 1

! ! 1

!lim 1

!! 1

x

x

n x n xp pxn

x n xn

x n nn

x x nn

x n n nn xn x x nn

nn

xx nnn x n

n

1 2 . . . 1 !lim 1

! !

x

x

nn n n n x n x

x nn x nn

1 2 . . . 1 !lim 1

!

x

x

nn n n n x

x nn n

1 2 . . . 1lim . 1

! . . .

x nn n n n x

x n n n nn

!

x

x

e

1 2 . . . 1sin 1

. . .1

nn n n n xce as n

n n n nand e

17

The tables below give an indication of the rate at which the Binomial distribution,

xnx ppx

npnB

1),( converges to the Poisson distribution, ,

!

)()(

x

enpP

npx

where = np = 1 in both cases.

(i) Table 4.1: 5, 0.2 5(0.2) 1n p and so np

x Binomial, 55

( ) 0.2 0.8x x

p xx

Poisson,11

( )!

x ep x

x

0

1

2

3

4

5

6+

0.328

0.410

0.205

0.051

0.006

0.000

0.000

0.368

0.368

0.184

0.61

0.015

0.003

0.001

(ii) Table 4.2: 100, 0.01 100(0.01) 1n p and so np

x Binomial, 100100

( ) 0.01 0.99x x

p xx

Poisson, 11

( )!

x ep x

x

0

1

2

3

4

5

6

7

8

9

0.366032

0.369730

0.184865

0.060999

0.014942

0.002898

0.000463

0.000063

0.000007

0.000001

0.367879

0.367879

0.183940

0.061313

0.015328

0.003066

0.000511

0.000073

0.000009

0.000001

We note from Table 4.1 (where n = 5) that for some x the agreement between the

Binomial probability and the Poisson approximation is not very good. If n is large as

100, (as indicated in Table 4.2), the agreement is remarkable very good for all .x

18

Example 5.6:

5.6(a) The number of telephone calls coming unto a switchboard of a building

averages 4 per minute. Find the probability that

(i) No call will arrive in a given one-minute period.

(ii) At least two calls will arrive in a given one-minute period.

(iii) At least three calls will arrive in a given two-minute period.

5.6(b) A book has 300 pages and the probability of finding misprints, x, in a page is

0.015. Find the probability of detecting misprints in at most one page of the

book using the Binomial distribution and Poisson Approximation to the

Binomial.

Solution:

(a) The random variable, x is the number of telephone calls received with an

average of = 4 per minute. Hence x will have the Poisson distribution,

...,2,1,0,!

4)(

44

xx

exp .

The required probabilities are;

(i) !0

4)0()0(

40

e

pxP

= e-4 = 0.018316

(ii) )1(1)2( xPxP

= )1()0(1 xPxP

=

!1

4

!0

41

4140 ee

= 1 – (e-4 + 4e-4) = 1 – 5e-4

0.908422

(iii) 824 Calls received in two minutes.

)2(1)3( xPxP

= )2()1()0(1 xPxPxP

=

!2

8

!1

8

!0

81

828180 eee

= 1 – (e-8 + 8e-8 + 32e-8)

= 1 – 41e-8 0.986246

19

(b) (i) By the Binomial distribution, n = 300 and p = 0.015. The required

probability,

)1()0()1( xPxPxP

= 0 300 1 299300 300

0.015 0.985 0.015 0.9850 1

= (0.985)300 + 300 (0.015) (0.985)299

= 0.010737 + 0.049051 0.059788

(ii) Using the Poisson Approximation, = np =300(0.015) = 4.5

)1()0()1( xPxPxP

= !1

5.4

!0

5.4 5.45.40

ee

= e-4.5 + 4.5 e-4.5

= 5.5 e-4.5 0.061099

5.6. (c) The probability that a car will breakdown after falling into a pot–hole on a road

is 0.00015. If 20,000 cars travel along the road, find the expected number of

break-downs and probability that at least one car will break down.

5.6(d) The number of serious accidents, y in a manufacturing plant has approximately

a Poisson distribution with a mean of 1.5 accidents per month. What is the

probability that:

(i) More than three accidents will occur within a period of one month?

(ii) Fever than three accidents will occur within a period six weeks?

Solution:

(c) (i) The expected number of break-downs,

= np = 20,000 (0.00015) = 3

(ii) The probability that at least one car will break down, using the Poisson

Approximation to Binomial,

0 3

3

1 1 0

31

0!

1 0.950213

P x P x

e

e

(d) Given that y approximately have the Poisson distribution,

...,2,1,0,!

5.1)(

5.1

yy

eyp

y

, then

20

(i) 1.52

0

1.53 1 2 1

!

y

y

eP y P y

y

= 1 1.5 2 1.50 1.5 1.5 1.51.5

10! 1! 2!

e ee

= 1 – (e-1.5 + 1.5 e-1.5 + 1.125 e-1.5)

= 1 – 3.25 e-1.5 0.191153

(ii) 25.22

35.1 accidents in six weeks,

2.252

0

2.253

!

y

y

eP y

y

= !2

25.2

!1

25.2

!0

25.2 25.2225.2125.20

eee

= e-2.25 + 2.25 e-2.25 + 2.53125 e-2.25

= 5.78125 e-2.25 0.609339

Exercise 5.1:

1. (a) The production of fuses are packaged in boxes after manufacturing. The

probability of a defective fuse in a box of 200 fuses from the manufacturing

process is 2 percent which is independent of the other fuses.

(i) Estimate the expected number of fuses and its standard deviation?

(ii) What is the approximate probability that there will be fewer than two

defective fuses in the box?

(b) It is conjectured that an impurity exists in 30% of all drinking wells in a certain

rural community. It is too expensive to test all the many wells in the area and so

a random sample of 10 wells was selected and subjected to testing. Is it likely to

find 7 or more impure wells for drinking?

(c) In a certain communication system, there is an average of 1 transmission error

per 10 seconds. What is the probability of observing a least 2 errors within a

duration of one-half minute?

(d) (i) Find the moment generating function for the Poisson random variable

x and use it to find its mean and variance.

(ii) It is established that the claim size for a certain class of accidents is a

random variable, y with moment generating function,

4

( ) 1 2500 .yM t t

Find the mean claim size and its variance.

21

(e) A sales person has found that the probability of a sale on a single contact is

approximately 0.03. If the salesperson contacts 100 prospects, what is the

approximate probability of making at least one sale?

2. (a) A random variable y is described by the Geometric distribution. State the

characteristics of y and drive its probability distribution. Also show that the

distribution of y is probability mass function. Hence derive the mean and

variance of the distribution of .y

(b) The employees of a firm that manufactures insulation are being tested for

indications of asbestos in their lungs. The firm is requested to send three

employees who have positive indication of asbestos to a medical centre for

further testing. If 40% of the employees have positive indication of asbestos in

their lungs

(i) Find the probability that ten employees must be tested in order to find

three positives.

(ii) Find the expected value and variance of the total cost of conducting the

tests necessary to locate the three positives if each test cost $20.

(c) From a large lot of new tyres n are randomly sampled by a potential customer

and the number of defectives x is observed. If at least one defective tyre is

observed in the sample of ,n the entire lot is rejected by the customer.

If 10% of the tyres are defective, find n such that the probability of rejecting

the lot is approximately 0.9. Find also the probability that at least three

defective tyres would be observed.

3. (a) In a particular departmental store customers arrive at a check-out counter

according to the Poisson distribution at an average of five per hour.

(i) During a particular one-hour period, determine the probability that at

least three customers will arrive.

(ii) What is the probability that in a given two-hour period, exactly two

customers will arrive?

(iii) In a thirty-minute period, determine the probability that at most two

customers will arrive.

(b) Many utility companies have begun to promote energy conservation by offering

discount rates to customers who keep their energy usage below certain

established subsidy standards. A recent EPA report notes that 70% of residents

22

in a town have reduced their electricity consumption sufficiently to qualify for

discount rates. Suppose 20 residential subscribers are randomly selected from

the town.

(i) Explain how this situation can be modelled by the binomial distribution.

(ii) Find the expected number of residents who qualify for the subsidy.

(iii) What is the probability that at least four qualify for the favourable rates?

(c) A warehouse contains10 printing machines 4 of which are defective. A

company randomly selects 5 of the machines for purchase. What is the

probability distribution of selecting number of defective machines? Use this

result to find the probability that at least 4 of the machines are non-defective.

4. (a) (i) State the properties of the Binomial Experiment.

(ii) Derive the mean and variance of the Binomial distribution.

(iii) Prove that the moment generating function of a binomial random

variable is ,ntp qe where 1q p

(b) The number of surface flaws in plastic panels used in the interior of automobiles

has a Poisson distribution with a mean of 0.05 per square foot of plastic panel.

Assume an automobile interior contains 10 square feet plastic panel.

(i) What is the probability that there are at least two surface flaws in an

auto’s interior?

(ii) If 10 cars are sold to a rental company, what is the probability that at

most one car has any surface flaws?

(c) A missile protection system consists of 10 radar units operating independently.

Suppose each has a probability of 0.70 of detecting a missile entering a zone

that is covered by all units.

(i) Describe how the operation of the systems fits into the binomial process.

(ii) How many radar units would be required if the probability of detecting a

missile by at least one unit is 0.998?

5. (a) A random sample of 6IC’s is taken from a large consignment and tested in two

independent stages. The probability that an IC will pass either stage is .p All

the 6IC’s are tested at the first stage. If 5 or more pass the test, those which

pass are tested at the second stage. The consignment is accepted if there is at

most one failure at each stage.

23

(i) What is the probability that the second stage of the test will not be

required?

(ii) Find the number of IC’s expected to undergo the second stage.

(b) A nuclear plant releases a detectable amount of radioactive gases twice a month

on the average. The number of emission of gases x has the probability

model, )(xf .

Identify )(xf and find the probability, ( 4)P x within the period of a month.

SESSION 2-5: CONTINUOUS PROBABILITY DISTRIBUTIONS

2-5.1 The Uniform Distribution

The uniform distribution provides a simple probability model to describe a continuous

random variable that can randomly assume any value between two points a and b (a <

b) on a line. The uniform probability density function has a rectangular shape over the

interval [ , ]a b with height ab

1 where the area under the density function is equal to

1.

)(xf

ab

1

a b x

Figure 5.4: Graph of Uniform Distribution

Thus, the uniform distribution provides a good model for a continuous random variable

whose values are uniformly distributed over an interval. For example:

If buses arrive at a given bus stop over 15 minutes and you arrive at the bus stop

at a random time, the time you must wait for the next bus to arrive could be

described by the uniform distribution over the interval from 0 to 15 or [0, 15] .

In a milling operation pieces of lumber less than 1 metre in length are

considered scrap. The distribution of the lengths of scrap lumber would have a

uniform distribution on the interval from 0 to 1 or[0, 1] .

24

Definition 5.7:

The random variable x uniformly distributed over the interval [ , ]a b has it probability

density function defined as:

elsewhere

bxaforabxf

,0

,1

)(

where the mean and variance are:

( )2

a bE x

and

2( )( )

12

b aVar x

Example 5.7:

5.7(a) Suppose that buses arrive at a bus stop every 15 minutes and that the waiting

time for the next bus to arrive has a uniform probability distribution on the

interval from 0 to 15 minutes.

(i) Compute the mean and standard of a person’s waiting time, x .

(ii) What the probability x will exceed 10 minutes?

5.7(b) The probability density function of time, x required to complete an assembly

operation is uniformly distributed for 30 x 40 seconds.

(i) Determine the proportion of assemblies that require less than 35 seconds

to complete.

(ii) What time is exceeded by 90% of the assemblies?

Solution:

(a) The probability distribution for the waiting time, x

1, 0 15

( ) 15

0 ,

xf x

elsewhere

(i) The mean and standard deviation of x are:

)(xE ) = 2

ba =

2

150 = 7.5 minutes

)(xVar = 12

)( 2ab = 2 =

12

)015( 2 = 18.75 = (4.33)2

Hence, = 4.33

(ii) The required probability,

)10( xP = 15

1015

1dx =

15

1015

x =

15

1(15 – 10) =

3

1

15

5

25

(b) The probability density function for the time required to complete the assembly

operation, x,

1 1, 30 40 0 10

( ) 40 30 10

0 ,

x or xf x

elsewhere

(i) ( 35) ( 5)P x P x = 5

0 10

1 dx =

5

010

x =

21

10

5

(ii) Let the required time be k seconds

( ) 0.90P x k = 0.90

10 1

10kdx = 0.90

10

10 k

x = 0.90 =

1010

10 k

10 – k = 9

k = 10 – 9 = 1

2-5.2 The Exponential Distribution

The length of time or space within which we have occurrence of an event in a Poisson

process results in a continuous random variable which is often of great interest. The

probability distribution of such a random variable is called Exponential distribution.

Typical situations resulting in this random variable are the length of time until a

machine or a component of it fails, length of time between arrivals at a car wash,

waiting time for a service line or in a queue, length between flaws on a price of

material, length of time between successive reports of an infectious disease spreading

in a random manner, length of time between successive filing of claims in an insurance

office and distance between major defects in a highway.

From the above examples, the Exponential distribution models situations in which the

random variable represents waiting time or measurements of length of time between

successive occurrences of an event. It plays important role in reliability theory where

we try to find the reliability of a system at time t.

Definition 5,8:

The Exponential probability distribution for the continuous random variable, which

represents length of time or space, is defined as:

26

, 0

( )0 ,

xe x

f xelsewhere

where the parameter > 0 is the mean number of events that occur in the given unit

of time or space. The mean value of x (i.e. the mean length of time or space

between successive occurrences of the event and the variance are:

)(xE =

1 )(xVar =

2

1

The shape of the distribution is determined by its simple parameter

( )f x ) ( )f x

= 2

= 10

= 0.5

A

0 1 2 3 4 x 0 a x

Figure 5.5: Exponential Curves

The area under the exponential curve, A gives the probability of the random variable,

x. That is,

0( ) xA P x a dxe

= x a

ae e

Example 5.8:

5.8(a) (i) Suppose the time in days between service calls on a photocopier

machine follows an exponential distribution with a mean call of 0.02 per day.

(ii) What is the probability that the time until the machine again requires

service exceeds 60 days?

(iii) What is the probability that the time until the machine again requires

service is less than 20 days?

(iv) Find the probability that the time until the machine again requires

service is longer than, 2 .

27

5.8(b) A new automated production process breaks down at an average of 2 per day.

(i) What is the mean time between breakdowns assuming 8 hours of

operations per day?

(ii) What is the probability that the process will run 3 hours or more before

another breakdown?

Solution:

(a) Let x represent the time (in days) between service calls. Then probability

distribution,

elsewhere

xexf

x

,0

0,)(

where = 0.02

(i) P(x > 60) = e-0.02(60)

= e-1.2 = 0.301194

(ii) P(x < 20) = 1 – P(x > 20)

= 1 – e-0.02(20)

= 1 – e-0.4 = 0.329680

(iii) The mean and standard deviation which are both equal is,

=

1

02.0

1 = 50 =

+ 2 = 50 + 2(50) = 150

2 150x P xP

= 1500.02 3 0.049787 0.05e e

Thus there is about 5% chance that the machine will not require servicing

during the next 150 days or approximately five months.

(b) (i) The expected time between breakdowns is given by,

1( )

2E x = 0.5

(ii) The probability distribution of x is

22 , 0( )

0 ,

x xf x

elsewhere

e

323xP e

) 6 0.002479e

5.8(c) Suppose the demand for domestic gas has approximately exponential

distribution with mean 100 cylinders per day.

28

(i) Find the probability that the demand will exceed zoo such cylinders on a

randomly selected day.

(ii) What supply capacity should the gas filling station maintain during the

day so that demand will not exceed its capacity with probability of only

0.01?

2-5.3 The Normal Distribution

2-5.3.1 Introduction

The Normal distribution is one of the most widely used probability distribution for

modelling random experiments. It provides a good model for continuous random

variables involving measurements such as time, heights/weights of persons, marks

scored in an examination, amount of rainfall, growth rate and many other scientific

measurements.

Definition 5.9:

The probability density function for the normal random variable, x which is simply

called normal distribution is defined by

)(xf =

2

2

1

21,

2

0 ,

x

x

elsewhere

e

where 0, and the mean and variance of the measurements, x are

)(xE and 2)( xVar

If a random variable is modelled by the Normal distribution with mean, and variance

2 , then it is simply denoted as 2( , )x N . The graph of the Normal probability

distribution is bell-shaped smooth curve, as illustrated in the diagram below.

( )f x

0 -3 +2 - + +2 +3 x

x

Figure 5.5: Normal Curve

29

2-5.3.2 Properties of Normal Curve:

The normal curve has the following desirable properties, accounting for the wide-

spread applications of the Normal distribution.

The normal curve is symmetrical about its mean, .

The mean, median and mode of x are the same.

The total area under the normal curve is equal to 1.

The probability distribution of the normal random variable, x is completely

determined by its two parameters and .

The curve is asymptotic to its horizontal axis

The probabilities of the normal random variable, x are given by the areas under

the curve. As an example, the probabilities that x lies within 1, 2 and 3

standard deviation(s) about the mean are approximately given as follows:

0.6826P x

2 2 0.9544P x

3 3 0.9980P x

2-5.3.3 Computation of Probability of Normal Random Variable:

To compute the probability that x lies within the interval [a, b], P a x b the

normal random variable, x is standardized using the transformation, Z =

x, called

the Z-score. The probability distribution function of the standardized random

variable, )(zf which has the same shape as )(xf is called the Standard (or Unit)

Normal distribution. It is defined as

212

1( ) ,

2

Zf z ze

, or (0, 1)NZ , where the mean and

variance are 0 and 1 respectively. Thus,

( )b

aP a x b f x dx

= 21

22

2

( )1

2

b

a

x

dxe

= ( )b

a

z

af z dz

= 21

21

2

b

a

z

z

Ze dz

30

= ( )a bP Z Z Z

= ( ) ( )abP Z Z P Z Z = ( ) ( )b az z

f z dz f z dz

= abz z =b

a

The evaluation of the integral of )(zf , (z) for various values of Z have been

tabulated in a table called the Normal table. In evaluating (z) the following are

noted.

(i) ( ) 1 ( ) 1 ( )k k P z k

(ii) ( ) 1 ( ) 1 ( )P z k P z k k

(iii) ( ) ( ) ( ) ( ) 1 ( ) 2 ( ) 1P k z k k k k k k

Example 5.9:

5.9(a) Find the following probabilities using the Normal table.

(i) 1.95P z

(ii) 1.18 0.48P z

(iii) 0 2.58P z

(iv) 2.63P z

(v) 2.35 2.35P z

5.9(b) Suppose that .6, 4y N what percentage will y fall between 5 and 10?

Solution:

(a) Using the Normal table, the required probabilities are as follows:

(i) 1.951.95P z

= 0.0256

(ii) 1.18 0.48P z

= 0.48 1.18P Pz z

= 0.48 1.18

= 0.6844 – 0.1190 = 0.4654

(iii) 0 2.58P z )

= 2.58 0.00P Pz z

31

= 2.58 0.00

= 0.9951 – 0.5000 = 0.4951

(iv) ( 1 (2.63) 2.63)P Pz z

= 1 2.63

= 1 – 0.9957 = 0.0043

(v) 2.35 2.35P z

= 2.35 2.35P Pz z

= 2.35 2.35 or 2 2.35 1

= 0.9906 – 0.0094 = 2(0.9906) – 1 = 0.9812

(b) Given that ,6, 4y N where 6 and 2

5 10P y

= 10 5P Py y

= 10 6 5 6

2 2

= 2 0.5

= 0.9772 – 0.3085

= 0.6687 or 66.87%

5.9(c) The weekly amount spent for maintenance and repairs in a certain company was

observed, over a long period of time, to be approximately normally distributed

with a mean of $400 and a standard deviation of $20.

(i) If $450 is budgeted for the week, what is the probability that the actual

costs will exceed the budgeted amount?

(ii) How much should be budgeted for weekly repairs and maintenance in

order for the budgeted amount is exceeded with a probability of 0.1?

5.9(d) The nicotine content of a brand of cigarettes is normally distributed with a mean

of 2.0mg and a standard deviation of 0.25mg. What is the probability that a

cigarette will have nicotine content?

(i) Of 1.65mg or less

(ii) Between 1.50 and 2.25mg

(iii) Of 2.18mg or more?

32

Solution:

(c) The amount spent on maintenance and repairs, 2 ,400, 20x N where

$400 and $20 ,

(i) ( 450) 1 ( 450)P x P x

450 4001

20

1 0.9938 0.0062

(ii) Let the budgeted amount be k dollars.

0.1P x k

= 1 0.1P x k

0.9P x k

4000.9

20

k

14000.9 1.28

20

k =-1 )9.0(

400 20 1.28 425.6k (or $425.60k )

(d) Given that the nicotine content of cigarettes, 2 ,2.0, 0.25y N

(i) 1.65 2.0

1.650.25

P y

= 1.40 00808

(ii) 1.50 2.25P y

= 2.25 2.0 1.50 2.0

0.25 0.25

= 1.04 1.92

= 0.8508 – 0.0274 = 0.8234

(iii) 2.18 1 2.18P y P y

= 2.18 2.0

10.25

= 1 0.72

= 1 – 0.7642 = 0.2358

33

2-5.3.4 The Normal Approximation to Binomial

The Normal distribution provides a good approximation to the binomial

distribution when the number of trials, n is large, probability of a success in a

trial, p not close to 0 or 1 and both np and )1( pnp are greater than 5. Thus

the binomial random variable, x becomes approximately normal random

variable with mean, np and variance, 2 (1 ).np p The theoretical

justification for this approximation is based on the Central limit Theorem which

is widely applied in inferential analysis of data. The approximation is said to be

adequate when the interval 2 falls between 0 and n and also said to be

very good if the interval 3 falls between 0 and n.

To improve upon the approximation, a continuity correction may be utilized by

adding or subtracting 0.5 to/from x to account for the fact that a discrete

distribution is being approximate by a continuous distribution. In this case the

standardized random variable thus becomes 0.5

(1 )

x npZ

np p

Example 5.10:

5.10(a) Suppose that x has a Binomial distribution with 200n and 0.4p . Using

the continuity correction use the Normal approximation to Binomial to find

each of the following probabilities

(i) ( 90)P x (ii) ( 95)P x

(iii) ( 65)P x (iv) ( 60)P x

(v) (70 100)P x

5.10(b) A manufacturer of components for electric motors has found that about 10% of

the production will not meet customer specifications. If 500 components are

examined,

(i) Find the expected number of components which did not meet customer

specifications.

(ii) Find the probability that exactly 52 components or more did not meet

customer specifications.

(iii) Find the probability that between 36 and 58 (inclusive) components did

not meet customer specifications.

34

Solution:

(a) Given that the binomial random variable, x is approximately normally

distributed, 2, ,x N where 200(0.4) 80 and

200(04)(06) 6.9282

(i) ( 90) (89.5 90.5)P x P x

= 90.5 80 89.5 80

6.9382 6.9382

= 2.81 1.37

= 0.9357 – 0.9147 = 0.0210

(ii) ( 95) ( 95.5)P x x

= 95.5 80

6.9282

= 2.24 = 0.9875

(iii) ( 65) 1 ( 65.5)P x P x

= 65.5 80

16.9282

= 1 2.09 = 1 – 0.0183 = 0.9817

(iv) ( 60) ( 59.5)P x P x

= 59.5 80

6.9282

= 2.96 = 0.0015

(v) (70 100) (70.5 99.5)P x x

= 99.5 80 70.5 80

6.9282 6.9282

= 2.81 1.37

= 0.9975 – 0.0853 = 0.9122

35

(b) The number of components that did not meet customer specifications, x is

approximately normally distributed where 500(0.1) 50np and

2 2(1 ) 500(0.1)(0.9) (6.71)np p

(i) The expected number of components that did not meet customer

specifications, ( ) 50E x np

(ii) ( 52) 1 ( 51.5)P x P x

= 51.5 50

16.71

= 1 0.22

= 1 – 0.5871 = 0.4129

(iii) (36 58) (35.5 58.5)P x P x

= 58.5 50 35.5 50

6.71 6.71

= 1.27 2.16

= 0.8980 – 0.0197 = 0.8783

2-5.3.5 Linear Combinations of Normal Random Variables:

Theorem: Let 1 2 3, , , . . ., nx x x x be independent normal random variables and

2 2 3, , , . . . , na a a a be constants. Then the linear combination given below:

1 1 2 2 3 3

1

. . .n

n n i i

i

y a x a x a x a x a x

is also a normal random variable with mean and variance given as follows.

1 1

( ) ( ),n n

i i i i

i i

E y E a x a E x and

2

1 1

( ) ( )n n

i i i i

i i

Var y Var a x a Var x

However, if the random variables, 1 2 3, , , . . ., nx x x x are not independent then

2

1

( ) ( ) 2 ( , )n

i i i j i j

i i j

Var y a Var x a a Cov x x

where the double sum is over all pairs ( , )i j with i j

36

The Sampling Distribution of the Sample Mean, ( ) :x Let 1 2 3, , , . . ., nx x x x be random

observations drawn from a normally distributed population with a mean of and a

variance of 2 . Then the sample mean

n

i

ixn

x1

1 is normally distribution with mean

and variance given as follows:

1

1 2 3

1( ) ( )

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

1( . . . )

1( )

n

i

i

n

E x E xn

E x E x E x E xn

n

nn

1

1 22

1( )

1( ) ( ) . . . ( )

n

i

i

n

Var x Var xn

Var x Var x Var xn

2 2 2

2

22

2

1. . .

1( )

n

nnn

The sampling distribution of the sample mean is approximately normally

distributed, by the Central Limit Theorem.

Central Limit Theorem: Let a random sample of size n observations be selected from

a

population with mean and variance,2 . The sampling distribution of the sample

mean )(x will be approximately normally distributed with mean, x and variance,

22

x n provided n is sufficiently large.

Example 5.11:

5.11(a) If x and y are independent normal random variables with ( ) 1,E x

( ) 4, ( ) 10Var x E y and ( ) 9,Var y determine the following:

(i) (2 3 )E x y and (2 3 )Var x y

(ii) (2 3 40)P x y

37

5.11(b) The mass of a biscuit is a normal random variable )(x with mean 50 grams

and a standard deviation of 4 grams. If a packet contains 20 biscuits and the

mass of the packaging material is also normal random variable with mean 100

grams and standard deviation, 3 grams, find the probability that the mass of the

total packet

(i) Will exceed 1,047 grams

(ii) Lies between 1,050 and 1,200 grams

Solution:

(a) Given that x N (1, 4) and y N (10, 9), we let 2 3 ,T x y then

(i) ( ) (2 3 )T E T E x y

2 ( ) 3 ( )

2(1) 3(10)

32

E x E y

2

2

( ) (2 3 )

4 ( ) 9 ( )

4(4) 9(9) 97 (9.85)

T Var T Var x y

Var x Var y

(ii) From (i) T N(32, 97)

40 32

( 40)9.85

P T

= (0.81) = 0.7910

(b) Let the mass of contents of the packet be T1 and the packaging material T2.

Then the total mass T = T1 + T2 is normally distributed, where the mean and

variance are given as:

1 2

1 2

2

1 2

1 2

2

1 2

2

2

( ) ( )

( ) ( )

(20 ) (100)

20 ( ) 10

20(50) 100

1,100

( ) ( )

(20 ) ( )

20 ( ) ( )

20 (16) 9

6, 409 (80.06)

T

T

E T E T T

E T E T

E x E

E x

grams

Var T Var T T

Var T Var T

Var T Var T

38

Hence, T N (1,100 + 6,409)

(i) ( 1,074) 1 ( 1,074)P T P T

1,074 1,1001

80.06

1 ( 0.32)

1 0.3745

0.6255

(ii) (1,050 1,200)P T

1,200 1,100 1.050 1,100

80.06 80.06

(1.25) ( 0.62)

0.8944 0.2676

0.6268

Exercise 5.2:

1(a) It is known that the weights of certain group of individuals are approximately

normally distributed with a mean of 140 pounds and a standard deviation of 25

pounds.

(i) What is the probability that a person picked at random from this group

will weigh between 100 and 170 pounds?

(ii) If the group is made up of 9,000 people, how many of them would you

expect to weigh more than 200 pounds?

(b) An experiment was conducted to test for the presence or absence of fungus on

tobacco plants. 400 plants were observed to have been inflected by fungus.

(i) Does this appear to meet the requirements of a binomial experiment?

(ii) Previous experience suggests that the fungus affect 60% of the planting

of tobacco seedlings. Is it probable that the observed number of infected

plants could be larger than 245? Explain.

(iii) Suppose the characteristics of a binomial experiment are satisfied, what

interpretation can you give to ?60.0p

2. (a) The prices of estate houses are assumed to be normally distributed with a mean

of ¢18 million. It is known that 90 percent of the houses are priced below ¢30

million.

39

(i) Find the standard deviation of the prices of the houses.

(ii) What percentage of houses will cost more than ¢20 million?

(iii) If the estate is made up of 2,500 houses how many would you expect to

be priced less than ¢10 million?

(b) It is believed that 45% of a large population of registered voters favour a

particular candidate for the constituency. A public opinion poll used a random

selected sample of voters and asked each person polled to indicate his/her

preference for the candidate. What is the probability that a weekly poll based on

150 responses of registered voters will show at least 65% of the voters

favouring the candidate?

3. (a) (i) State the normal distribution and its properties

(ii) Under what conditions and how would you use the Normal

Approximation to the Binomial?

(b) The specification for the length of engine parts of equipment is a minimum of

99 mm and a maximum of 104 mm. A batch of parts is produced that is

normally distributed with a mean of 102 and a standard deviation of 2 mm.

Parts cost ¢9,000 to make. Those that are too short have to be scrapped and

those too long are shortened at a further cost of ¢8,000.

(i) What percentage of the parts is undersize or oversize?

(ii) What is the expected cost of producing 1,500 usable parts?

(iii) Calculate and explain the implications of changing the production

method so that the mean is half way between the upper and lower

specification limits. Assume the same standard deviation.

4. (a) The reaction time of a driver to visual stimulus is normally distributed with a

mean of 0.35 second and a standard deviation of 0.15.

(i) What is the probability that a reaction time requires more than 0.5

second?

(ii) What is the probability that a reaction time requires between 0.28 and 62

second?

(iii) What reaction time is required to exceed 90 percent of the time?

(b) A machine fills millet flour into 500-gram bags. The actual weights of the filled

bags vary being approximately normally distributed with a variance of 100

grams.

40

(i) Find the mean weight of the filled bags, if 15% of the filled bags are

underweight.

(ii) Calculate the proportion of the bags whose weight is between 495 and

530 grams.

(iii) If the mean weight is adjusted to 518.8 grams and the standard deviation

remains unchanged, what percentage of bags would be sold

underweight?

5. (a) The manufacturing of semi-conductor chips produces 2% defective chips.

Assume that the chips are independent and that a lot contains 1000 chips. Use

the continuity correction to approximate the probability that

(i) Exactly 20 chips are detective, and

(ii) Between 20 and 30 chips in the lot are defective.

(iii) Determine the number of defective chips, x such that the probability of

obtaining x defective chips is greatest.

(b) Limit gauges are used to reject all components in which a certain dimension is

greater than 60.4 mm or less than 59.6 mm. It is found that about 5% are

rejected under size and 5% rejected oversize. Assuming the dimensions are

normally distributed,

(i) Find the mean, µ and the standard deviation, , and

(ii) Estimate the percentage of the rejects that world is between 59.7 and

60.3 mm.

(c) A chartered airliner agency is asked to carry regular loads of 100 cartons of an

item. The plane available for this work has a carrying capacity of 5,000 kg . If

is known that the weight of a carton of the item is normally distributed with a

mean of 40 kg and standard and 9 kg . Can the agency take the order?

6. (a) The systolic blood pressure of a group of female industrial workers are

normally distributed with a mean of 130 mmHg and a standard deviation of 15

mmHg .

(i) If 500 female workers are randomly selected, how many would you

expect to have systolic blood pressure )( of at least 100 mmHg and )(

between 110 and 150 mmHg ?

(ii) If 68% of the workers have a systolic blood pressure of at most

,mmHgk find the value of .k

41

(b) The manager of a firm takes an average of 25 minutes to travel to this office

while his secretary takes an average time of 24 minutes. The travelling time of

both are normally distributed with equal standard deviation of 3 minutes.

(i) If the manager leaves home 2 minutes before the secretary, what is the

probability that he arrives first?

(ii) If the secretary knows that her boss leaves home promptly at 7.30 a.m.

everyday, what time should she leave if she wants to be 97.5% sure that

she would arrive first

(b) In a collection of plants, it is found that 20% have heights greater than 36.3 cm

and 67% have heights greater than 29.9cm. Suppose the heights are normally

distributed in this collection.

(i) Find the mean and the standard deviation of the heights of the plants.

(ii) If the collection is made up 500 plants, how many of them would you

expect to have heights exceeding 25.8 cm but less than 37.5 cm?

(c) A cigarette manufacturer claims that the mean nicotine content in cigarettes is 2

mg with a standard deviation of 0.3 mg .

(i) If this claim is valid what is the probability that a sample of size 900

cigarettes will yield a mean nicotine content exceeding 2.02 mg ?

(ii) What role does the Central Limit Theorem play in identifying the

distribution used in (i)?

7(a) The lifetime of a mechanical assembly in a vibration test is exponentially

distributed with a mean of 400 hours. What is the probability that

(i) an assembly on test fails in less 100 hours?

(ii) an assembly operates more than 500 hours before failure?

(iii) at least one assembly fails in less than 100 hours if 10 assemblies are

placed on test? Assume that assemblies fail independently.

(b) In a certain city, the daily consumption of water (in million of litres) is a

random variable with probability density function given by

91 , 0

9( )0 ,

x

x xf x

elsewhere

e

(i) Identify the distribution and use its parameters to find the city’s

expected water consumption for any given day.

42

(ii) What is the probability that the water supply on a given day is

inadequate, if the daily capacity of the city is 9 million litres?

(b) The operator of a pumping station has observed that the demand for water

during early afternoon hours has approximately exponential distribution with

mean of 100 cubic feet per second.

(i) Write down the exponential probability distribution.

(ii) Find the probability that the demand will exceed 200 cubic feet per

second during its early afternoon on a randomly selected day.

(iii) What pumping capacity should the station maintain during early

afternoons so that the demand exceeds the capacity with probability

of 0.01?