CHAPTER2.pdf

Chapter 2: Special Probability Distributions

CHAPTER 2 : SPECIAL PROBABILITY DISTRIBUTIONS Sub-Topic  Introduction  The binomial distribution.  The Poisson distribution.  Poisson approximation to binomial distribution.  The normal distribution.  Normal approximation to binomial distribution. Chapter Learning Outcome  Identify and solve the problems of probability using binomial, Poisson, and normal distribution. Learning Objective By the end of this chapter, students should be able to  State the probability distribution whether it is binomial, Poisson or normal distribution.  Find the probability by applying the formula for each distribution in problem solving.  Find the probability by using the probability table for each distribution to solve the problem.  Apply Poisson and normal approximation to binomial distribution.

Key Term (English to Bahasa Melayu)

English

Bahasa Melayu

1. Approximation

→

Penghampiran

2. Binomial distribution

→

Taburan binomial

3. Success

→

Kejayaan

4. Failure

→

Kegagalan

5. Poisson distribution

→

Taburan Poisson

6. Normal distribution

→

Taburan normal

7. Standard normal distribution

→

Taburan normal piawai

8. Z score

→

Skor Z

9. Continuity correction

→

Pembetulan keselanjaran

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Chapter 2: Special Probability Distributions

2.1 Introduction

In previous chapter, we have defined discrete and continuous random variables and have introduced discrete probability distributions. In this chapter, we present two important discrete distribution, which is the binomial and Poisson distributions. Furthermore, we will discuss the general idea of a continuous probability distribution and normal distribution.

2.2 The Binomial Distribution

Perhaps the most widely known of all discrete distributions is the binomial distribution. The binomial distribution has been used for hundreds of years. In this section, we will discuss the most important discrete probability with the situation such as below.

Situation 1 Suppose that historical sales records indicate that 40 percent of all customers who enter a discount department store make a purchase. What is the probability that two of the next three customers will make a purchase?

In order to find this probability, first we note that the experiment of observing three customers making a purchase decision has several distinguishing characteristics, which is  The experiment consists of a sequence of three identical trials, where each trial consists of a customer making a purchase decision.  Two outcomes are possible on each trial where the customer makes a purchase (we call it success) or the customer does not make a purchase (we call it failure).  Since 40 percent of all customers make a purchase, it is reasonable to assume that PS  , the probability that a customer makes a purchase, is 0.4 and it

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Chapter 2: Special Probability Distributions

constant for all customers. Otherwise P F  is the probability that a customer does not make a purchase which is 0.6 and it constant for all customers.  We assume the customers make independent purchase decisions. Beside, we assume that the outcomes of the three trials are independent of each other.

From the explanation above, we know that the sample space of the experiment consists eight sample space outcomes which is S   SSS , SSF , SFS , FSS , FFS , FSF , SFF , FFF .

The sample space outcome SSS represents all three customers making purchases. It means that FFS represent the first customer not making a purchase, the second customer also not making a purchase and the third customer making a purchase. Furthermore, since the trials (purchase decision) are independent, we can simply multiply the probabilities associated with the different trial outcomes to find the probability of a sequence of outcomes. Therefore, P SSF   P S P S P F   0.4 0.4 0.6  0.096 P SFS   P S P F P S   0.4 0.60.4   0.096 P FSS   P F P S P S   0.60.4 0.4   0.096

It follows that the probability that two out of the next three customers make a purchase is P SSF   P SFS   P FSS   0.096  0.096  0.096

 0.288 With this, we can find the probability of the next n customers who will make a purchase. Here, we assume that p is the probability that a customer makes a purchase, while q  (1  p) is the probability that a customer does not make a purchase and we assume that the purchase decisions are independent. Therefore, the probability that two of the next three customers make a purchase is  The number of ways    2 32  to arrange 2 successes  p  q  among 3 trials   

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Chapter 2: Special Probability Distributions

In general, each of the sample space outcomes describing the occurrence of x successes (purchases) in n trials represents a different arrangement of x successes in n trials. However, each of these sample space outcomes consists of x successes and n  x failures. The probability of each sample space outcome is p x  q n x . It follows

with the probability that x of the next n trials are successes (purchases) is  The number of ways    x n x  to arrange 2 successes  p  q  among 3 trials   

It means, the binomial random variable, X is the number of success in n trials, that is 0, 1, 2, ..., n. . The probability distribution of X is called a binomial distribution (or

binomial probability distribution) with X ~ Bn, p  .

Definition 1 A Bernoulli trial can result in a success with probability p and a failure with probability, q  (1  p) . Then, the probability distribution of X is called a binomial distribution (or binomial probability distribution) with the number of success in n independent trial is written as X ~ Bn, p  . Theory 1 Several assumptions underlie the use of the binomial distributions are  The experiment involves n identical trials.  Each trial has only two possible outcomes denoted as success or as failure.  Each trial is independent of the previous trials.  The terms p and q remain constant throughout the experiment, where the term p is the probability of getting a success on any one trial and the term q  (1  p) is the probability of getting a failure on any one trial.

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Chapter 2: Special Probability Distributions

Theory 2 Binomial formula P X  x  

n!  p x  q n x  n C x  p x  q n x x !  n  x  !

where p = probability of success in a single trial

q = 1  p , probability of failure in a single trial

n = number of trials

x = number of success in n trials, so x can be any number between 0 and n

(inclusive) with X ~ B n, p 

Example 1 Consider the discount department store in the situation 1 which is ‘Suppose that historical sales records indicate that 40 percent of all customers who enter a discount department store make a purchase’. Find the probability that three of the next five customers will make purchases.

Answer Example 1 Here we know that, p  0.4 , q  0.6 and n  5 . Px  3  5 C3  0.4  0.6 3

53

 100.064 0.36 

 0.2304 Example 2 A coin is tossed three times. Find the probability of getting exactly (a)

no head.

(b)

one head.

(c)

two heads.

(d)

three heads.

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Chapter 2: Special Probability Distributions

Answer Example 2 In this case, n  3 p  0.5 and q  0.5 . Substitute in the formula, then (a)

Px  0  3 C0  0.5  0.5 0

3

 110.5

3

 0.125 (b)

Px  1  3 C1  0.5  0.5

31

1

 30.50.25

 0.375 (c)

Px  2   3 C2  0.5  0.5

32

2

 30.250.5

 0.375 (d)

Px  3  3 C3  0.5  0.5 3

33

 10.5 1 3

 0.125

Theory 3 The usual inequalities that we will use in Binomial are Name

Symbol

Greater than

>

Less than

<

At least



At most



Not more than



Not less than



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Chapter 2: Special Probability Distributions

Example 3 A survey from the Teenage Research in UTHM found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that (a)

none has a part-time job.

(b)

exactly two have part time jobs.

(c)

not more than three of them will have part-time jobs.

(d)

greater than three will have part-time jobs.

(e)

at least three of them will have part-time jobs.

Answer Example 3 (a)

Px  0  5 C0  0.3  0.7 0

50

 1  1  0.1681 

 0.1681

(b)

Px  2 5 C2  0.3  0.7

52

2

 10  0.09   0.343 

 0.3087

(c)

Px  3  P( x  0)  P( x  1)  P( x  2)  P( x  3)

 0.1681 5 C1  0.31  0.7 51  0.3087  0.1323

 0.1681  0.3602  0.3087  0.1323  0.9693

(d)

P  x  3  P ( x  4)  P ( x  5)

 5 C4  0.3  0.7 4

54

 5 C5  0.3  0.7  5

 0.0284  0.0024

 0.0308

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55

Chapter 2: Special Probability Distributions

(e)

P  x  3  P ( x  3)  P ( x  4)  P ( x  5)

 5 C3  0.3  0.7 3

53

 0.0284  0.0024

 0.1323  0.0284  0.0024  0.1631

Theory 4 Although the binomial distribution is easily evaluated by calculators, sometimes it is more suitable to use binomial probability table especially when the trials more than ten.

Example 4 Solve the problem in Example 3(e) by using binomial probability table.

Answer Example 4 From Example 3(e) : at least three of them will have part-time jobs. We know that n  5 , p  0.3 and q  0.7 . Then, to find the probability, P  x  3 from the table statistics (J. Murdoch) below:

The value of 0.1631 is the answer, then P  x  3  0.1631 .

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Chapter 2: Special Probability Distributions

Theory 5 Some ideas in finding the value of probability by using table such as below. (a)

P( X  k )  from table

(b)

P ( X  k )  1  P( X  k )

(c)

P( X  k )  1  P( X  k  1)

(d)

P( X  k )  P( X  k  1)

(e)

P( X  k )  P( X  k )  P( X  k  1)

(f)

P(k  X  l )  P( X  k )  P( X  l  1)

(g)

P(k  X  l )  P( X  k  1)  P( X  l )

(h)

P( k  X  l )  P( X  k )  P( X  l )

(i)

P(k  X  l )  P( X  k  1)  P( X  l  1)

Example 5 Given that X ~ B 5, 0.5 , by using binomial probability table, find (a)

P( X  5) .

(b)

P( X  2) .

(c)

P( X  3) .

(d)

P( X  1) .

(e)

P( X  2) .

(f)

P(2  X  4) .

(g)

P(1  X  4) .

(h)

P(1  X  3) .

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Chapter 2: Special Probability Distributions

(i)

P(1  X  3) .

Answer Example 5 (a)

P( X  5)  0.0313

(b)

P( X  2)  1  P( X  3)

 1 0.5000  0.5000 (c)

P( X  3)  P( X  4)

 0.1875 (d)

P( X  1)  1  P( X  1)

 0.0312 (e)

P( X  2)  P( X  2)  P( X  3)

 0.8125  0.5000  0.3125 (f)

P(2  X  4)  P( X  2)  P( X  5)

 0.8125  0.0313  0.7812

(g)

P(1  X  4)  P( X  2)  P( X  4)

 0.8125  0.1875  0.6250

(h)

P(1  X  3)  P( X  2)  P( X  4)

 0.8125  0.1875  0.6250

(i)

P(1  X  3)  P( X  1)  P( X  3)

 0.9688  0.5000  0.4688

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Chapter 2: Special Probability Distributions

Example 6 Public Opinion reported that 5% of Malaysians are afraid of being alone in a house at night. If a random sample of 20 Malaysian is selected, find these probabilities by using the binomial table. (a)

There are exactly five people in the sample who are afraid of being alone at night.

(b)

There are at most three people in the sample who are afraid being alone at night.

(c)

There are at least three people in the sample who are afraid being alone at night.

Answer Example 6 Since 20 people are selected at random, then n  20 . Let probability of success (afraid of being alone in the night), p  0.05 . Therefore, the Binomial random variable, X ~ B 20, 0.05

(a)

P  X  5  P  X  5  P  X  6 

 0.0026  0.0003  0.0023 (b)

P  X  3  1  P  X  4 

 1 0.0159  0.9841 (c)

P  X  3  0.0755

Example 7 Assume that when a certain hunter shoots at a bird from five, the probability of hitting it is 0.6. Find the probability that the hunter (a)

will hit at least four birds.

(b)

will hit at least one bird.

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Chapter 2: Special Probability Distributions

Answer Example 7 Since then hunter shoots at a bird from five, then n  5 , p  0.6 and q  0.4 . Therefore, the binomial random variable, X can be written as X ~ B 5, 0.6 . (a)

P( X  4)  P( X  4)  P( X  5)

5 C4  (0.6) 4  (0.4)1  5 C5  (0.6) 5  (0.4) 0  0.2592  0.0778  0.3370

(b)

P( X  1)  1  P( X  1)  1  P( X  0)

 1 5C0  (0.6) 0  (0.4) 5  1 0.0102  0.9898 Because of the limited value of p (just until p  0.5 ) in binomial probability table, so we need to use q  0.4 as the alternative value of p . (a)

P hit at least four birds   1  P (doesn' t hit at least two birds )

 1  P( X  2)  1 0.6630  0.3370 **Note : We assume that p  0.4 .

(b)

P hit at least one bird   1  P (doesn' t hit at least five birds )

 1  P( X  5)  1 0.0102  0.9898 **Note : We assume that p  0.4 .

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Chapter 2: Special Probability Distributions

Theory 6 The mean and variance of the binomial distribution X ~ B(n, p) with n trials and probability of success, p are Mean,   np Variance,  2  npq ** Note : We know that q  (1  p)

Example 8 A coin is tossed four times. Find the mean, variance and standard deviation of the number of heads that will be obtained.

Answer Example 8 Use the formula for the binomial distribution with n  4 , p  0.5 and , q  0.5 . The value of mean, variance and standard deviation are such as below. Mean,   np  4  0.5  2 Variance,  2  npq  4  0.5  0.5  1 Standard deviation,   npq  1  1

Example 9 A general hospital recorded that 75% of the cancer patients are died after the chemotherapy treatment. Assume that the distribution of the cancer patients who died after the chemotherapy treatment is binomially distributed. If five patients are selected randomly, find the probability that (a)

all of them are died.

(b)

only two of the patients are recovered.

Answer Example 9 (a)

Let X = no of cancer patients who died after the chemotherapy

n  5 , p  0.75 , q  0.25

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Chapter 2: Special Probability Distributions

P( X  5)5C5  (0.75) 5  (0.25) 0  0.2373

(b)

Only two recovered, so three are died.

P( X  3) 5C3  (0.75) 3  (0.25) 2  0.2637

Example 10 The probability that a man is late in a day for work is 0.1. Assume that each day is independent of each other. Find the probability that he will be late for work on at least three days during five days.

Answer Example 10 Let X be the number of days in a week in which he is late, then X ~ B 5, 0.4  . P(at least three days he is late)  P  X  3  0.3174 or  5  5  5 P  X  3     (0.4) 3  (0.6) 2     (0.4) 4  (0.6)1     (0.4) 5  (0.6) 0  3  4  5

 0.3174

Exercise 2.2 1.

Let X be a binomial random variable with X ~ B 10, 0.1 . Calculate the following probabilities. (a)

P X  2 .

(b)

P  X  8 .

(c)

P X  4 .

(d)

P 5  X  7  .

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Chapter 2: Special Probability Distributions

2.

3.

Evaluate these binomial probabilities. (a)

8

C2  0.3  0.7 .

(b)

4

C0  0.05  0.95 .

(c)

10

2

6

0

4

C3  0.5  0.5 . 3

7

Let X be a binomial random variable with n  7 and p  0.3 . Find the value of

4.

(a)

P x  4  .

(b)

P x  1 .

(c)

P x  1 .

(d)

mean.

(e)

variance.

The phone lines to an airline reservation system are occupied 40% of the time. Assume that the events that the lines are occupied on successive calls are independent. Assume that ten calls are placed to the airline. (a)

What is the probability that for exactly three calls the lines are occupied ?

(b)

What is the probability that for at least one call the lines are not occupied ?

(c)

What is the expected number of calls in which the lines are all occupied ?

5.

A particularly long traffic light on your morning commute is green 20% of the time that you approach it. Assume that each morning represents an independent trial. (a)

Over five mornings, what is the probability that the light is green on exactly one day ?

(b)

Over 20 mornings, what is the probability that the light is green on

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Chapter 2: Special Probability Distributions

exactly four days ? (c)

Over 20 mornings, what is the probability that the light is green on more than four days ?

6.

The TV10 television shows 60 minutes has been successful for many years. That show recently had a share of 20, meaning that among the TV sets in use, 20% were tuned to 60 minutes (based on data from Aryssa Media Research). Assume that an advertiser wants to verify that 20% share value by conducting its own survey, and a pilot survey begins with 10 households having TV sets in use at the time of a 60 minutes broadcast. (a)

Find the probability that none of the households are tuned to 60 minutes.

7.

(b)

Find the probability that at least one household is tuned to 60 minutes.

(c)

Find the probability that at most one household is tuned to 60 minutes.

An experiment involving a gender-selection method includes a control group of 15 couples who are not given any treatment intended to influence the genders of their babies. Each of the 15 couples has one child. (a)

Find the mean and standard deviation for the numbers of girls in such groups of 15.

(b)

If the couples have 10 girls and 5 boys, is that unusual? Why or why not?

8.

The Haniss Department Store has experienced a 3.2% rate of customer complaints and attempts to lower this rate with an employee training program. After the program, 850 customers are tracked and it is found that only 7 of them filed complaints. (a)

Assuming that the training program has no effect, find the mean and standard deviation for the number of complaints in such groups of 850 customers.

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Chapter 2: Special Probability Distributions

(b)

Based on the results from part (a), is the result of seven complaints unusual? Does it seem that the training program was effective in lowering the rate of complaints?

9.

In a study of 420,000 cell phone users in Johor, it is found that 135 developed cancer of the brain or nervous system. If we assume that such cancer is not affected by cell phones, the probability of a person having such a cancer is 0.000340. (a)

Assuming that cell phones have no effect on cancer, find the mean and standard deviation for the numbers of people in groups of 420,000 that can be expected to have cancer of the brain or nervous system.

(b)

Based on the results from part (a), is it unusual to find that among 420,000 people, there are 135 cases of cancer of the brain or nervous system? Why?

10.

A recent Gallup poll consisted of 1012 randomly selected adults who were asked whether “cloning of humans should or should not be allowed”. Results showed that 89% of those surveyed indicated that cloning should not be allowed. (a)

Among the 1012 adults surveyed, how many said that cloning should not be allowed?

(b)

If we assume that people are indifferent so that 50% believed that cloning of humans should not be allowed, find the mean and standard deviation for the numbers of people in groups of 1012 that can be expected to believe that such cloning should not be allowed.

11.

A regimen consisting of a daily dose of vitamin C was tested to determine its effectiveness in preventing the common cold. Ten people who are were following the prescribed regimen were observed for a period of 1 year. Eight survived the winter without a cold. Suppose the probability of surviving the

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Chapter 2: Special Probability Distributions

winter without a cold is 0.5 when the vitamin C regimen is not followed. What is the probability of observing eight or more survivors, given that the regimen is ineffective in increasing resistance to colds?

12.

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events:

13.

(a)

at least one of the alarms is triggered.

(b)

More than seven of the alarms are triggered.

(c)

Eight or fewer alarms are triggered.

Car color preferences change over the years and according to the particular model that the customer selects. In a recent year, suppose that 10% of all luxury cars are randomly selected. Let the sample of luxury cars are 20. Find the following probabilities:

14.

(a)

at least five cars are black.

(b)

At most six cars are black.

(c)

Between three and five cars (inclusive) are black.

Records shows that 30% of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose n = 4 new patients represent a random selection from the large set of prospective patients served by the clinic. Find these probabilities:

15.

(a)

all the patients’ bills will eventually have to be forgiven.

(b)

One will have to be forgiven.

(c)

None will have to be forgiven.

The taste test for PTC (phenylthiocarbamide) is a favorite exercise for every human genetics class. It has been established that a single gene determines the characteristic, and that 70% of Malaysian are “tasters”, while 30% are

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Chapter 2: Special Probability Distributions

“nontasters”. Suppose that 20 Malaysian are randomly chosen and are tested for PTC. (a)

What is the probability that 17 or more are “tasters”?

(b)

What is the probability that 15 or fewer are “tasters”?

Answer Exercise 2.2 1.

(a)

0.9298 (b)

0

(c)

2.

(a)

0.2965

(b)

0.8145

(c)

0.1172

3.

(a)

0.097

(b)

0.329

(c)

0.671

(e)

1.212

4.

(a)

0.215

(b)

0.999

(c) 4

5.

(a)

0.410

(b)

0.218

(c) 0.37

6.

(a)

0.107

(b)

0.893

(c) 0.375

7.

(a)

  7.5 ,   1.9

(b)

0.0112 (d)

0.0016

(d)

2.1

No. Because 10 is within 2 std dev of the

mean and its showing that it is easy to get 10 or more girls by chance. 8.

(a)

  27.2   5.1

(b)

Yes,

it

appears

that

the

training

(b)

No, 135 is not unusual because it is

programmed had an effect. 9.

(a)

  142.8   11.9

within 2 standard deviation of the mean. 901

(b)

  506   15.9

(a)

1

(b)

0.997

13.

(a)

0.0432

(b)

14.

(a)

0.0081 (b)

0.4116 (c)

15.

(a)

0.107

(b)

10.

(a)

11.

0.055

12.

(c) 0.9976

(c) 0.2401

0.762

73

0.086 0.3118

Chapter 2: Special Probability Distributions

2.3 The Poisson distribution

Another discrete random variable that has numerous practical applications is the Poisson random variable. Its probability distribution provides a good model for data that represent the number of occurrences of a specified event in a given unit of time or space. Here are some examples of experiments for which the random variable x can be modeled by the Poisson random variable.  The number of calls received by a switchboard during a given period of time.  The number of bacteria per small volume of fluid.  The number of customer arrivals at a checkout during a given minute.  The number of machine breakdowns during a given day.  The number of traffic accidents at a given intersection during a given time period.

In each example above, X represents the number of events that occur in a period time or space during with an average of mean such events can be expected to occur.

Definition 2 The probability distribution of X is called a Poisson distribution and the distribution is write as X ~ P   . O

Theory 7 Some properties of Poisson distribution that we must know are  The experiment consists of counting the number X of times for a particular event occurs during a given unit of time, or in a given area or volume (or weight, distance, or any other unit of measurement).  The probability that an event occurs in a given unit of time, area, or volume is the same for all the units.  The number of event is independent.  The Poisson random variable is the number of success event.

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Chapter 2: Special Probability Distributions

Theory 8 Formula of Poisson distribution is P X  x  

e   x x!

,

x  0, 1, 2, ..., 

where

  mean or average number of event x  number of success event

with X ~ P   O

The symbol e  2.71828  is evaluated using our scientific calculator, which should have function e x . For each value of X, we can obtain the individual probabilities for the Poisson random variable, just like binomial random variable. Otherwise, we can use cumulative Poisson table. The way how to find the values are similar with Binomial random variable. It is important to use consistent units in the calculation of probabilities, means and variances involve Poisson random variables. The following example illustrates unit conversions. For example, if the  Average number of flaws per millimeter of wire is 3.4.  Average number of flaws in 10 millimeters of wire is 34.  Average number of flaws in 100 millimeter of wire is 340. If a Poisson random variable represents the number of events in some interval, the mean of the random variable must equal to the expected number of events in the same length of interval.

Example 11 In the case of the thin copper wire, suppose that the number of flaws follows a Poisson distribution with a mean of 2.3 flaws per millimeter. Find the probability (a)

exactly two flaws in one millimeter of wire.

(b)

ten flaws in five millimeter of wire.

(c)

at least one flaws in two millimeters of wire.

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Chapter 2: Special Probability Distributions

Answer Example 11 Let X denote the number of flaws in 1 millimeter of wire. Then X ~ Po 2.3 flaws e 2.3  2.32  0.2650 2!

(a)

P X  2 

(b)

Since 1 millimeter = 2.3, so for 5 millimeter, the new average is

  5 mm  2.3 flaws/mm  11.5 flaws , then, X ~ Po 11.5 . P  X  10  

(c)

e 11.5  11.510  0.1130 10 !

Since 1 millimeter = 2.3, so for 2 millimeter, the new average is

  2 mm  2.3 flaws/mm  4.6 flaws , then X ~ Po 4.6 . e 4.6  4.60 P X  1  1  P  X  0  1   0.9899 0!

Theory 9 Sometimes we use Poisson table or Poisson probability table regarding the value of success event small or large. We will use the same ideas as in Binomial distribution.

Example 12 Given a Poisson random variable with X ~ Po 4 . Find the value of P  X  3 . Answer Example 12

From the statistical table (J. Murdoch), we know that P  X  3  0.7619

76

Chapter 2: Special Probability Distributions

Theory 10 The formula mean, variance and standard deviation of the Poisson random variable X are Mean,    Variance,  2   Standard deviation,    **Note : If the variance of count data is much greater than the mean of the same data, the Poisson distribution is not a good model of the distribution for the random variable.

Example 13 The number of incorrectly dialed numbers made per hour by a new telephone operator has a Poisson distribution. The probability she dials all the number correctly is 0.90. Find the mean number of incorrectly dialed number per hour.

Answer Example 13 Let X be the number of incorrectly dialed numbers. Then, we have X ~ Po   , with  as the mean incorrectly dialed number. The probability that the telephone sales representative dials all the numbers correctly is 0.90. That means she got no incorrectly dialed number is P X  0   0.90

P X  0 

e   0  0.90 0!

e    0.90

  ln 0.90  0.1054 Therefore, the mean number of incorrectly dialed numbers per hour is 0.1054.

77

Chapter 2: Special Probability Distributions

Exercise 2.3 1.

Which of the following are qualifying for the application of the Poisson probability distribution? (a) The number of customers coming to a grocery store during a one-hour interval. (b) The arrival of commercial airplanes at Senai Airport. (c) The number of television sets sold at a department store during a given week. (d) The arrival of patients at a physician’s office. (e) The number of defective items in the next 100 items manufactured on a machine. (f) The number of defects in a five-meter-long ion rod.

2.

Suppose X has a Poisson distribution with a mean of 4. Determine the following probabilities: (a) P  X  0 

3.

(b) P  X  2 

(c) P  X  4 

(d) P  X  8

Suppose that the number of customers that enter a bank in an hour is a Poisson random variable, and suppose that P  X  0   0.05 . Determine the mean and variance of X.

4.

Astronomers treat the number of stars given volume of space as Poisson random variable. The density in the Mickey Way Galaxy in the vicinity of our solar system is one star per 16 cubic light years. (a)

What is the probability of two or more stars in 16 cubic light years?

(b)

What many cubic light years of space must be studied so that the probability of one or more stars exceeds 0.95?

5.

When a computer disk manufacturer tests a disk, it writes to the disk and then tests it using a certifier. The certifier counts the number of missing pulses or

78

Chapter 2: Special Probability Distributions

errors. The number of errors on a test area on a disk has a Poisson distribution with   0.2 .

6.

(a)

What is the expected number of errors per test area?

(b)

What percentage of test areas has two or fewer errors?

If there are 200 typographical errors randomly distributed in a 500-page manuscript, find the probability that a given page contains exactly three errors.

7.

A sales firm receives, on the average three calls per hour on its toll-free number. For any given hour, find the probability that it will receive the following: (a)

8.

at most three calls

(b) at least three calls (c)

five or more calls

A recent study of robberies for a certain geographic region showed an average of one robbery per 20,000 people. In a city of 80,000 people, find the probability of the following (a) No robberies

9.

(b) One robberies

(c) Three or more robberies

If approximately 2% of the people in a room of 200 people are left-handed, find the probability that exactly five people there are left-handed.

10.

If 3% of all cars fail the emissions inspection, find the probability that in a sample of 90 cars, at least 4 will fail.

11.

The average number of traffic accidents on a certain section of highway is two per week. Assume that the number of accidents follows a Poisson distribution with   2 . (a)

Find the probability of no accidents on this section of highway during a 1-week period.

(b)

Find the probability of at most three accidents on this section of

79

Chapter 2: Special Probability Distributions

highway during a 2-week period.

12.

Parents who are concerned that their children are “children prone” can be reassured, according to a study conducted by the Department of Pediatric at the University of Malaysia. Children who are injuries two or more times tend to sustain these injuries during a relatively limited time, usually 1 year or less. If the average number of injuries per year for school-age children is two, what are the probabilities of these events?

13.

(a)

A child will sustain two injuries during the year.

(b)

A child will sustain two or more injuries during the year.

In a food processing and packaging plant, there are, on the average, two packaging machine breakdowns per week. Assume the weekly machine breakdowns follow a Poisson distribution. (a)

What is the probability that there are no machine breakdowns in a given week?

(b)

Calculate the probability that there are no more than two machine breakdowns in a given week.

14.

In average, 3 electric bulbs manufactured by a company are defective. Find the probability that in a sample bulbs have (a) None is defective (b) three are defective

15.

According to the research of some lecturers at UTHM, the average number of accidental drowning per year in Parit Raja is 6. Find the probability there will be (a)

between 4 and 8 accidental drowings per year.

(b)

Fewer than 3 accidental drawings per year.

80

Chapter 2: Special Probability Distributions

Answer Exercise 2.3 1.

(a) yes

(b) no

(c) yes

2.

(a) 0.183

3.

E  X   Var  X   2.996

4.

(a) 0.264

(b) 48

5.

(a) 0.2

(b) 99.89%

6.

0.0072

7.

(a) 0.6472

(b) 0.5768

(c) 0.1848

8.

(a) 0.0183

(b) 0.0733

(c) 0.7619

9.

0.1563

10.

0.2859

11.

(a) 0.135335

12.

(a) 0.2706

13.

(a) 0.135335

(b) 0.676676

14.

(a) 0.04979

(b) 0.1494

15.

(a) 0.4589

(b) 0.062

(b) 0.2381

(d) no

(e) yes

(c) 0.1954

(f) yes

(d) 0.0298

(b) 0.433471 (b) 0.405

2.4 Poisson Approximation to Binomial Distribution

The Poisson probability distribution provides a simple, easy to calculate and accurate approximation to Binomial probabilities when n is large and p is small, preferable with, n  30 , p tend to 0  p  0.1 . That is mean binomial distribution can be approximated to Poisson distribution where   np .

Definition 3 If random variable X is binomially distributed with its parameter n and p and fulfill the condition n is large ( n  30 ) and p  0.1 , then the binomial distribution can be approximated to Poisson distribution with   np , X ~ Bn, p   X ~ Po   .

81

Chapter 2: Special Probability Distributions

Example 14 Suppose a life insurance company insures the lives of 5000 men aged 42. If actuarial studies show the probability that any 42-year-old man will die in a given year to be 0.001, find the exact probability that the company will have to pay x  4 claims during a given year.

Answer Example 15 The exact probability is given by the binomial distribution as Px  4  5000 C4 0.0014 0.999 4996  0.1755

By the way, Poisson distribution can be used to provide a good approximation to P x  4  . Compute,   np  5000 0.001  5 and substitute into the formula for the Poisson probability distribution, we have e 5 5 4 Px  4   0.17546 4!

The value of P x  4  could also be obtained by using Poisson probability table with

  5.

Example 16 A manufacturer of power lawn mowers buys 1-horsepower, two-cycle engines in lot of 1000 from a supplier. She then equips each of the mowers produced by her plant with one of the engines. History shows that the probability of any one engine from that supplier proving unsatisfactory is 0.001. In a shipment of 1000 engines, what is the probability that none is defective ?

Answer Example 16 This is a binomial distribution with X ~ B 1000 ,0.001 . Since n is large and   np  1 the probability of x defective engines in the e 1  10  0.3678 shipment may be approximated by X ~ Po 1 . Therefore P x  0  0!

82

Chapter 2: Special Probability Distributions

Example 17 On average, the number of prepaid card was sold at the telecommunication shop is 8 cards per hour. Assume that the number of prepaid card was sold at the shop is approximated by a Poisson distribution, what is the probability that (a)

there are seven cards are sold in an hour.

(b)

more than three cards are sold in 15 minutes.

Answer Example 17 (a)

Let X = no of prepaid cards sold per hour. P ( X  7) 

(b)

e 8  87  0.1396 7!

60 min  8 cards 15 min  2 cards X ~ P0 (2) , P( X  3)  P( X  4)  0.1428

Example 18 Diabetes is a chronic health condition where the body is unable to produce insulin. Medical Science Center report that the probability of a patient who suffer from diabetes will lead to blindness is 0.002. Assume that the distribution of the diabetic patients that will lead to blindness is binomial distributed. If 2000 patients diagnosed with diabetes are selected at random and examined, find the probability that (a)

fewer than three of the 2000 patients diagnosed will lead to blindness.

(b)

exactly seven of the 2000 patients diagnosed will lead to blindness.

Answer Example 18 (a)

Let, X = no of patients who suffer from diabetes that will lead to blindness.

  np  2000 (0.002)  4 P( X  3)  P(0)  P(1)  P(2)  0.0183  0.0733  0.1465

= 0.2381 (b)

Let, X = no of patients who suffer from diabetes that will lead to blindness.

83

Chapter 2: Special Probability Distributions

e 4  47 P ( X  7)   0.0595 7!

Exercise 2.4 1.

A bag contains 1 red and 7 white marbles. A marbles is drawn from the bag and its color is observed. Then the marble is put back into the bag and the contents are thoroughly mixed. By using the Poisson approximation to the binomial distribution, find the probability that in 8 such drawing a red ball is selected exactly 3 times.

2.

Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in the sample of 10 tools chosen at random exactly two will be defective.

3.

If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001, determine the probability that out of 2000 individuals (a) exactly 3

4.

(b) more than 2 individuals will suffer a bad reaction.

Suppose a life insurance company insures the lives of 5000 men aged 42. If actuarial studies show the probability that any 42-year-old man will die in a given year to be 0.001, find the exact probability that the company will have to pay 4 claims during a given year.

5.

A manufacturer of power lawn mowers buys 1-hosepower, two-cycle engines in lots of 1000 from a supplier. She then equips each of the mowers produced by her plant with one of the engines. History shows that the probability of any one engine from that supplier proving unsatisfactory is 0.001. In a shipment of 1000 engines, what is the probability that none is defective?

6.

Evidence shows that the probability that a driver will be involved in a serious automobile accident during a given year is 0.01. A particular corporation

84

Chapter 2: Special Probability Distributions

employs 100 full-time traveling sales reps. Based on this evidence, use the Poisson approximation to the binomial distribution to find the probability that exactly two of the sales reps will be involved in a serious automobile accident during the coming year.

7.

Suppose that 1000 vaccinations are to be administered by a medical team from the local health department. The probability of an allergic reaction is 0.001. A bad reaction can be counteracted with drugs. What is the probability that there will be at most 4 bad reactions if 1000 shots are given?

8.

It has been reported that a medicine called Norvasc, used in treating hypertension and angina, has a discontinuation rate of 1.5%, and it will discontinued when the patients exhibit adverse effects. (a)

If a cardiologist prescribes this medicine to 100 patients, what is the probability that more than 5 will suffer adverse effects?

(b)

What is the probability no more than 3 will suffer adverse effects if there is 100 patients were given this medicine?

(c)

9.

Find the standard deviation of this distribution.

If 3% of the electric bulbs manufactured by a company are defective, find the probability that in the sample of 100 bulbs have at most 4 bulbs will be defective.

10.

Haneffs Imprint buys T-shirts (to be imprinted with an item of the customer’s choice) from manufacturers who are guarantees that the shirts have been inspected and that no more than 1% are imperfect in any way. The shirts arrive in boxes of 12. Find the probability that: (a)

any one box has no imperfect shirts?

(b)

Any box has no more than one imperfect shirt?

85

Chapter 2: Special Probability Distributions

11.

Consider 8 packs of 60 watt bulbs and let x be the number of bulbs in a pack that “fail” the first time they are used. If 0.02 of all bulbs of this type fail on their first use and each 8-pack is consider a random sample, what is the probability (a) that any one 8-pack has no bulbs that fail on first use? (b)

12.

That any one 8-pack has no more than one bulb that fails on first use?

According to a recent article, one in four college students are aged 30 or older. Many of these students are women updating their job skills. Assume that the 2.5% figure is accurate, and the sample size they are select are 200 students. Recording x, is the number of students age 30 or older, what is the mean and standard deviation of x?

13.

The 10-year survival rate for bladder cancer is approximately 5%. If 200 people who have bladder cancer are properly treated for the disease, what is the probability that:

14.

(a)

no one will survive for 10 years?

(b)

At least 1 will survive for 10 years?

Despite reports that dark chocolate is beneficial to the heart, 7% of adults’ stills prefer milk chocolate to dark chocolate. Suppose a random sample of 100 adults is selected and asked whether they prefer milk chocolate to dark chocolate. (a)

What is the probability that five adults say that they prefer milk chocolate to dark chocolate?

(b)

What is the exactly three adults say that they prefer milk chocolate to dark chocolate?

15.

The president of a company specializing in public opinion claims that 5% of all people to whom agency send questionnaires respond by filling out and returning the questionnaires. 80 such questionnaires are sent out, and assume

86

Chapter 2: Special Probability Distributions

that the president’s claim is correct. What is the probability that exactly ten of the questionnaires are filled out and returned?

Answer Exercise 2.4 1.

0.06131

2.

0.1937

3.

(a) 0.1804

4.

0.175

5.

0.368

6.

0.1839

7.

0.9963

8.

(a)

9.

0.81539

10.

(a)

0.8869 (b)

0.9933

11.

(a)

0.8521 (b)

0.9884

12.

  5   2.2079

13.

(a)

0.0000454

(b)

14.

(a)

0.1277 (b)

0.05213

15.

0.00529

(b) 0.6767

0.0046 (b)

0.9342 (c)

1.224

0.9999

2.5 The normal distribution

Random variables can be either discrete or continuous. Recall that a discrete random variable cannot assume all values between any two given values of the variables. On the other hand, a continuous variable can assume all values between any two given values of the variable. Undoubtedly, the most widely used model for the distribution of a random variable is a normal distribution. Whenever a random experiment is replicated, the random variable that equals the average (or total) result over the replicates tends to have a normal distribution as the number of replicates becomes large. The normal distribution, which may already be familiar to us as the curve with bell shape, is sometime called as Gauss distribution.

87

Chapter 2: Special Probability Distributions

Definition 4 The random variable, X with a mean  and a standard deviation of  is written as

X ~ N   ,  2  . The probability density function of X is   f x  

1

 2



e

1  x 



2 

2

 ,

x 

where e  2.718

 = 3.14  = population mean

 = population standard deviation is called a normal distribution. Normal distribution is frequently applied in finding the probability of certain range for continuous measurement such as height, weight and time.

Theory 11  The value of x can take any numerical value.  A normal distribution curve is bell-shaped.

 The total area under a normal distribution curve is equal to 1.00 or 100%.





1  f x  dx    2  

88

 x    e 2   

2

1

1

Chapter 2: Special Probability Distributions

 The curve is symmetric, which means that its shape is the same on both sides.

0.5

0.5 

 The mean, mode and median are equal and are located at the center of the distribution.  The curve never touches the x axis.  For the calculation, we will standardized the normal distribution. x   The formula for standard normal distribution is P Z  .   

Example 19 By using the normal distribution table, find the value of P Z  1.98  .

Answer Example 19 Using J.Murdoch statistical table,

The answer for P Z  1.98  = 0.0239

89

Chapter 2: Special Probability Distributions

Example 20 Calculate the probabilities for the standard normal distribution. (a)

PZ  3.15  .

(b)

PZ  2.85  .

(c)

P Z  2.2  .

(d)

PZ  1.54  .

(e)

P 1.72  Z  2.76  .

(f)

P  1.72  Z  2.3 .

(g)

P 0.24  Z  1.89  .

(h)

P  Z  1.53 .

(i)

P Z  1.91 .

Answer Example 20 (a)

P Z  3.15   0.00082 .

(b)

P Z  2.85   1  P Z  2.85   1  0.00219  0.9978 .

(c)

P Z  2.2   PZ  2.2   0.0139 .

(d)

P Z  1.54   1  PZ  1.54   1  0.0618  0.9382 .

(e)

P1.72  Z  2.76   P Z  1.72   P Z  2.76   0.0427  0.00289  0.0398 .

(f) P 1.72  Z  2.3  1  PZ  1.72   PZ  2.3  1  0.0427  0.01072  0.9466 . (g) P 0.24  Z  1.89   PZ  0.24   PZ  1.89   0.4052  0.0294  0.3758 . (h) P Z  1.35  2PZ  1.35  0.0885  0.1770 . (i) P Z  1.91  1  2PZ  1.91  1  20.0281  0.9438 .

90

Chapter 2: Special Probability Distributions

Example 21 Suppose the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 mill amperes and a variance of 4 (mill amperes)2. What is the probability that a measurement will (a)

exceed thirteen mill amperes ?

(b)

between nine and eleven mill amperes ?

Answer Example 21 Let X denote the current in mill amperes, X ~ N 10, 4  . So   10 and   4  2 . (a)

13  10   P X  13  P Z    PZ  1.5  0.0668 2  

(b)

11  10   9  10 P9  X  11  P Z  2   2  P  0.5  Z  0.5  1  2P Z  0.5

 0.3829 Example 22 The diameter of a shaft in an optical storage drive is normally distributed with a mean 0.2508 inch and standard deviation 0.0005 inch. The specifications on the shelf are 0.2500  0.0015 inch. What is the probability of shaft conforms to specifications ?

Answer Example 22 Let X denote the shaft diameter in inches, then X ~ N  0.2508 , 0.0005 2  .   0.2515  0.2508   0.2485  0.2508 P0.2485  X  0.2515   P Z  0.0005 0.0005    P  4.6  Z  1.4   1  P Z  4.6  P Z  1.4 

 1  0.0000  0.0808

 0.9192

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Chapter 2: Special Probability Distributions

Theory 13 We can find the normal distribution probabilities by using Casio scientific calculator (FX-570W and FX-570MS). Step 1 : Press MODE

MODE

1

to enter SD Mode.

Step 2 : Press SHIFT

DISTR to produce the screen shown below P( Q( R( →t 1 2 3 4

Step 3: Input a value from

1 calculation we want perform.

to 3

to select the probability distribution

P(t)

Q(t)

R(t)

Example 23 Find the following probabilities, by using Casio Scientific Calculator fx-570W or fx570MS. (a)

P( Z  0.75).

(b)

P( Z  0.95).

(c)

P( Z  1.42) .

(d)

P( Z  1.43).

(e)

P( | Z |  1.23).

(f)

P( | Z |  0.99).

(g)

P(0.67  Z  1.34) .

(h)

P(1.74  Z  0.75).

(i)

P(1.54  Z  2.32) .

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Chapter 2: Special Probability Distributions

Answer Example 23

(a)

SHIFT

DISTR

3

0.75

)

=

0.2266

(b)

SHIFT

DISTR

1

0.95

)

=

0.8289

(c)

SHIFT

DISTR

3

1.42

)

=

0.0778

(d)

SHIFT

DISTR

1

1.43

)

=

0.9236

(e)

SHIFT

DISTR

3

1.23

)

X

2

=

0.2187

(f)

SHIFT

DISTR

2

0.99

)

X

2

=

0.6778

(g)

SHIFT

DISTR

3

0.67

)

1.34 0.75

)

1.74

)

1.54

)

2.32

)

-

(h)

SHIFT

SHIFT

DISTR

-

(i)

1

-

3 SHIFT

SHIFT

-

3

DISTR

DISTR

SHIFT

3

DISTR 3

DISTR

3

=

0.1613

=

0.1901

=

0.9281

)

Example 24 Let exam marks for 200 students are normally distributed with   50 and  2  16 . Find the number of student that passed the exam (the marks are 40).

Answer Example 25 Let X be the exam marks, X ~ N 50, 16  . 40  50   P  X  40   P Z    PZ  2.5 `16    1  P Z  2.5

 1 0.0062  0.9938 So the number of student that passed the exam is 200  0.9938  198.758  199 .

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Chapter 2: Special Probability Distributions

Exercise 3.5 1.

If the height of 300 students is normally distributed with mean 68 in and standard deviation 0.0025 in, how many students have height greater than 72 in?

2.

Studies show that gasoline use for compact cars sold in the Batu Pahat is normally distributed, with a mean of 25.5 miles per gallon (mpg) and a standard deviation of 4.5 mpg. What percentage of compacts gets 30 mpg or more?

3.

A study of lease rates for a selection of 2006 cars revealed that the average monthly rate for a vehicle was RM 220.67. The standard deviation was RM 59.63. Assume that the rates follow a normal distribution. (a)

What is the probability that a random car will lease for less than RM 150?

(b)

Find the probability that a random car will lease for more than RM 350?

4.

Based on data from ACT in 2007, the average science reasoning test score was 20.9, with a standard deviation of 4.6. Assuming that the scores are normally distributed: (a)

Find the probability that a randomly selected student has a science reasoning ACT score of least 25.

(b)

Find the probability that a randomly selected student has a science reasoning ACT score between 20 and 26.

5.

The average annual charges per credit card in 2006 were RM 9600 according from one research. Assuming that the annual charges per card are approximately normally distributed with a standard deviation of RM 2100, what is the probability that a credit card customer’s annual charges are (a)

less than RM 4000

(b)

between RM 5000 and RM 10,000

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Chapter 2: Special Probability Distributions

6.

The grades on an examination whose mean is 525 and whose standard deviation is 80 are normally distributed. What is the probability that the person will score the examination below than 350 ?

7.

For a car traveling 30 miles per hour (mph), the distance required to brake to a stop is normally distributed with a mean of 50 feet and a standard deviation of 8 feet. Suppose you are traveling in a residential area and a car abruptly into your path at a distance of 60 feet. What is the probability that you will

8.

(a)

brake a stop within 40 feet or less ?

(b)

Brake a stop between 40 feet and 45 feet ?

Suppose that you must establish regulations concerning the maximum number of people who can occupy an elevator. A study of elevator occupancies indicates that if eight people occupy the elevator, the probability distribution of the total weight of the eight people has a mean equal to 1200 pounds and a standard deviation of 99 pounds. What is the probability that the total weight of eight people exceeds 1300 pounds ?

9.

Suppose that the unsupported stem diameters at the base of a particular species of sunflower plant have a normal distribution with an average diameter of 35 millimeter and a standard deviation of 3 millimeter. (a)

What is the probability that a sunflower plant will have a basal diameter of more than 40 mm ?

(b)

What is the probability that a sunflower plant will have a basal diameter between 30 mm and 40 mm ?

10.

The meat department at a Parit Raja supermarket specifically prepares its “1 kg” packages of ground beef so that there will be a variety of weights, some slightly more and more slightly less than 1 kg. Suppose it is normally distributed with a mean of 1 kg and a standard deviation of 0.15 kg. What is

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Chapter 2: Special Probability Distributions

the probability of the packages will weight between 0.95 kg and 1.05 kg ?

11.

The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter (mg/l) and a standard deviation of 14 mg/l. What is the probability if the daily discharge exceed 50 mg/l ?

12.

Suppose the individual 2008 January prime interest rate forecasts of economic analysts are approximately normally distributed with the mean equal to 8.5 and a standard deviation equal to 0.2. If a single analyst is randomly selected from among this group, what is the probability that the analyst’s forecast of the prime rate will take on these values ? (a)

13.

exceed 8.75

(b)

less than 8.375

Suppose the numbers of a particular type of bacteria in samples of 1 milliliter of drinking water tend to be approximately normally distributed, with a mean of 85 and a standard deviation of 9. What is the probability it will more than 100 bacteria?

14.

How does the income tax bureau decide on the percentage of income tax returns to audit for each state? Suppose it is approximately normally distributed with a mean 1.55% and a standard deviation 0.45%. (a)

What is the probability that a particular state will have more than 2.5% of its income tax returns audited ?

(b)

What is the probability that a state will have less than 1% of its income tax returns audited ?

15.

A study have been done to look at the amount of money spent at shopping complex between 4 pm and 6 pm on Sundays and had a normal distribution with a mean RM 85 and with a standard deviation of RM 20. A shopper is

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Chapter 2: Special Probability Distributions

randomly selected on a Sunday between 4 pm and 6 pm and asked about their spending patterns. (a)

What is the probability that they spent more than RM 95 at the mall ?

(b)

What is the probability that they spent between RM 95 and RM 115 at the mall ?

Answer Exercise 2.5 1.

20

2.

15.87%

3.

(a)

0.1170

(b)

0.0150

4.

(a)

0.1867

(b)

0.4458

5.

(a)

0.0038

(b)

0.5610

6.

0.01435

7.

(a)

0.10565

(b)

0.16034

8.

0.1562

9.

(a)

0.0475

(b)

0.90448

10.

0.26086

11.

0.05021

12.

(a)

0.2676

13.

0.04779

14.

(a)

0.01738

(b)

15.

(a)

0.3085 (b)

0.2417

0.1056 (b)

0.11085

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Chapter 2: Special Probability Distributions

2.6 Normal approximation to binomial distribution

It should not be surprise to learn that the normal distribution can be used to approximate binomial probabilities for cases in which n is large. Normal approximation should be used only when np  5 and nq  5 . Figure (a) shows the probability for a binomial distribution. Figure (b) shows an approximation of binomial probabilities by normal probabilities.

Because binomial distribution is discrete and the normal distribution is continuous, then we must use the continuity correction (adding and subtracting 0.5). For example, to find the probability of 12 successes, we will use P 11.5  X  12.5 .

Theory 14 Binomial Distribution

Normal Distribution

(i)

P(X = a)

P(a – 0.5 < X < a + 0.5)

(ii)

P(X  a)

P(X > a – 0.5)

(iii)

P(X > a)

P(X > a + 0.5)

(iv)

P(X  a)

P(X < a + 0.5)

(v)

P(X < a)

P(X < a – 0.5)

(vi)

P(a  X  b)

P(a – 0.5 < X < b + 0.5)

(vii)

P(a < X < b)

P(a + 0.5 < X < b – 0.5)

For all cases,  = E(x) = np,   npq , np  5 , and nq  5 .

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Chapter 2: Special Probability Distributions

Theory 15 There are five steps to use normal approximation to binomial Distribution Step 1 : Check whether the normal approximation can be used. Step 2 : Find the value of mean and standard deviation. Step 3 : Write the problem in probability notation, using X . Step 4: Rewrite the problem by using the continuity correction factor, and show the corresponding area under the normal distribution. Step 5 : Find the corresponding Z value and solution.

Example 26 In a digital communication channel, assume that the number of bits received in error can be modeled by a binomial random variable, and assume that the probability that a bit is received in error is 1  10  5 . If 16 million bits are transmitted, what is the probability that 150 of fewer errors occur ?

Answer Example 26 Let X denote the number of errors. Then X ~ B16000000 , 1  10 5  .   Step 1 : np  16000000 1  10  5   160 , the normal distribution can be use   since np  5 and nq is much larger. Step 2 :   np  16000000 1  10  5   160  

  npq  16000000 1  10  5 1  1  10  5  





  Step 3 : X ~ N 160, 1601  10 5       

Step 4 : P X  150   PZ  150  0.5

 150.5  160 Step 5 :  P Z   160 1  10 5 



   PZ  0.751  0.2263  

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Chapter 2: Special Probability Distributions

Example 27 A magazine reported that ten percent of Malaysian drivers use hand phone while on the motorcycle. If two hundred drivers are selected at random, find the probability that exact thirty of them use hand phone while riding the motorcycle.

Answer Example 27 Firstly, thought as a binomial experiment: Since three hundred drivers are selected at random, then n = 200. Let : probability of success, i.e. use hand phone while driving,

p = 0.10.

probability of failure, i.e. not use hand phone while driving, q = 0.90. Therefore the binomial random variable, X can be written as X ~ B(200, 0.10) . Step 1 :

np = (200)(0.10) = 20 and nq = (200)(0.90) = 180 Since np  5 and nq  5 , the normal distribution can be used.

Step 2 :

 = np = (200)(0.10) = 20

  npq  (200)(0.10)(0.90)  18  4.2426 Step 3 :

X ~ N(20, 4.2426) P(X = 30)

Step 4 :

Using the continuity correction, we find the probability 45.5  20   44.5  20 P(44.5 < X < 45.5)  P  Z  4.2426   4.2426

Step 5 :

= P(2.2392 < Z < 2.4749) = 0.0126 – 0.0067 = 0.0059.

Example 28 In a photographic process, the developing time of prints may looked upon as random variable having the normal distribution with mean 15.40 and standard deviation 0.48 second. Find the probabilities that the time it takes to develop one of the prints will be (a)

at least 16.00 seconds.

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Chapter 2: Special Probability Distributions

(b)

at most 14.20 seconds.

(c)

anywhere from 15.00 to 15.80 seconds.

Answer Example 28 Let X be the time it takes to develop one of the prints X ~ N 15.40,0.482  (a)

16.00  15.40   P X  16.00   P Z   0.48    PZ  1.25

 0.10565

(b)

14.20  15.40   P X  14.20   P Z   0.48  

 PZ  2.50 

 0.00621

(c)

15.80  15.40   15.00  15.40 P15.00  X  15.80   P Z  0.48 0.48  

 P 0.833  Z  0.833  1  20.20327   0.5935

Example 29 Based upon past experience at TT Tires Manufacturing, 8% of certain branded tires produced are defective during ongoing production process. If a random sample of 1,600 tires s selected, what is the approximate probability that (a)

exactly 125 tires will be defective ?

(b)

at least 150 tires will be defective ?

(c)

not more than 110 tires will be defective ?

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Chapter 2: Special Probability Distributions

Answer Example 29

  np  128,   npq  1280.92   10.8517 (a)

P X  125  P124.5  X  125.5 125 .5  128   124 .5  128  P Z  10.8517   10.8517  P 0.3225  Z  0.2304   PZ  0.2304   PZ  0.3225   0.40889  0.37354  0.03535

(b)

P X  150   P X  149.5  X  128 149 .5  128   P   10.8517   10.8517  PZ  1.9813   0.02378

(c)

P X  110   P X  109.5  X  128 109 .5  128   P   10.8517   10.8517  P Z  1.7048   1  P Z  1.7048   1  0.04412  0.95588

Exercise 2.6 1.

Evaluate (a)

P(Z  1.02).

(b)

P( Z  1.26).

(c)

P( Z  1.52).

(d)

P(0.91  Z  0.91).

(e)

P( 1.97  Z  1.26).

(f)

P(0  Z  1.74).

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Chapter 2: Special Probability Distributions

2.

Assume that X is normally distributed with a mean of 5 and a standard deviation of 4. Determine the following.

3.

(a)

P X  11 .

(b)

P X  0 .

(c)

P 3  X  7  .

(d)

P  2  X  9  .

The comprehensive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. (a)

What

is

the

probability

that

a

sample

strength

is

less

than 6250 Kg/cm2 ? (b)

What is the probability that sample strength is between 5800 Kg/cm2 and 5900 Kg/cm2 ?

4.

The line width for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer. (a)

What is the probability that a line width is greater that 0.62 micrometer ?

(b)

What is the probability that a line width is between 0.47 and 0.63 micrometer ?

5.

The speed of a file transfer from a server on a campus to a personal computer at a student’s college on a weekday evening is normally distributed with a mean of 60 kilobits per second and a standard deviation of 4 kilobits per second. (a)

What is the probability that the file will transfer at a speed of 70 kilobits per second or more ?

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Chapter 2: Special Probability Distributions

(b)

What is the probability that the file will transfer at a speed of less than 58 kilobits per second ?

6.

In an accelerator center, an experiment needs a 1.41 cm thick aluminum cylinder. Suppose that the thickness of a cylinder has a normal distribution with a mean 1.41 cm and a standard deviation of 0.01 cm. What is the probability that a thickness of a cylinder is greater than 1.42 cm ?

7.

The average time for a mail carrier to cover his route is 380 minutes and the standard deviation is 16 minutes. If one of these trips is selected at random, find the probability that the carrier will have the following route time. Assume the variable is normally distributed.

8.

(a)

At least 350 minutes.

(b)

At most 395 minutes.

An electronic office product contains 5000 electronic components. Assume that the probability that each component operated without failure during the useful life of the product is 0.999, and assume that the components fail independently. Approximate the probability that 10 or more of the original 5000 components fail during the useful life of product.

9.

The reliability of an electrical fuse is the probability that a fuse, chosen at random from production, will function under its designed conditions. A random sample of 1000 fuses was tested and x = 27 defectives were observed. Calculate the approximate probability of observing 27 or more defectives, assuming that the fuse reliability is 0.98.

10.

A producer of soft drinks was fairly certain that her brand had a 10% share of the soft drink market. In a market survey involving 2500 customers of soft drinks, x = 211 expressed a preference for her brand. If the 10% figure is

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Chapter 2: Special Probability Distributions

correct, find the probability of observing 211 or fewer consumers who are prefer her brand of soft drink.

11.

Airlines and hotels often grant reservations in excess of capacity to minimize losses due to no-shows. Suppose the records of a hotel show that, on the average, 10% of their prospective guests will not claim their reservation. If the hotel accepts 215 reservations and there are only 200 rooms in the hotel, what is the probability that all guests who are arrive to claim a room will receive one?

12.

13.

Find the probability that 200 tosses of a coin will result in (a)

between 80 and 120 heads inclusive

(b)

less than 90 heads

Find the probability that on a true-false examination a student can guess correctly the answers to

14.

(a)

12 or more out of 20 questions.

(b)

24 or more out of 40 questions.

Ten percent of the bolts that a machine produces are defective. Find the probability that in a random sample of 400 bolts produced by this machine

15.

(a)

at most 30 bolts will be defective

(b)

between 35 and 45 bolts will be defective.

Find the probability of getting more than 25 sevens in 100 tosses of a pair of fair dice.

105

Chapter 2: Special Probability Distributions

Answer Exercise 2.6 1.

(a)

0.8461 (b)

0.8962 (c)

(d)

0.6372

(e)

0.0794

(f)

0.4591

(a)

0.93319

(b)

0.89435

(c)

0.38292

(d)

0.80128

3.

(a)

0.99379

(b)

0.13591

4.

(a)

0.0082

(b)

0.72109

5.

(a)

0.00621

(b)

0.308538

6.

0.1587

7.

(a)

0.9699

(b)

0.8264

8.

0.02203

9.

0.0708

10.

0.0051

11.

0.9441

12.

(a)

0.9962 (b)

0.0687

13.

(a)

0.2511 (b)

0.1342

14.

(a)

0.0567 (b)

0.6404

15.

0.0089

2.

0.0643

106

Chapter 2: Special Probability Distributions

EXERCISE CHAPTER 2 1.

Because not all airline passengers show up their reserved seat, an airline sells 125 tickets for a flight that holds only 120 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently. (a)

What is the probability that every passenger who shows up can take the flight ?

(b)

2.

What is the probability that the flight departs with empty seats ?

A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events

3.

(a)

at least one of the alarms is triggered.

(b)

more than seven of the alarms are triggered.

(c)

eight or fewer alarms are triggered.

Car color preferences change over the years and according to the particular model that the customer selects. In a recent year, suppose that 10% of all luxury cars sold were black. If 25 cars of that year and type are randomly selected, find the following probabilities

4.

(a)

at least five cars are black.

(b)

at most six cars are black.

(c)

exactly four cars are black.

(d)

between three and five cars (inclusive) are black.

(e)

more than 20 cars are not black.

Records shows that 30% of all patients admitted to a medical clinic fail to pay their bills and that eventually the bills are forgiven. Suppose the large set of prospective patients served by the clinic. Find these probabilities (a)

All the patients’ bills will eventually have to be forgiven.

(b)

One will have to be forgiven.

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Chapter 2: Special Probability Distributions

(c)

5.

None will have to be forgiven.

A safety engineer claims that only sixty percent of all workers wear safety helmets when they eat lunch at the workplace. Assume that his claim is correct, find the probability that four out of six workers chosen randomly will wear their helmets while having lunch at the workplace.

6.

A recent study of robberies for a certain geographic region showed an average of one robbery per 20,000 people. In a city of 80,000 people, find the probability of the following

7.

(a)

no robberies.

(b)

one robberies.

(c)

three or more robberies.

If approximately 2% of the people in a room of 200 people are left-handed, find the probability that exactly five people there are left-handed.

8.

If 3% of all cars fail the emissions inspection, find the probability that in a sample of 90 cars, at least four will fail.

9.

If a baseball player’s batting average is 0.320 (32%), find the probability that the player will get at most 26 hits in 100 times at bat.

10.

The percentage of Malaysians 25 years or older who have at least some college education is 50.9%. In a random sample of 300 of Malaysian 25 years old and older, what is the probability that more than 175 have at least some college education ?

11.

In a large shipment of automobile tires, 10% have a certain flaw. Ten tires are chosen at random to be installed on a car.

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Chapter 2: Special Probability Distributions

12.

(a)

Find the mean and variance of the flaw.

(b)

What is the probability at least three tires has flaw ?

Shafts manufactured for use in optical storage devices have diameters that are normally distributed with mean 0.652 cm and standard deviation 0.003 cm. The specification for the shaft diameter is 0.650  0.005 cm . (a)

What is the probability that the shafts manufactured by this process meet the specification?

(b)

If the mean is set to 0.650 cm, what must the standard deviation be so that 99% of the shafts will meet specification?

13.

When a customer places an order with Haniss’s On-Line Office Supplies, a computerized accounting information system (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past record indicated that the probability of customers exceeding their credit limit is 0.35. Suppose that, on a given day, 20 customers place orders. Assume that X is the number of customers that AIS detects as having their credit limit. (a)

What are the mean and standard deviation of the number of customers exceeding their credit limits ?

(b)

What is the probability that none customers will exceed their limits ?

(c)

What is the probability that at least thirteen customers will exceed their limits ?

14.

Let say that another supplier, Umar’s On-Line Service Supplies, is doing same as Haniss’s company. Past record indicated that the probability of customers exceeding their credit limit is 0.5 and on a given day, and 35 customers place their orders. (Hint: Use suitable approximation) (a)

What are the mean and standard deviation of the number of customers exceeding their credit limits ?

(b)

What is the probability that 15 of the customers will exceed their

109

Chapter 2: Special Probability Distributions

limits ? (c)

What is the probability that at least 13 customers will exceed their limits ?

15.

The number of messages received by a computer bulletin board is Poisson random variable with a mean rate of five messages per hour. What is the probability that (a)

three message are received in a given hour ?

(b)

six messages are received in 1.5 hours ?

(c)

fewer than four messages are received in 1.5 hours ?

ANSWER EXERCISE CHAPTER 2 1.

(a)

0.9961

(b)

0.9886

2.

(a)

1.00

(b)

0.997

(c)

0.086

3.

(a)

0.098

(b)

0.991

(c)

0.098

(d)

0.138

(e)

0.430

(f)

0.902

4.

(a)

0.0081

(b)

0.4116

(c)

0.2401

5.

0.31104

6.

(a)

(b)

0.0733

(c)

0.7619

7.

0.1563

8.

0.2859

9.

0.119

10.

0.00423

11.

(a)

 1 

(b)

0.0702

12.

(a)

0.8314

(b)

0.001519

13.

(a)

2.1331

(b)

0.0002

(c)

0.006

14.

(a)

  17.5   2.958 (b)

0.09425

(c)

0.04549

15.

(a)

0.1404

0.1367

(c)

0.05914

0.0183

2

 1.9

(b)

110

Chapter 2: Special Probability Distributions

SUMMARY CHAPTER 2

Formula Mean Variance

Formula Mean Variance

Formula

Condition Mean

Condition Mean Variance

Binomial Distribution n! P X  x    p x  q n x  n C x  p x  q n x x !  n  x  !   np

 2  npq Poisson Distribution e   x

P X  x  

 2 

x!

,

x  0, 1, 2, ..., 

Normal Distribution x  P Z      Poisson Approximation to the Binomial Distribution Use if n  30 and p  0.1   np Normal Approximation to the Binomial Distribution Use if n is large and np  5 and nq  5   np



2

 npq

111

Chapter 2: Special Probability Distributions

CORRECTION PAGE CHAPTER 2

112