Chapter 4. Probability Distributions

Chapter 4

Probability Distributions

Lesson 4-1/4-2

Random Variable

Probability Distributions

 This chapter will deal the construction of probability distribution.

 By combining the methods of descriptive

statistics in Chapter 2 and those of probability presented in Chapter 3.

 Probability Distributions will describe what will probably happen instead of what

actually did happen.

Combining Descriptive Methods and Probabilities

In this chapter we will construct probability distributions by presenting possible outcomes along with the relative

frequencies we expect.

Definitions

 A random variable is a variable (typically represented by X) that has a single

numerical value, determined by chance, for each outcome of a procedure.

 A probability distribution is a graph, table, or formula that gives the probability for each value of the random variable.

2 Types of Random Variables

 Discrete Random Variable has either a

finite number of values or countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they result from a counting process.

 Continuous Random Variable has infinitely many values, and those values can be

associated with measurements on a

continuous scale in such way that there are no gaps or interruptions.

Example – Page 192, #2

Identify the given variable as being discrete or continuous.

A. The cost of making a randomly selected movie.

B. The number of movies currently being shown in U.S.

theaters.

C. The exact running time of a randomly selected movie.

Discrete

Discrete

Continuous

Example – Page 192, #2

D. The number of actors appearing in a randomly selected movie.

E. The weight of the lead actor in a randomly selected movie.

Discrete

Continuous

Graphs

The probability histogram is very similar to a relative

frequency histogram, but the vertical scale shows probabilities.

Requirements for a Discrete Probability Distribution

1.

2.

P x ( )  1



P x

0  ( )  1

where x assumes all possible values

for every individual value of x

Describing a Distributions

 Center

 Value that indicates where the middle of the data set is located.

 Variation

 Measures the amount that the values vary among themselves

 Distribution

 Shape of the distribution of data

 Outliers

 Sample values that lie very far away from the vast majority of the other sample values

Mean, Variance and Standard Deviation of a Probability

Distribution

[ ( )]







 

2 



 

2 2 2

 ( )





 

2 2

 ( )





 

Mean

Variance

Variance (shortcut) Standard Deviation

Roundoff Rule for μ, σ, and σ²

 Round results by carrying one more decimal place than the number of decimal places used for the random variable x.

 If the values of x are integers, round μ, σ, and σ² to one decimal place.

Example – Page 192, #4

Determine whether a probability distribution is given. If probability distribution is described, find its mean and standard deviation.

When manufacturing DVDs for Sony, batches of DVD’s are randomly selected and the number of defects x is found for each batch.

x P x( )

0 0.502

1 0.365

2 0.098

3 0.011

4 0.001

P x( )  0.977  1



Not a probability distribution

Example – Page 192, #6

Determine whether a probability distribution is given. If probability distribution is described, find its mean and standard deviation.

The Telektronic Company provides life insurance policies for its top four executives, and the random variable x is the number of those employees who live through next year.

x P x( )

0 0.0000

1 0.0001

2 0.0006

3 0.0387

4 0.9606

P x( )  1



A probability distribution

Example – Page 192, #6

0 0.0000 0 0

1 0.0001 0.0001 0.0001

2 0.0006 0.0012 0.0024

3 0.0387 0.1161 0.3483

4 0.9606 3.8424 15.37

x P x( ) x P x ( )

3.9598



x² P x( )

15.7204



Example – Page 192, #6

3.9598  4





μ 



2 2 2

σ 



^

^

Example – Page 192, #6

Use the TI to find the mean, variance, and standard deviations.

x P(x) StatCALC

Identifying Unusual Results

 According to the range rule of thumb, most values should lie within 2 standard deviation of the mean.

 We can therefore identify “unusual”

values by determining if they lie outside these limits:

 Maximum usual value =  + 2

 Minimum usual value =  – 2

Identifying Unusual Results Probabilities

 Rare Event Rule

 Given the assumption that boys and girls are equally likely the probability of a particular observed event (such as 13 girls in 14 births) is extremely small, we conclude that the assumption is probably not correct.

 Unusually high

 x successes among n trials is an unusually high number of successes if P(x or more) is very small (such as 0.05 or less).

 Unusually low

 x successes among n trials is an unusually low

number of successes if P(x or fewer) is very small (such as 0.05 or less).

Example – Page 194, #18

x (girls) P(x)

0 0.000

1 0.001

2 0.006

3 0.022

4 0.061

5 0.122

6 0.183

7 0.209

8 0.183

9 0.122

10 0.061

11 0.022

12 0.006

13 0.001

14 0.000

Assume that in a test of gender-selections

technique, in clinical trial results in 12 girls and 14 births. Refer to Table 4-1 and find the

indicated probabilities.

A. Find the probability of exactly 12 girls in 14 births.

P x(  12)  0.006

Example – Page 194, #18

x (girls) P(x)

0 0.000

1 0.001

2 0.006

3 0.022

4 0.061

5 0.122

6 0.183

7 0.209

8 0.183

9 0.122

10 0.061

11 0.022

12 0.006

13 0.001

14 0.000

P x(  12)

P x( 12) P x( 13) P x( 14)

     

B. Find the probability of 12 or more girls in 14 births.

0.006 0.001 0.000 0.007

  



Example – Page 194, #18

C. Which probability is relevant for determining whether 12 girls in 14 births is unusually high: the results from part (A) or part (B)

Part (B), the occurrence of 12 girls among 14 would be unusually high if the probability of 12 or more

girls is very small 0.007 ≤ 0.05

Example – Page 194, #18

D. Does 12 girls in 14 births suggest that the gender- selection technique is effective? Why or why not?

Yes, because the probability of 12 or more girls is

very small (0.007). The result of 12 or more girls could not easily occur by chance.

Definition

 The expected value of a discrete

random variable is denoted by E, and it represents the average value of the outcomes.

 It is obtained by find the value of E

 

E 



 The expected value of a discrete

random variable is denoted by E, and it represents the average value of the outcomes.

 It is obtained by find the value of E

 

E 



Example – Page 192, #12

When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the

probability of losing $5 is 20/38.

Example – Page 192, #12

When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the probability of losing $5 is 20/38.

A. If you bet $5 on the number 7, what is your expected value?

x P x( ) x P x( )

1 175

175 38 38

37 185

5 38 38

10 38



 



0.2631 38

   

The expected loss is $0.263 for a $5 bet.

Example – Page 192, #12

When you give the casino $5 for a bet on the number 7 in a roulette, you have 1/38 probability of winning $175 and 37/38 probability of losing $5. If you bet $5 that outcome is an odd number, the probability of winning $5 is 18/38, and the probability of losing $5 is 20/38.

B. If you bet $5 that the outcome is an odd number, what is your expected value?

x P x( ) x P x( )

18 90

5 38 38

20 100

5 38 38

10 38



 



0.2631 38

   

The expected loss is $0.263 for a $5 bet.

Lesson 4-3

Binomial Probability Distribution

Binomial Probability Distribution

 Specific type of discrete probability distribution

 The outcomes belong to two categories

 pass or fail

 acceptable or defective

 success or failure

Example of a Binomial Distribution

Suppose a cereal manufacturer puts pictures of famous athletes on cards in boxes of cereal, in the hope of

increasing sales. The manufacture announces that 20%

of the boxes contain a picture of Tiger Woods, 30% a picture of Lance Armstrong, and the rest a picture of Serena Williams.

You buy 5 boxes of cereal. What’s the probability you get exactly 2 pictures of Tigers Woods?

Requirements for a Binomial Probability Distribution

1) The procedure has a fixed number of trials

2) Trials are must be independent. (The

outcome of any individual trial doesn’t affect the probabilities in the other trials.)

3) Each trial must have all outcomes classified into two categories

4) Probabilities must remain constant for each trial.

Example – Page 203, #4

Determine whether the given procedure results in a binomial distribution.

Rolling a loaded dice 50 times and find the number of times that 5 occurs.

Binomial

Notation for a Binomial Distribution

 There are n number of fixed trials

 Let p denote the probability of success

 Let q denote the probability of failure

 q = 1 – p

 Let x denote the number of success in n independent trials:

 Let P(x) denotes the probability of getting exactly x successes among the n trials

x n 0  

Important Hints

 Be sure that x and p both refer to the same category being called a success.

 When sampling without replacement, the events can be treated as if they were

independent if the sample size is no more than 5% of the population size.

 Where n  0.05N

Mathematical Symbols

Phrase Math Symbols

“at least” or “no less than” ≥

“more than” or “greater than” >

“fewer than” or “less than” <

“no more than” or “at most” ≤

“exactly” =

Methods for Finding Probabilities of a Binomial Distribution

 Using the Binomial Probability Formula

 Using Table A-1 in Appendix A

 Using the TI

Binomial Probability Formula

 ^  ^

 n 

n x x

P x ! p q

( ) ! !

Number of outcomes with exactly x

successes among n trials

Probability of x successes among

n trials for any particular order







P x ( ) C p q

Binomial Probability Formula







P x ( ) C p q

for x = 0, 1, 2, ……,n

where

n = number of trials

x = number of success among n trials

p = probability of success in any one trial

q = probability of failure in any one trial (q = 1 – p)

Example 1 – Cereal

Suppose you buy 5 boxes of cereal. Where n = 5 and p = 0.2. What’s the probability you get exactly 2 pictures of Tiger Woods?

 

5 2

5 5!

2 C 2! 5 2 ! 10

    

  

 

First find the total number of outcomes.

There are 10 ways to get 2 Tiger pictures in 5 boxes.

Example 1 – Cereal

Suppose you buy 5 boxes of cereal. Where n = 5 and p = 0.2.

What’s the probability you get exactly 2 pictures of Tiger Woods?

2 3

( 2) 10(0.20) (0.80) 0.2048

P X   

There are 10 ways to get 2 Tiger pictures in 5 boxes.

2^nd Vars

Example – Cereal

The following table show the probability distribution function (p.d.f) for the binomial random variable, X.

X = Tiger P(X)

0 0.32768

1 0.4096

2 0.2048

3 0.0512

4 0.0064

5 0.00032

Sum 1

(5, 0.20, 0) 0.32768 (5, 0.20,1) 0.4096 (5, 0.20, 2) 0.2048 (5, 0.20,3) 0.0512 (5, 0.20, 4) 0.0064 (5, 0.20,5) 0.00032 Binompdf

Binompdf Binompdf Binompdf Binompdf Binompdf













Example – Cereal

The following table show the cumulative distribution function (c.d.f) for the binomial random variable, X.

0 1 2 3 4 5

( ) 0.32768 0.4096 0.2048 .0512 .0064 .00032

( 0) ( 1) ( 2) ( 3) ( 4) ( 5)

( )

0.32768 0.73728 0.94208 0.99328 0.99968 1

pdf

cdf

X P X

P X P X P X P X P X P X

P X      

(5, 0.20, 0) 0.32768 (5, 0.20,1) 0.73728 (5, 0.20, 2) 0.94208 (5, 0.20,3) 0.99328 Binomcdf

Binomcdf Binomcdf Binomcdf









(5, 0.20, 4) 0.99968 (5, 0.20,5) 1

Binomcdf Binomcdf





Example – Cereal

0 1 2 3 4 5

( ) 0.32768 0.4096 0.2048 .0512 .0064 .00032

( 0) ( 1) ( 2) ( 3) ( 4) ( 5)

( )

0.32768 0.73728 0.94208 0.99328 0.99968 1

pdf

cdf

X P X

P X P X P X P X P X P X

P X      

Construct a histogram of the pdf and cdf using X[0, 6]₁ and Y[0, 1]_0.01

pdf cdf

TI – Binomial Probability

 Computing exact probabilities

 2^nd/Vars/binompdf

• binompdf(n, p, x)

 pdf: probability distribution function

 Computing less than or equal to probabilities

 2^nd/Vars/binomcdf

• binomcdf(n, p, x)

 cdf: cumulative distribution function

Example – Page 204, #18

 



P x ( )

C

p q

Use the Binomial Probability Formula. Assume that a procedure yields a binomial distribution.

n  6, x  2, p  0.45

 

 

P x ( 2) 15(0.45) 0.55

0.2779 0.278

 

Binomial Probability Table

Example – Page 204, #12

Using Table A-1. Assume that a procedure yields a binomial distribution.

n  7, x  2, p  0.01

P x(  2)  0.002

Example – Page 204, #14

Using Table A-1. Assume that a procedure yields a binomial distribution.

n  6, x  5, p  0.99

P x(  5)  0.057

Example - Nielson Media

According to Nielson Media research, 75% of US households have cable TV.

A). In a random sample of 15 households, what is the probability that 10 have cable?



P x( 10)

 binompdf 0.2361

(15,0.75,10)

x = households, p = 0.75, n = 15

Example - Nielson Media

The probability of getting exactly 10 households out 15

with cable is 0.1651. In 100 trials of this experiment we would expect (100)(0.1651) = 17 trials to result in 10 households

with cable.

Example - Nielson Media

According to Nielson Media research, 75% of US households have cable TV.

B). In a random sample of 15 households, what is the probability that fewer than 13 have cable?

P x( 13)  P x(  0)  P x(   1) ... P x( 12)

binomcdf(15, 0.75, 12)  0.7639 P x(  12) 

Example - Nielson Media

There is 0.7639 probability that in a random sample of 15 households, less than 13 will have cable.

In 100 trials of this experiment, we would expect (0.7639)(100) = 76 trials to result in fewer than 13 households that have cable.

Example – Nielson Media

According to Nielson Media research, 75% of US households have cable TV.

C). In a random sample of 15 households, what is the probability that at least 13 have cable?

P x(  13)  P x(  13)  P x(  14)  P x( 15)



  binomcdf 0.2361

1 (15,0.75,12)

 1 P x(  12)

Example – Nielson Media

There is 0.2361 probability that in a random sample of 15 households, at least 13 will have cable. In 100 trials at

of this experiment we would expect about (100)(0.2361) = 24 trials to result in at least 13 households have cable

Example – Page 205, #28

An article in USA Today stated that “Internal surveys paid by the directory assistance providers show that even the most accurate companies give out wrong numbers 15% of the time.” Assume that you are

testing such a provider by making 10 requests and also assume that the provider gives the wrong telephone

number 15% of the time.

A). Find the probability of getting one wrong number.



P x( 1) 

 binompdf 0.347

(10,0.15,1)

x = the number of wrong answers, n = 10, p = 0.15

Example – Page 205, #28

B). Find the probability of getting at most one wrong number.

P x(  1) P x(  0)  P x(  1)

0.544

 binomcdf (10,0.15,1) 

C). If you do get at most one wrong number, does it appear that the rate of wrong numbers is not 15%, as claimed?

No, since 0.544 > 0.05 “at most one wrong number” is not an unusual occurrence when the error rate is 15%

Example – Page 206, #32

A study was conducted to determine whether there were significant

difference between medical students admitted through special programs (such as affirmative action) and medical students admitted through the regular admission criteria. It was found that the graduation rate was 94% for the medical students admitted through special programs.

A). If 10 students from the special programs are randomly selected, find the probability that at least nine of them graduated.

P x(  9)  P x(  9)  P x( 10)  1 P x(  8)

 1 binomcdf(10,0.94,8) 0.8824



x = the number of special program student who graduate n = 10, p = 0.94

Example – Page 206, #32

B). Would it be unusual to randomly select 10 students from the special programs and get only seven that graduate?

Why or why not?

 

P x( 7) binompdf (10,0.94,7) 0.01681



Yes, since 0.017 < 0.05 getting only 7 graduates would be unusual result.

Lesson 4-4

Binomial Distribution: Mean,

Variance and Standard Devation

Any Discrete

Probability Distribution Formulas

Std. Dev   = [  ^x

^{P(x) ] – µ}

Mean µ =



Variance





Binomial Distribution

Std. Dev.   = n

p

q

Where

n = number of fixed trials

p = probability of success in one of the n trials q = probability of failure in one of the n trials

Mean µ = n ^• p

Variance



Interpretation of Results

Maximum usual values = µ ^{+ 2} 

Minimum usual values = µ ^{– 2} 

The range rule of thumb suggests that values are unusual if they lie outside of these limits:

Example – Page 210, #2

Finding , , and Unusual Values. Assume that a procedure yields a binomial distributions with n trials and the probability of success for one trial is p.

n = 250, p = 0.45

250(0.45) 112.5

  np





(250)(0.45)(1 0.45) 7.86

  npq

 



min 2 112.5 2(7.86) 96.8 max 2 112.5 2(7.86) 128.2

 

 

    

    

Example – Page 210, #6

Several students are unprepared for a multiple-choice quiz with 10 questions, and all of their answers are guesses.

Each question has five possible answers, and only one of them is correct.

A). Find the mean and standard deviation for the number of correct answers for such students.

n  10 p 1

5 0.20

 

q  1 0.20  0.80

μ  np

10(0.20) 2





σ  npq

(10)(0.20)(0.80) 1.264

1.3







Example – Page 210, #6

B). Would it be unusual for a student to pass by guessing and getting at least 7 correct answers? Why or why not?

Yes, since 7 is greater than or equal to the maximum usual value, it would be unusual for a student to

pass by getting at least 7 correct answers.

max   2  2 2(1.3)  4.6

Example – Page 210, #10

The Central Intelligence Agency has specialists who analyze the frequencies of letters of the alphabet in an attempt to decipher intercepted messages. In standard English text, the letter r is used at a rate of 7.7%

A). Find the mean and standard deviation for the number of times the letter r will be found on a typical page of 2600 characters.

n  2600 p  0.077

q 1 0.077 0.923

 



μ  np

2600(0.077) 200.2





σ  npq

(2600)(0.077)(0.923) 13.6





Example – Page 210, #10

B). In the intercepted message sent to Iraq, a page of

2600 characters is found to a have the letter r occurring 175 times. Is this unusual?

μ 2σ unusual values are outside of these boundaries.

200.2  2(13.6)  173 and 227.4

No, since 175 is within the above limits it would not be considered an unusual result.

Example – Page 212, #16

Car Crashes: For drivers in the 20 – 24 age bracket, there is a 34% rate of car accidents in one year (based on the data from the National Safety Council). An insurance investigator finds that in a group of 500 randomly selected drivers age

20 – 24 living in New York City, 42% had accidents in the last year.

A) How many drivers in the New York City group of 500 had accidents in the last year?

(500) (0.42)  210

Example – Page 212, #16

B) Assuming that the same 34% rate applies to New York City, find the mean and standard deviation for the number of people in groups of 500 that can expected to have

accidents

500(0.34) 170.00

  np





 

 



500(0.34)(1 0.34) 10.6

npq

Example – Page 212, #16

C) Based on the preceding results, does 42% result for the New York City drivers appear to be unusually high when compared to the 34% rate for the general population?

Does it appear that higher insurance rates for New York City drivers justified?

max 2

170 2(10.6) 191.2

 

 

  

Yes, the result of 210 accidents for NYC drivers are

unusually high if the 34% rate for the general population applies. Yes, it appears that the higher rates for NYC drivers are justified.

Lesson 4-5

The Poisson Distribution

Examples of Poisson Distribution

 Planes arriving at an airport

 Arrivals of people in a line

 Cars pulling into a gas station

Definition

The Poisson distribution is a discrete probability distribution that applies to occurrences of some event over a specified interval. The random variable x is the number of occurrences of the event in an interval. The interval can be time, distance, area, volume, or some similar unit.

P(x) =

^e

x!

Formula

Requirements for a Poisson Distribution

 Random variable (x) is the number of occurrences of an event over some interval.

 Occurrences must be random

 Occurrences must be independent of each other

 Occurrences must be uniformly distributed over the interval being used.

 mean is μ

 Standard deviation is ^σ  ^μ

Difference from a Binomial Distribution

 The binomial distribution is affected by the

sample size n and the probability p, whereas the Poisson distribution is affected only by

the mean .

 In a binomial distribution the possible values of the random variable x are 0, 1, … n, but a Poisson distribution has possible x values of 0, 1, …., with no upper limit.

Example – Page 216, #2

Use a Poisson Distribution to find probability.

If μ = 0.5, find P(2).

x e

P x x

μ μ

( ) !

 

 

P e

2 0.5

(2) 0.5

2!

0.0758

 

 



Example – Page 216, #6

Currently, 11 babies are born in the village of Westport (population 760) each year.

A). Find the mean number of births per day.

μ 11 0.0301

 365 

Example – Page 216, #6

B). Find the probability that on a given day, there is no births.

P x poissonpdf P x

( 0) (11/ 365,0)

( 0) 0.970312

 

 

2^ndvars/C:poissonpdf(μ,x) 11

  365 x = births

Example – Page 216, #6

C). Find the probability that on a given day, there is at least one birth.

P x(  1) P x(  1) P x(  2) ...  P x(  11)

  

 



P x

poissonpdf

1 ( 0)

1 (11/ 365,0)

0.0297

Example – Page 216, #6

D). Based on the preceding results, should medical birthing personnel be on permanent standby, or should they be called in as needed? Does this mean that Westport mothers might not get the immediate medical attention they would be likely to get in a more populated area?

The personnel should be called as needed. Yes; this does mean that women giving birth might not get

the immediate attention available in more populated areas.

Example – Page 216, #8

Homicide Deaths: In one year, there were 116 homicide deaths in Richmond, Virginia (Based on “A Classroom Note on the

Poisson Distribution: A Model for Homicidal Deaths in Richmond, VA for 1991,” in Mathematics and Computer

Education, by Winston A. Richards). For a randomly selected day, find the probability that the number of homicide deaths is a. 0 b. 1 c. 2 d. 3 e. 4

Compare the calculated probabilities to these actual results:

268 days (no homicides); 79 days (1 homicides); 17 days (2 homicides); 1 day (3 homicides); no days with more than 3 homicides.

Example – Page 216, #8

Let x = the number of homicides per day

116 0.3178

  365 

a. P(x = 0) = poissonpdf(0.3178, 0) = 0.7277 b. P(x = 1) = poissonpdf(0.3178, 1) = 0.2313 c. P(x = 2) = poissonpdf(0.3178, 2) = 0.0368 d. P(x = 3) = poissonpdf(0.3178, 3) = 0.0038 e. P(x = 4) = poissonpdf(0.3178, 4) = 0.0003

Example – Page 216, #8

x frequency Relative frequency P(x)

0 268 0.7342 = 268/365 0.7277

1 79 0.2164 0.2313

2 17 0.0466 0.0368

3 1 0.0027 0.0038

4 or more 0 0.0000 0.0004 (by subtraction)

365 1.000 1.0000

The agreement between the observed relative frequencies and the probabilities predicted by the Poisson formula

is very good.

Poisson as Approximation to Binomial

 The Poisson distribution is sometimes used to approximate the binomial distribution when n is large and p is small.

 Rule of thumb

 n  100

 np  10

 If the two conditions are satisfied we can use the Poisson distribution as an approximation to the binomial distribution where

  = np