Chapter 4 - StartLogicwellsmat.startlogic.com/.../sitebuilderfiles/honstat_ch4_cn.pdf · Probability Distributions This chapter will deal the construction of probability distribution

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


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 x0 ( ) 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

[ ( )] x P x

22 ( )

x P x

2 2 2( ) x P x

2 2( ) x P x

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.


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


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


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

x P x2 ( )

15.7204


3.9598 4 x P xμ ( )

x P x2 2 2σ ( ) μ 2

15.7204 3.9598 .04038

2σ σ 0.04038 0.2009 0.2


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

x P(x) StatCALC

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


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


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

B. Find the probability of 12 or more

girls in 14 births.

0.006 0.001 0.000

0.007


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


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 x P x( )

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 x P x( )


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.


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 175175

38 38

37 1855

38 38

10

38

E10

0.263138

The expected loss is $0.263

for a $5 bet.


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( ) ( )

18 905

38 38

20 1005

38 38

10

38

E10

0.263138

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.


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 n0

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 N0.05

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

x n xn

n x xP x p q

!

! !( )

Number of outcomes

with exactly x

successes among

n trials

Probability of x

successes among

n trials for any

particular order

x n x

n xP x C p q( )

Binomial Probability Formula

x n x

n xP 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!10

2 2! 5 2 !C

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) 0.2048P X

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

2nd 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 XP 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 XP 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

2nd/Vars/binompdf

• binompdf(n, p, x)

pdf: probability distribution function

Computing less than or equal to probabilities

2nd/Vars/binomcdf

• binomcdf(n, p, x)

cdf: cumulative distribution function


x n xP x C p q6 2( )

Use the Binomial Probability Formula. Assume that a

procedure yields a binomial distribution.

n x p6, 2, 0.45

P x6 22( 2) 15(0.45) 0.55

0.2779 0.278

Binomial Probability Table


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


n x p7, 2, 0.01

P x( 2) 0.002


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


n x p6, 5, 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


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.



have cable TV.

B). In a random sample of 15 households, what is the

probability that fewer than 13 have cable?

P x P x P x P x( 13) ( 0) ( 1) ... ( 12)

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


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


have cable TV.

C). In a random sample of 15 households, what is the

probability that at least 13 have cable?

P x P x P x P x( 13) ( 13) ( 14) ( 15)

binomcdf

0.2361

1 (15,0.75,12)

P x1 ( 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


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


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

P x P x P x( 1) ( 0) ( 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, #32A 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 P x P x( 9) ( 9) ( 10) 1 ( 8)P x

binomcdf1 (10,0.94,8)

0.8824

x = the number of special program student who graduate

n = 10, p = 0.94


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 2 • P(x) ] – µ 2

Mean µ = [x • P(x)]

Variance 2= [ x 2 • P(x) ] – µ 2

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 2 = n • p • q

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:


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


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

p1

0.205

q 1 0.20 0.80

npμ

10(0.20)

2

npqσ

(10)(0.20)(0.80)

1.264

1.3


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


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

npμ

2600(0.077)

200.2

npqσ

(2600)(0.077)(0.923)

13.6


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.


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


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


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

DefinitionThe 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) = where e 2.71828µ x • e -µ

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.


Use a Poisson Distribution to find probability.

If μ = 0.5, find P(2).

x eP x

x

μμ( )

!

eP

2 0.50.5(2)

2!

0.0758


Currently, 11 babies are born in the village of Westport

(population 760) each year.

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

11μ 0.0301

365


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

2ndvars/C:poissonpdf(μ,x)

11

365 x = births


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

one birth.

P x P x P x P x( 1) ( 1) ( 2) ... ( 11)

P x

poissonpdf

1 ( 0)

1 (11/ 365,0)

0.0297


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.


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.


Let x = the number of homicides per day

1160.3178

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


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

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Chapter 4 - StartLogicwellsmat.startlogic.com/.../sitebuilderfiles/honstat_ch4_cn.pdf · Probability Distributions This chapter will deal the construction of probability distribution