Chapter 4 Discrete Probability Distributions. PROBABILITY AND STATISTICS CHAPTER 4 NOTES DISCRETE...

Chapter 4

• Discrete Probability Distributions

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS I. INTRODUCTION TO RANDOM VARIABLES AND PROBABILITY DISTRIBTIONSA. Random Variable x.

1. x represents a numerical outcome which could be a count or a

measurement resulting from a statistical experiment.a) For example:

Take a multiple choice test with 20 questions. Your raw score will be the number of

correct responses.1)

x = number of correct responses.2)

We can list the values for x: 0, 1, 2, 3,….20.3)

A possible value for the random variable is x = 18. b) For example:

Measure the mileage you drive between your home and your job.

1)x = number of miles you drive

2)We cannot list ALL of the possible responses here, since all

decimal numbers are possible.

3)A possible value for the random variable is x = 26.835 miles.

2. The term random is used because the particular value it takes in one

trial of an experiment occurs by chance.

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS

B. Types of random variables.1. Discrete Random Variables.

a) Values can be listed

b) Usually determined by a counting process.

c) A discrete random variable takes on only certain values.

d) The only values are usually integers, however, an exception would

be a shoe size, which could include half sizes. However, a shoe size

would not be 7.23.2. Continuous Random Variables

a) Possible values cannot be put in a list since there are infinitely

many.b) Usually

determined by a measuring process.c) All measures in an

interval (including decimals and fractions) are candidates for replacement.

d) No values are skipped.


C. Probability distribution for a discrete random variable1. List each value x that may occur in the

experiment and assign the probability of its occurrence P(x).

a) P(x) = number of times x has occurred divided by total number

of times the experiment is performedb) The sum of all the

simple probabilities is 1. ∑ P(x) = 1c) The expected

value of a discrete probability distribution (also called the mean) is denoted μ.

1)μ = ∑xP(x)

a. To calculate μ, make a table.

1. List the values of x in a vertical column I.

2. List the corresponding probabilities in vertical

column II.

3. In vertical column III, calculate the product of

each value of x times its probability.

4. The expected value of x is the sum of the

products listed in column III.


d) The standard deviation of an probability distribution is denoted σ.

1)Since we use the same table in which we have calculated the

expected value, columns I and II already are formed and have x and P(x)

entries.2)

Subtract the mean μ from each value of x, thus computing the

deviations (x – μ). Form a vertical column IV for these deviations.

3)Square each deviation and form vertical column V for

(x – μ)2, the deviations squared.4)

Multiply each deviation squared by the corresponding

probability from column II.

Put these products into vertical column VI.5)

Add the entries in column VI.6)

Find the square root of the sum of column VI.


D. Probability distribution of a continuous random variable.1. At this time we will not be required to

evaluate these probabilities, but will deal with them in later chapters.

Section 4-1 Example: Page 203 #30Cats 0 1 2 3 4 5

Households 1941 349 203 78 57 40

a) Construct a frequency distribution

Find the total number of experimentsDivide the frequency of each x by the total.

x Freq/Total P(x)

0

1

2

3

4

5

.728

.029

.076

.131

.021

.015

1941/2668

40/2668

203/2668

78/2668

57/2668

349/2668

CHECK!!!Are all the P(x) values between 0 and 1?Do all the P(x) values add up to 1? (Or within a hundredth or so of 1?)

1.000

2668


Households 1941 349 203 78 57 40

b) Find the mean.Multiply each x value by its probability.

x P(x) x P(x)

0

1

2

3

4

5

.728

.029

.076

.131

.021

.015

Add these up to get the mean (µ)

.000

.087

.152

.131

.084

.075

The Round Off Rule (Used for the Mean, Variance, & Standard Deviation)Round to one decimal place more than is used for the x-values.

.529

Since these x values were all whole numbers, we round to the nearest tenth.The mean is .5

x =x =

x =

x =

x =x =


Households 1941 349 203 78 57 40

c) Find the variance.

x P(x) x-µ (x-µ)2 P(x)(x-µ)2

0 .728

1 .131

2 .076

3 .029

4 .021

5 .015

Subtract the mean from each x value (remember that the mean is .5)

0 – 0.5

1 - 0.5

-0.5

0.5

2 - 0.5

3 - 0.5

4 - 0.5

5 - 0.5

1.5

2.5

3.5

4.5

Square the answers of (x - µ) to make all values positive.

(-.5)2

(.5)2

(1.5)2

(2.5)2

(3.5)2

(4.5)2

.25

.25

20.25

2.25

6.25

12.25

(.728)(.25)

(.131)(.25)

(.076)(2.25)

(.029)(6.25)

(.021)(12.25)

(.015)(20.25)

.182

.181

.304

.171

.257

.033

Multiply the probability of each x by the square of x - µ -- (P(x)(x-µ)2)The sum of this column (rounded to the nearest 10th) is the variance.

1.128

The variance is 1.1

Section 4-1 Example: Page 203 #30d) Find the standard deviation.

The standard deviation is the square root of the variance.

The square root of 1.1 is 1.05, which rounds to 1.1 (remember the round off rule).So, in summary, the mean was .5, the variance was 1.1, and the standard deviation was 1.1What this means is that in this one small town, the average household has .5 cats, with a standard deviation of 1.1.

𝜎=√𝜎2

Identify each of the following as either a discrete or continuous random variable.A. The number of people who are in a car B. The number of miles you drive in one week. C. Weight of a box of cereal. D. The number of boxes of cereal you buy in a year. E. Length of time you spend for lunch. F. The number of patients on a psychiatric ward in one day. G. The volume of blood which is transfused during an operation.

Discrete -- countable

Continuous -- measured






In a personality inventory test for passive-aggressive traits, the possible scores are 1= extremely passive; 2 = moderately passive; 3= neither; 4 = moderately aggressive; and 5 = extremely aggressive. The test was administered to a group of 110 people and the results were as follows:

Construct a probability distribution table and compute the expected value (the mean) and the standard deviation.

X (score) 1 2 3 4 5F (Freq.) 19 23 32 26 10

x f P(x) x * P(x) x - µ (x - µ)2 (x - µ)2 * P(x)

1 19

2 23

3 32

4 26

5 10

Σ 110

X (score) 1 2 3 4 5F (Freq.) 19 23 32 26 10

.173

.209

.291

.236

.091

1.000

.173

.418

.873

.944

.455

2.863 = 2.9

-1.9 3.61 .625

-.9 .81 .169

.1 .01 .003

1.1 1.21 .286

2.1 4.41 .401

1.484 =1.5

The Mean (Expected Value) is 2.9, and the Variance is 1.5Remember that the standard deviation the square root of the

variance.The standard deviation is 1.2

Your assignment is pages 201 – 203, # 9 - 33 Odd

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS SECTION 4-2 I. BINOMIAL PROBABILITIES

In the previous chapter, we looked at statistical experiments and computed the probabilities of specified events. We will now examine a particular type of statistical experiment called a binomial experiment.

A. Characteristics of a binomial experiment.1. The same action is

repeateda)

Conditions for repetition must be identical.b)

One trial must be independent of all others. (the results of one trial

cannot affect another.)c)

The number of trials is n.2. A binomial

experiment must have exactly two outcomes.a)

Success (defined in the problem)b)

Failure (all outcomes that do not qualify as successes.)3. On an individual

triala)

P (success) = pb)

P (failure) = qc)

q = 1 – p (They are complementary events)

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS SECTION 4-2

4. The number of trials that are successful is denoted r where r ≤ n.

B. The basic problem is to find the probability of getting exactly r successes out

of n trials1. In a binomial

experiment we use r as the random variable.a)

r counts the number of successful outcomes.b)

r is a discrete random variable.c)

For each experiment, the possible values of r can be listed, r = 0, 1,

2, 3, … n. The values range from r = 0 (no successful outcomes) to r

= n (all outcomes are successful).2. The probability of

getting exactly 0 successes (all failures) is denoted P(0), the probability of getting exactly 1 success is

denoted P(1), and so on.a)

Since the categories of no successes, exactly one success, exactly

two successes, etc. are mutually exclusive events, we can add their

probabilities to answer a question using the OR combination.b)

The sum of all probabilities = 1. P(0) + P(1) + P(2) + P(3) + .. P(n) = 1.


C. Methods of determining the probability of r successes out of n trials.

1. Formula

a) P(r) = Cn,r pr qn-r

2. Table of Binomial Probabilities found in Appendix II of your text.

3. The TI-84 can solve a binomial distribution problem for us, so that’s the

way we are going to go.

a) Press 2nd VARS (DISTR).

b) This takes you to a screen that lists 16 different types of statistical

distributions. We will be returning to this screen A LOT!!

1) You are looking for the 11th and/or 12th type of distribution

listed.

a). Pressing the Alpha and A keys takes you to the

binompdf function.

1. binompdf gives you the probability that you will

receive PRECISELY (exactly) r successes.


b) Pressing the Alpha and B keys takes you to the binomcdf

function.

1.binomcdf gives you the CUMULATIVE probability for up

to r successes.

c) Once you have selected the type of distribution you want,

either binompdf or binomcdf, you enter the parameters in

identical fashion.

1. The calculator screen will prompt you for trials, p, and

x.

a. enter the values asked for, highlight Paste, and

press Enter.

b. Press the Enter key again to receive your answer.


D. THE MEAN AND STANDARD DEVIATION OF THE BINOMIAL DISTRIBUTION

1. The binomial distribution’s special properties allow for an easier formula

to compute the expected value (mean), the variance, and the standard

deviation.a.

To compute the mean of a binomial distribution, simply multiply

the number of trials by the probability of success on each trial.

Use b.

The variance of a binomial distribution is found by .

The standard deviation of a binomial distribution is 2. Application of the

binomial distributiona.

Useful in determining the probability of meeting certain quotas or

specifications under success/failure conditions.b.

For these problems, you’ll be determining the value of n (the

number of trials) necessary to meet the given quota.

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS SECTION 4-2 EXAMPLE As an example, let’s go through Question Number 2 on Page 235.

A. n = 10 (number of questions)p = 1/5, or .20 (probability of guessing the

correct answer on one question)r = 10 for part a, 0 for part b, is greater than or

equal to 1 for part c, and greater than or equal to 5 for part d.

B. Let’s do part a.1. This is asking us

for the probability of getting PRECISELY 10 correct, so we use the binompdf.

2. 2nd VARS Alpha Math (A) gives us binompdf.

You can also just scroll until you see binompdf in the list.

3. Fill in the values 10,.2,10, highlight Paste and press Enter.

a. Your screen should have binompdf(10,.2,10) showing.4. Press Enter. The

screen will tell you that the probability of getting all ten questions correct is 1.024 x 10-7.

That means that you have a 0.0000001024 chance of getting all 10

correct.

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS SECTION 4-2 EXAMPLE

C. Part b asks for the probability of getting none of them correct.

1. This is also asking for a precise probability, so we again go to binompdf.

2. This time, we enter 10,.2,0 into the calculator and get the answer of

.1073741824. That means that

you have a 10.74% chance of getting a zero on that quiz.D. Part c asks for the probability of getting at least one right.

This is a CUMULATIVE probability, as we need to add the probability of 1 to the probability of 2, all the way up to the probability of 10.

1. The smart way to do this is to realize that the complement of “at least

1” is “0”. Since we have already found P(0) to be .107, we can subtract

that from 1 and get the answer of .893

E. Part d asks for the cumulative probabilities for r = 5,6,7,8,9, and 10. The

calculator can’t do that, but it can find the probability of r = 0,1,2,3, and 4. This is the complement of 5 and up, so we can subtract this from 1 to get our

answer.

PROBABILITY AND STATISTICSCHAPTER 4 NOTESDISCRETE PROBABILITY DISTRIBUTIONS SECTION 4-2 EXAMPLE

1.binomcdf(10,.2,4) gives the cumulative probabilities for r = 0,1,2,3, and

4. This

number is .9672. When

this is subtracted from 1, we get .0328. So, you

have about a 3% chance of getting at least a 50 on that quiz.Maybe

we should try studying!!F. The calculator will ALWAYS give you the

probability of less than or equal to the number you enter for x.

If you want greater than x, you need to use the complement rule.

PROBABILITY AND STATISTICSCHAPTER 4 NOTES SECTION 4-2 EXAMPLES For each of the following:

State whether or not it is a binomial experiment.If it is a binomial experiment, describe “success” and “failure”.Identify values for n, q, and the range of values for r.

U.S.A. Today reported on July 27, 1990, that 70% of the people questioned said that they watched less T.V. than they did a year ago, 22% said they watched the same amount, and 8% said they watched more. Find the probability that exactly 3 out of a randomly selected group of 5 will say they watch less T.V. this year than last.

This a binomial experiment, with success being “watch less T.V. this year than last” and failure being “watch the same or more T.V. than last year”.

n = 5p = .70 (probability that one person will say they watch

less T.V. this year)q = .30 (1-.7)r = 3 (number of successes)



There are 20 M&M candies in a dish. 8 are brown, 3 are red, 5 are green and 4 are yellow. Two candies are picked from the dish at random. What is the probability that both are red?

This is not a binomial experiment, since the trials are not independent.The probability that the first one is red is 3/20.The probability of the second being red, given that the first one was, is 2/19.

(3/20)*(2/19) = 6/380 = 3/190 = .0158



A ten question multiple choice test is given. Each question has four choices. You did not study and have no clue as to any of the answers, so you have to randomly guess each answer. What is the probability you guess exactly 6 correctly (and pass)?

This is a binomial experiment.Success is “guessing the answer correctly” and failure is “guessing incorrectly”n = 10 (the number of questions)p = .25 (since there is one chance in four of guessing each answer correctly)q = .75 (1 - .25)r = 6



U.S.A. Today reported on July 27, 1990, that 70% of the people questioned said that they watched less T.V. than they did a year ago, 22% said they watch the same amount, and 8% said they watch more. Find the probability that exactly 3 out of a randomly selected group of 5 will say they watch less T.V. this year than last.

On the calculator, use binompdf, since we want exactly 3 successes.2nd VARS binompdf, then enter 5, .7, and 3.Paste and Enter will give you .3087, which rounds to .309.)



U.S.A. Today reported on July 27, 1990, that 70% of the people questioned said that they watched less T.V. than they did a year ago, 22% said they watch the same amount, and 8% said they watch more. Find the probability that between 2 and 4 (inclusive) people out of 5 will say they watched less T.V. this year than last year.

Use the binompdf function to find P(2), P(3), and P(4).2nd VARS binompdf (5, .7, 2), (5, .7, 3) and (5, .7, 4)

P(2) = .132P(3) = .309P(4) = .360

P(between 2 and 4 inclusive) = P(2) + P(3) + P(4) = .132 + .309 + .360 = .801



U.S.A. Today reported on July 27, 1990, that 70% of the people questioned said that they watched less T.V. than they did a year ago, 22% said they watch the same amount, and 8% said they watch more. Find the probability that at least one will say they watched less T.V. this year than last year.

P(at least one) could be computed by adding the probabilities for r = 1, 2, 3, 4, and 5.

However, since r =0 is the complement of “at least one”, it is easier to use the complement rule.

Compute P(at least one) by 1- P(0).P(0) can be found by using the binompdf distribution, and is .002.P(at least one) = 1 - .002 = .998.

Assignments:Classwork: Pages 215-216 #1-14

AllHomework: Pages 216-219 #16-

32 Evens

PROBABILITY AND STATISTICSCHAPTER 4 NOTES SECTION 4-3

More Discrete Probability Distributions.I. The Geometric Distribution

We use the Geometric distribution when we want to know how many times we may have to try something to get a success.

A. A Geometric Distribution is a discrete probability distribution of a random

variable x that satisfies the following conditions:1. A trial is

repeated until a success occurs.

a) You will take the English SOL test until you pass it, no matter how

many tries it takes!!!2. The

repeated trials are independent of each other.3. The

probability of success, p, is constant for each trial (which is just another

way of saying that they are independent).B. The good news is that the calculator

will also find the probability of a geometric distribution.

1. 2nd VARS will again get us to the list of distributions.

geometpdf and geometcdf are the last two on the list.2. Enter

the probability of success, the number of trials you are interested in, and press Enter.

PROBABILITY AND STATISTICSCHAPTER 4 NOTES SECTION 4-3

More Discrete Probability Distributions.II. The Poisson Distribution

We use the Poisson distribution when we want to know the probability that a specific number of occurrences takes place within a given unit of time or space.

(How likely is it that an employee will miss 15 days of work in a year?)

A. The Poisson Distribution is a discrete probability distribution of a random

variable x that satisfies the following conditions:1. The

experiment consists of counting the number of times x, and event, occurs in a given interval.2. The

probability of the event occurring is the same for each interval.3. The

number of occurrences in one interval is independent of the

number of occurrences in other intervals.B. The Poisson distributions are also on

the calculator, between the binom and geomet distributions.

1. Enter the average number of occurrences for the desired interval, and

the number you are interested in.

Paste and Enter to find the probability.

PROBABILITY AND STATISTICSCHAPTER 4 NOTESSECTION 4-3 EXAMPLES

You are a telemarketer. From experience, you know that the probability that you will make a sale on any given call is 0.23.

Find the probability that your first sale on any given day will occur on the fourth or fifth call.

A. This is a geometric distribution question, since you are being asked how

many times you will have to call to make a sale.1. We need

to add the probability of the first sale happening on the fourth

call to the probability of the first sale occurring on the fifth call.(This is an OR

question, so we add).2. 2nd

VARS geometpdf, .23, 4 gives us .105, and 2nd VARS geometpdf, .23, 5

gives us .081. Add

those up to get .186.

a) We have an 18.6% chance of the first sale occurring on the

fourth or fifth call.


You are a telemarketer. From experience, you know that the probability that you will make a sale on any given call is 0.23.

Find the probability that the first sale will occur before your fourth sales call.A) This is still a geometric distribution question,

since you are being asked how many times you will have to call to make a sale.

1. We need to add the probability of the first sale happening on the first

call to the probability of the first sale occurring on the second call to

the probability of the first sale happening on the third call.

2. We can use the geometcdf option here, using 3 as the upper limit.

2nd VARS geometcdf, .23, 3 gives us .543.

a) We have a 54.3% chance of the first sale occurring before the

fourth call.


Two thousand brown trout are introduced into a small lake. The lake has a volume of 20,000 cubic meters.

What is the probability that three brown trout are found in any given cubic meter of the lake?

A) This is a Poisson distribution, since we are interested in how many brown

trout occur within a specified area.1. To find

how many trout would be expected to be found in each cubic meter of

water, divide the 2,000 fish by the 20,000 cubic meters of water in

the lake.

This gives us an expected value of 0.1 trout per cubic meter.2. On the

TI-84, go to 2nd VARS C (poissonpdf). We use the pdf function because

we are interested in PRECISELY 3 brown trout.Enter

0.1 (what we expect to find) for the first value, and 3 (what we want to

find) for the second value. 3. We have

about a 0.00015 probability of finding 3 brown trout in any

randomly selected cubic meter of the lake.

Assignments:Classwork: Page 226

#1-10 AllHomework: Pages

226-227 #11-24 All

Chapter 4 Discrete Probability Distributions. PROBABILITY AND STATISTICS CHAPTER 4 NOTES DISCRETE...

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