# Document

## Chapter Contents

6.1 Discrete Distributions

6.2 Uniform Distribution

6.3 Bernoulli Distribution

6.4 Binomial Distribution

6.5 Poisson Distribution

6.6 Hypergeometric Distribution

6.7 Geometric Distribution

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## Chapter Learning Objectives (LO’s)

1: Define a discrete random variable.

2: Solve problems using expected value and variance.

3: Define probability distribution, PDF, and CDF.

4: Find Poisson probabilities using tables, formulas, or Excel.

5: Find hypergeometric probabilities using Excel.

6: Calculate geometric probabilities.

7 : Find binomial probabilities using tables, formulas, or Excel.

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1 Discrete Distributions

1: Define a discrete random variable.

Random Variables

• A random variable is a function or rule that assigns a numerical value to each outcome in the sample space of a random experiment.

• Nomenclature:

- Upper case letters are used to represent random variables (e.g., X , Y ).

- Lower case letters are used to represent values of the random variable (e.g., x , y ).

• A discrete random variable has a countable number of distinct values.

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Discrete Distributions

Probability Distributions

• A discrete probability distribution assigns a probability to each value of a discrete random variable X .

• To be a valid probability distribution, the following must be satisfied.

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Discrete Distributions

Example: Coin Flips Table 6.1)

When a coin is flipped 3 times, the sample space will be

If

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

X is the number of heads, then is a random variable whose probability

X distribution is as follows:

Possible Events

TTT

HTT, THT, TTH

HHT, HTH, THH

HHH

Total

1

2

3 x

0

P ( x )

1/8

3/8

3/8

1/8

1

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Discrete Distributions

Example: Coin Flips

Note that the values of X need not be equally likely.

However, they must sum to unity.

Note also that a discrete probability distribution is defined only at specific points on the X -axis.

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Discrete Distributions

2: Solve problems using expected value and variance.

Expected Value

• The expected value E ( X ) of a discrete random variable is the sum of all X -values weighted by their respective probabilities.

• If there are n distinct values of X , then

• E ( X ) is a measure of central tendency .

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Discrete Distributions

Example: Service Calls

E(X) = μ = 0(.05) + 1(.10) + 2(.30) + 3(.25) + 4(.20) + 5(.10) = 2.75

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Discrete Distributions

Example: Service Calls

0.30

0.25

0.20

0.15

0.10

0.05

0.00

This particular probability distribution is not symmetric around the mean m

= 2.75.

However, the mean is still the balancing point, or fulcrum.

0 1 2 3 m

= 2.75

4 5

E ( X ) is an average and it does not have to be an observable point.

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Discrete Distributions

Variance and Standard Deviation

• If there are n distinct values of X , then the variance of a discrete random variable is:

• The variance is a weighted average of the dispersion about the mean, denoted either as s

2 or V ( X ). It is a measure of variability .

• The standard deviation is the square root of the variance and is denoted s

.

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Discrete Distributions

Example: Bed and Breakfast

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Discrete Distributions

Example: Bed and Breakfast

The histogram shows that the distribution is skewed to the left and bimodal.

The mode is 7 rooms rented but the average is only

4.71 room rentals.

0.30

0.25

0.20

0.15

0.10

0.05

0.00

0 1 2 3 4 5

Num ber of Room s Rented

6 s

= 2.06 indicates considerable variation around m

.

7

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Discrete Distributions

3: Define probability distribution, PDF, and CDF.

What is a PDF or CDF?

• A probability distribution function (PDF) is a mathematical function that shows the probability of each X -value.

• A cumulative distribution function (CDF) is a mathematical function that shows the cumulative sum of probabilities, adding from the smallest to the largest X -value, gradually approaching unity.

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Discrete Distributions

What is a PDF or CDF?

Consider the following illustrative histograms:

0.25

0.20

0.15

0.10

0.05

PDF = P(X = x)

0.00

0 1 2 3 4 5 6 7 8

Value of X

9 10 11 12 13 14

CDF = P(X ≤ x)

1.00

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0.10

0.00

0 1 2 3 4 5 6 7 8

Value of X

9 10 11 12 13 14

Illustrative PDF

(Probability Density Function)

Cumulative CDF

(Cumulative Density Function)

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Binomial Distribution

LO6-5: Find binomial probabilities using tables, formulas, or Excel.

Characteristics of the Binomial Distribution

• The binomial distribution arises when a Bernoulli experiment is repeated n times.

• Each trial is independent so the probability of success p remains constant on each trial.

• In a binomial experiment, we are interested in X = number of successes in n trials. So, X = X

1

+ X

2

+ ... + X n

• The probability of a particular number of successes P ( X ) is determined by parameters n and p

.

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Bernoulli Distribution

Bernoulli Experiments

• A random experiment with only 2 outcomes is a Bernoulli experiment .

• One outcome is arbitrarily labeled a “success” (denoted X = 1) and the other a “failure” (denoted X = 0).

p is the P (success), 1 – p is the P (failure).

• “Success” is defined as the less likely outcome so that p

< .5 for convenience.

• Note that P (0) + P (1) = (1 – p

) + p = 1 and 0 ≤ p ≤ 1.

• The expected value (mean) and variance of a Bernoulli experiment is calculated as:

E(X) = p and V(X) = p

(1 p

)

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Binomial Distribution

Characteristics of the Binomial Distribution

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It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education.

If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education.

How many adult workers do you expect to have a high school diploma but do not pursue any further education?

Let X = the number of workers who have a high school diploma but do not pursue any further education.

X takes on the values 0, 1, 2, ..., 20 where n = 20, p = 0.41, and q = 1 – 0.41 =

0.59.

X ~ B (20, 0.41)

Find P ( x ≤ 12). P ( x ≤ 12) = 0.9738.

Binomial Distribution

Compound Events

• Individual probabilities can be added to obtain any desired event probability.

• For example, the probability that the sample of 4 patients will contain at least 2 uninsured patients is

• HINT: What inequality means “at least?”

P ( X

2) = P (2) + P (3) + P (4)

= .1536 + .0256 + .0016 = .1808

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Binomial Distribution

Compound Events

• What is the probability that fewer than 2 patients have insurance?

• HINT: What inequality means “fewer than?”

P ( X < 2) = P (0) + P (1) = .4096 + .4096 = .8192

• What is the probability that no more than 2 patients have insurance?

• HINT: What inequality means “no more than?”

P ( X

2) = P (0) + P (1) + P (2)

= .4096 + .4096 + .1536 = .9728

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Binomial Distribution

Compound Events

It is helpful to sketch a diagram:

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Binomial Distribution

Using a TI-83 or Ti-84 Calculator

Go into 2nd DISTR. The syntax for the instructions are as follows:

To calculate (x = value): binompdf(n, p, number) if "number" is left out, the result is the binomial probability table.

To calculate P(x ≤ value): binomcdf(n, p, number) if "number" is left out, the result is the cumulative binomial probability table.

For this problem: After you are in 2nd DISTR, arrow down to binomcdf.

Press ENTER.

Enter 20,0.41,12). The result is P(x ≤ 12) = 0.9738.

If you want to find P ( x = 12), use the pdf (binompdf).

If you want to find P ( x > 12), use 1 - binomcdf(20,0.41,12).

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Binomial Distribution

Example: Quick Oil Change Shop

• It is important to quick oil change shops to ensure that a car’s service time is not considered “late” by the customer.

• Service times are defined as either late or not late .

• X is the number of cars that are late out of the total number of cars serviced.

• Assumptions:

- cars are independent of each other

- probability of a late car is consistent

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Binomial Distribution

Example: Quick Oil Change Shop

• What is the probability that exactly 2 of the next n = 12 cars serviced are late ( P ( X = 2))?

• P (car is late) = p

= .10

• P (car is not late) = 1 p

= .90

If you want to find P ( x = 2), use the pdf (binompdf) (12,0. 1,2).

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Binomial Distribution

Application: Art Supplies

In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages.

Eight of the pages feature signature artists. Suppose we randomly sample 100 pages.

Let X = the number of pages that feature signature artists.

Find the following probabilities:

1. the probability that two pages feature signature artists

2. the probability that at most six pages feature signature artists

3. the probability that more than three pages feature signature artists.

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Binomial Distribution

Application: Art Supplies

Find the following probabilities:

1.

the probability that two pages feature signature artists

P ( x = 2) = binompdf (100, 8/560, 2 )= 0.2466

2. the probability that at most six pages feature signature artists

P ( x ≤ 6) = binomcdf ( 100, 8/560, 6 ) = 0.9994

3. the probability that more than three pages feature signature artists.

P ( x > 3) = 1 – P ( x ≤ 3) = 1 – binomcdf (100, 8/560, 3 ) = 1 – 0.9443 =

0.0557

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6.5 Poisson Distribution

6: Poisson Events Distributed over Time.

• Called the model of arrivals , most Poisson applications model arrivals per unit of time .

• The events occur randomly and independently over a continuum of time or space:

Each dot ( •) is an occurrence of the event of interest.

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Poisson Distribution

• Let X = the number of events per unit of time.

• X is a random variable that depends on when the unit of time is observed.

• For example, we could get X = 3 or X = 1 or

X = 5 events, depending on where the randomly chosen unit of time happens to fall.

The Poisson model’s only parameter is l (Greek letter “lambda”) where l represents the mean number of events per unit of time or space .

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Poisson Distribution

Characteristics of the Poisson Distribution

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Poisson Distribution

Example: Credit Union Customers

• On Thursday morning between 9 A.M. and 10 A.M. customers arrive and enter the queue at the Oxnard University Credit Union at a mean rate of 1.7 customers per minute.

• Find the PDF, mean and standard deviation:

Note the unit for the mean and standard deviation is customers/minute.

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Poisson Distribution

Compound Events

• Cumulative probabilities can be evaluated by summing individual X probabilities.

• What is the probability that two or fewer customers will arrive in a given minute?

P ( X

2) = P (0) + P (1) + P (2)

= .1827 + .3106 + .2640 = .7573

• What is the probability of at least three customers (the complimentary event)?

P ( X

3) = 1 P ( X

2) = 1 - .7573 =.2427

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Hypergeometric Distribution

8: Find hypergeometric probabilities using Excel.

Characteristics of the Hypergeometric Distribution

• The hypergeometric distribution is similar to the binomial distribution.

• However, unlike the binomial, sampling is without replacement from a finite population of N items.

• The hypergeometric distribution may be skewed right or left and is symmetric only if the proportion of successes in the population is

50%.

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Hypergeometric Distribution

Characteristics of the Hypergeometric Distribution

NOTE

Currently, the TI-83+ and TI-84 do not have hypergeometric probability functions. There are a number of computer packages, including Microsoft

Excel, that do.

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Hypergeometric Distribution

Example: Damaged iPods

• In a shipment of 10 iPods, 2 were damaged and 8 are good.

• The receiving department at Best Buy tests a sample of 3 iPods at random to see if they are defective.

• Let the random variable X be the number of damaged iPods in the sample.

N = 10 (number of iPods in the shipment) n = 3 (sample size drawn from the shipment) s = 2 (number of damaged iPods in the shipment (“successes” in population))

N

– s = 8 (number of non-damaged iPods in the shipment) x = number of damaged iPods in the sample

(“successes” in sample) n – x = number of non-damaged iPods in the sample

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Hypergeometric Distribution

Example: Damaged iPods

• This is not a binomial problem because p is not constant.

• What is the probability of getting a damaged iPod on the first draw from the sample?

p

1

= 2/10

• Now, what is the probability of getting a damaged iPod on the second draw?

p

2

= 1/9 (if the first iPod was damaged) or

= 2/9 (if the first iPod was undamaged)

• What about on the third draw?

p

3

= 0/8 or

= 1/8 or

= 2/8 depending on what happened in the first two draws.

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Hypergeometric Distribution

Using the Hypergeometric Formula

Since there are only 2 damaged iPods in the population, the only possible values of x are 0, 1, and 2. Here are the probabilities:

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Hypergeometric Distribution

Using Software: Excel

Since the hypergeometric formula and tables are tedious and impractical, use

Excel’s hypergeometric function to find probabilities.

Figure 6.27

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Geometric Distribution

9: Calculate Geometric probabilities.

Characteristics of the Geometric Distribution

• The geometric distribution describes the number of Bernoulli trials until the first success.

• X is the number of trials until the first success.

• X ranges from {1, 2, . . .} since we must have at least one trial to obtain the first success. However, the number of trials is not fixed.

• p is the constant probability of a success on each trial.

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Geometric Distribution

Characteristics of the Geometric Distribution

Example: Telefund Calling

• At Faber University, 15% of the alumni make a donation or pledge during the annual telefund.

• What is the probability that the first donation will occur on the 7 th call?

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Assume that the probability of a defective computer component is 0.02. Components are randomly selected.

Find the probability that the first defect is caused by the seventh component tested.

How many components do you expect to test until one is found to be defective?

Let X = the number of computer components tested until the first defect is found.

X takes on the values 1, 2, 3, ... where p = 0.02. X ~ G(0.02)

Find P ( x = 7). P ( x = 7) = 0.0177.

To find the probability that x = 7,

• Enter 2 nd , DISTR

• Scroll down and select geometpdf(

• Press ENTER

• Enter 0.02, 7); press ENTER to see the result: P ( x = 7) = 0.0177

To find the probability that x ≤ 7, follow the same instructions EXCEPT select

E:geometcdf (as the distribution function.

The probability that the seventh component is the first defect is 0.0177.

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