ch5: discrete probability distributions - uc...

Ch5: Discrete Probability Distributions Santorico - 47

Ch5: Discrete Probability Distributions Section 5-1: Probability Distribution

Recall:

A variable is a characteristic or attribute that can assume different values. o Various letters of the alphabet (e.g. X, Y, Z) are used to

represent variables. A random variable is a variable whose values are

determined by chance. Discrete variables are countable.

Example: Roll a die and let X represent the outcome so X = {1,2,3,4,5,6}

Ch5: Discrete Probability Distributions Santorico - Page 148

Discrete probability distribution - the values a random variable can assume and the corresponding probabilities of the values. The probabilities may be determined theoretically or by

observation.

They can be displayed by a graph or a table. How does this connect to our frequency distributions, tables and graphs from Chapter 2?


Example: Create a probability distribution for the number of girls out of 3 children.

We previously used a tree diagram to construct the sample space which consisted of 8 possible outcomes:

BBB X=0 BBG, BGB, GBB X=1 BGG, GBG, GGB X=2 GGG X=3

The corresponding (discrete) probability distribution is:

Number of Girls X 0 1 2 3

Probability P(X) 1/8 3/8 3/8 1/8

Check of calculations in table: MUST SUM TO 1!


Graph the probability distribution above.

0.0 0.5 1.0 1.5 2.0 2.5 3.0

0.1

50

.20

0.2

50

.30

0.3

5

Number of Girls

Pro

ba

bility


Example: The World Series played by Major League Baseball is a 4 to 7 game series won by the team winning four games. The data shown consists of the number of games played in the World Series from 1965 through 2005. The number of games played is represented by the variable X.


Construct the corresponding discrete probability distribution and graph the probability distribution above.


Two Requirements for a Probability Distribution

1. The sum of the probabilities of all the outcomes in the sample space must be 1; that is

P(X ) 1.

2. The probability of each outcome in the sample space

must be between or equal to 0 and 1; that is

0 P(X)1. These are good checks for you to use after you have computed a discrete probability distribution! The “sums to 1” check will often find a calculation error!


Example: Determine whether each distribution is a probability distribution. Explain.


Section 5-2: Mean, Variance, Standard Deviation, and Expectation

The mean, variance, and standard deviation for a probability distribution are computed differently from the mean, variance, and standard deviation for sample.

Recall that a parameter is a numerical characteristic of a population.

The mean of a probability distribution is denoted by the symbol,

.

The mean of a probability distribution for a discrete random variable is where the sum is taken over all possible values of X.

Rounding Rule: Round to one more decimal place than the outcome X when finding the mean, variance, and standard deviation for variables of a probability distribution.

( )X P X


Example: Find the mean number of girls in a family with two children using the probability distribution below.

1 1 1( ) 0 1 2 1

4 2 4X P X

X


Example: Find the mean number of trips lasting five nights or longer that American adults take per year using the probability distribution below.

X P(X )


Variance and Standard Deviation The variance of a probability distribution, σ², for a discrete random variable is found by multiplying the square of each outcome, X, by its corresponding probability, summing those products, and subtracting the square of the mean.

2 X2 P(X) 2

The standard deviation, σ, of a probability distribution is:

2


Example: Calculate the variance and standard deviation for the number of girls in the previous example:

2 2 2

22

2 2 2

[ ( )]

[ ( )] 1

1 1 1 3 10 1 2 1 1

4 2 4 2 2

X P X

X P X

2 10.707

2


Example: Calculate the variance and standard deviation for the number of trips five nights or more in the previous example.

2 [X2 P(X )]2

2


Expectation Another concept closely related to the mean of a probability distribution is the concept of expectation. Expected value has wide uses in the insurance industry, gambling, and other areas such as decision theory. The expected value of a discrete random variable of a probability distribution is the theoretical average of the variable.

E(X ) X P(X )

Does this look familiar?

Ch5: Discrete Probability Distributions Santorico - 2

Example: Suppose one thousand tickets are sold at $10 each to win a used car valued at $5,000. What is the expected value of the gain if a person purchases one ticket? The person will either win or lose. If they win which will happen with probability 1/1000, they have gained $5000-$10. If they lose, they have lost $10. Gain, X Probability, P(X) $4990 1/1000 -$10 999/1000

1 999

$4990 ( $10) $51000 1000

E X


Example: Suppose one thousand tickets are sold at $1 each for 3 prizes of $150, $100, and $50. After each prize drawing, the winning ticket is then returned to the pool of tickets. What is the expected value if a person purchases 3 tickets? Gain, X Probability, P(X) E(X) =


When gambling:

If the expected value of the game is zero, the game is said to be fair.

If the expected value of a game is positive, then the game is in the favor of the player.

If the expected value of the game is negative, then the game is said to be in the favor of the house. o This means you lose should expect to lose money in the

long run. o Every game in Las Vegas has a negative expected

value!!!


Section 5-3: The Binomial Distribution A binomial experiment is a probability experiment that satisfies the following four requirements:

1. Each of the n trials has two possible outcomes or can be reduced to two outcomes: “success” and “failure”. The outcome of interest is called a success and the other outcome is called a failure.

2. The outcomes of each trial must be independent of each other.

3. There must be a fixed number of trials. 4. Each trial has the same probability of success, denoted by p.


The acronym BINS may help you remember the conditions:

B – Binary outcomes

I – Independent outcomes

N – number of trials is fixed

S – same probability of success Examples:


Notation:

P(S), probability of success

P(F), probability of failure

p, the numerical probability of success q, The numerical probability of failure

P(S) p and

P(F)1P(S)1pq

n, the number of trials. X, the number of successes in n trials.

NOTE:

0 X n and Binomial distribution – the outcomes of a binomial experiment along with the probabilities of these outcomes.

0,1,2,3,...,X n


Probabilities for a Binomial Distribution In a binomial experiment, the probability of exactly X successes in n trials is

P(X )n!

X!(n X )!pX (1 p)nX

.

Note:

x! stands for x factorial where x is a non-negative integer.

! ( 1)( 2)...(2)(1)x x x x when x > 0

0!1 You can use your calculator or Table C at the back of the book

to solve binomial probabilities for selected values of n and p.


Examples: 5! = 5*4*3*2*1 = 120 8! = 5! = 5*4=20 3! 25! =


Example: Dionne Warwick claims to possess ESP. An experiment is conducted to test her. A person in one room picks one of the integers 1, 2, 3, 4, 5 at random. In another room, Dionne identifies the number she believes was picked. The experiment is done with eight trials. Dionne gets the correct answer four times. If Dionne does not actually have ESP and is actually guessing the number, what is the probability that she’d make a correct guess in four of the eight trials? We have a Binomial experiment here since with each guess she will either be right (success) or wrong (failure). If she does not have ESP, then the probability of a correct guess is 1/5. Hence, we would like to know P(X=4) given we have a Binomial distribution with n=8 and p=1/5.

4 8 48! 1 4

( 4) 0.04594!8! 5 5

P X


Example: Consider a family with six children and suppose there is a 25% chance that each child will be a carrier of a particular mutated gene, independent of the other children. What is the probability that exactly 2 of the children will carry the mutated gene? What is the probability that 2 or less children will carry the mutated gene?


Binomial Mean and Standard Deviation The binomial probability distribution for n trials with probability p of success on each trial has mean

, variance

2, and standard deviation

given by:

np

2 npq

npq


Example: You will take a 10 question multiple-choice test with 4 possible answers for each question. Find the mean, variance, and standard deviation if you simply guess the answer for each question. Each question represents a success/failure. We have:

X=number of correct answers n=10 p=0.25

10 0.25

2.5

np

2

10 0.25 0.75

1.875

npq

1.875

1.369

npq


Example: In the U.S., 85% of the population has Rh positive blood. Suppose we take an independent random sample of 10,000 persons and count the number with Rh positive blood. Find the mean, variance, and standard deviation for the number of Rh positive individuals in the sample.

ch5: discrete probability distributions - uc...

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