Chapter 5

Nutan S. Mishra

Department of Mathematics and Statistics

University of South Alabama

Discrete VariablesA random variable is a variable whose value is

determined by the outcome of a random experiment.

A random variable assigns a numerical value to an event in S

Example: In tossing a coin S = {H,T}Define random variable as follows:X = 1 when H occursX = 0 when T occurs.Here X is a discrete random variable.

Discrete Random Variable

In some cases the outcomes of the experiment are themselves numerical values, in such a case we may not have to define a random variable separately.

Example: rolling a dieS ={1,2,3,4,5,6}. All outcomes are numerical thus

we do not have to define the random variable separately.

Define X = # dots showed upX takes values 1,2,3,4,5,6

Discrete Random variables

Toss three coins and note the number of heads showed up.

Let X = # heads occurred. Then x can take the values 0,1,2,3

Number of vehicles owned by a family

Number of dependents in a family

Number of goals scored by a player

Continuous Random Variable

• When outcomes of an experiment are numerical values in an interval then X is continuous

• Example: Recording the GPA of students• X = GPA of a student• Then X [0,4] and X can take any value in this

interval.• X= highest temperature on a given day• X= Time taken by a runner to complete a race.

Probability Distribution

Probability distribution of a discrete random variable is the set of all values of X and the set of corresponding probabilities P(X=x).

Example: toss a single coin and observe the number of heads occurred. Define X = # heads

Then X can take two values 0 or 1

X P(X=x)

0 .5

1 .5

These two columns together are called probability distribution of X

Probability distribution

Toss three coins and define

X = # heads then X=0,1,2,3

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

The probability distribution x is :

X P(X=x)

0 1/8

1 3/8

2 3/8

3 1/8

Properties of distribution

P(X=x) is denoted by f(X=x) or just f(x) .

f(x) is called probability mass function.

Two properties

1. f(x) 0

2. f(x) = 1

X P(X=x) = f(x)

0 .5

1 .5

X P(X=x)=f(x)

0 1/8

1 3/8

2 3/8

3 1/8

Exercise 5.8

x P(x)

0 .10

1 .05

2 .45

3 .40

ΣP(x) = 1.00

x P(x)

2 .35

3 .28

4 .20

5 .14

ΣP(x) = .97

x P(x)

7 -.258 .85

9 .40

ΣP(x) = 1.00

Exercise 5.10

x 0 1 2 3 4 5 6

P(x) .11 .19 .28 .15 .12 .09 .06

a. P(x=3) = .15

b. P(x ≤ 2) = P(x=0)+P(x=1)+P(x=2) =

c. P(x ≥ 4) = P(x=4)+P(x=5)+P(x=6) =

d. P(1 ≤x ≤4) = P(x=1)+P(x=2)+P(x=3)+P(x=4)

e. P(x<4) = 1- P(x ≥ 4)

f. P(x>2) = 1- P(x ≤ 2)

g. P(2 ≤x ≤5) = P(x=2)+P(x=3)+P(x=4)+p(x=50)

Exercise 5.14X = # TV sets owned by a family

Size of the dataset is 2500 (This is a population)

Use classical approach to find probability distribution of x.

# TV sets owned (x) # families (f) P(x)

0 120 .048

1 970 .388

2 730 .292

3 410 .164

4 270 .108

sum 2500 1.00

a. The two highlighted columns together form the probability distribution of X

b. These probabilities are exact because this is a population data and we are using classical approach to compute probabilties.

C. P(x=1) = .388, P(x≥ 3) = P(x=3)+P(x=4) = .064+.108

P(2 ≤ x ≤ 4) = P(x=2)+P(x=3)+P(x=4) = .292+.064+.108

P(x<4) = 1-P(x≥4) = 1-P(x=4) = 1- .108

Mean of a discrete random variable

Mean of a discrete random variable x is the value that is expected to occur per repetition of the experiment. )(xxP

# TV sets owned (x) P(x) xP(x)

0 .048 0

1 .388 .388

2 .292 .584

3 .164 .492

4 .108 .432

sum 1.00 ΣxP(x)=1.896

Mean = 1.896 TV sets

Interpretation : If we repeat this experiment number of times then on the average a family owns 1.896 TV sets.

Variance of discrete random variable

22

222

)(

)(

xPx

xPx

# TV sets owned (x) P(x) xP(x) x2 x2P(x)

0 .048 0 0 0

1 .388 .388 1 .388

2 .292 .584 4 1.168

3 .164 .492 9 1.476

4 .108 .432 16 1.728

sum 1.00 ΣxP(x)=1.896 4.76

σ2 = 4.76 – (1.896)2 = 1.165184

σ = 1.0794

Exercise 5.28Machines sold/day (x)

P(x) xP(x) x2 x2P(x)

4 .08 0.32 16 1.28

5 .11 0.55 25 2.75

6 .14 0.84 36 5.04

7 .19 1.33 49 9.31

8 .20 1.6 64 12.8

9 .16 1.44 81 12.96

10 .12 1.2 100 12

7.28 56.14

Mean of x = 7.28 σ (Standard deviation) = 7.4927

Interpretation: if experiment of collecting data on machines sales is collected for a large number of days then the on the average 7.28 machines would be sold per day.

Counting revisited

Useful links: http://www.unc.edu/~knhighto/econ70/lec4/lec4.htm

http://pavlov.psyc.queensu.ca/~flanagan/202_1999/lecture9/lecture9.html

FactorialsExample: three students with names A B C and

three chairs red blue and yellow.Question: in how many ways students can be

allocated to the chairs?

In general n distinct objects can be arranged in n places in n! (factorial n) ways

n! = n*(n-1)*…*13! = 3*2*1 = 6 ways

A CB B A C B C AA C B C A B C B A

PermutationsExample: four students A B C D and two chairs

Question: How many ways a team of two students can occupy the chairs?

# ways we can arrange n items in r places

Is called permutation

nPr = n*(n-1)*…(n-r+1)

4P2 = 4*3 = 12

A B C DB DB CA C A D

B A D CD BC BC A D A

CombinationsExample: four students A B C D and two identical chairs.

Question: In how many ways we can allocate students to the chairs?

Here order does not matter (that is it does not matter which student occupies which chair because chairs are identical)

# ways choosing r items out of n distinct items is

4C2 = 6 Table III on page C7 lists the values of combinations

A B A C A D B C B D C D

)!(!

!

rnr

nCrn

Bernoulli trialsAn experiment which has only two possible

outcomes is a Bernoulli trial with probability of one outcome p and that of the other as 1-p.

Example: toss a coin : has only two outcomes H and T if P(H) =.5 then P(T) = 1-.5 = .5

If the coin is not fair and P(H) = .7 then P(T) = 1-.7 = .3

Bernoulli distribution

More examples of Bernoulli trials:

Inspecting a car at an assembly plant and declaring it as lemon or good.

If P(L) = .001 then P(G) = .999

People entering into a football stadium. Classifying them into one of the genders. M or F if P(M) = .62 then P(F) = .38

Binomial Experiment

A binomial experiment consists of n independent Bernoulli trials.

That is repeating the Bernoulli experiment n times. Each repetition is independent of the other.

Example: Toss three fair coins.

n=3 Each coin is a Bernoulli trial. Outcome of one toss does not affect that of the others i.e. all tosses are independent. And p=.5 for all the three tosses.

Binomial experimentDefinition:

1. The experiment consists of n identical trials.

2. Each trial has only two possible outcomes

3. The probability of outcomes remain constant at each trial.

4. The trials are independent.

Binomial probability distributionIn a Binomial experiment define a random variable x asX =# heads occurred in n tosses orX = # successes in n trials. Then x takes values 0, 1,2 …,nSuch an X is called Binomial random variable with n and p

specified.If we repeat the binomial experiment a large number of

times then we are interested probabilities of x assuming different values

i.e in tossing n coins what is P(X=0) or P(X=1) and so onIn other words we are interested in probability distribution of

x.

Binomial distributionConsider a binomial experiment with n trials and p

as probability of success in a single trial then

X = # successes in n independent trials is a binomial variable and its probability distribution is called Binomial distribution with parameters n and p

P(x=k) = nCk pk (1-p)n-k for x = 0,1, …n

Alternatively replace 1-p =q

P(x=k) = nCk pk qn-k for x = 0,1, …n

Exercise 5.51a. Rolling a die many times and observing # spots is not

a binomial experiment because there are more than two possible outcomes of a roll

b. Rolling a die many times and observing whether the number is even or odd is a binomial experiment because at each roll there are only two possible outcomes: even or odd. Besides all the rolls are independent.

c. Yes this a binomial experiment because we are selecting a few people, each person has only two possible answers: in favor or not in favor. The probability of an individual being in favor is known i.e. .54 and all persons answers are independent.

Exercise 5.53a. x= binomial vairate with n=8 and p=.7

P(x=5) = 8C5 *p5 (1-p)8-5 = 8C5 *p5 (1-p)3 = 8C5 *(.7)5 (.3)3

= .2441 (from table iv)

b. Given n=4, p= .40

P(x=3) = 4C3 *p3 (1-p)4-5 = 4C3 *(.4)3 (.6) =

Mean and Variance of binomial

Let x be a binomial variable with parameters n and p. then mean of x

µ = Σx. nCk pk (1-p)n-k = np

µ = np

Variance σ2 = npq

Exercise 5.58Question asked: Do you eat home cooked food three or more times a

week?

Yes: 85% No: 15%

P(Y) = p = .85 P(N) = 1-p =q = .15

a. n=12 (a random sample of 12 Americans) is selected

X = # Americans from the sample of 12 who say Yes

Then X has binomial distribution with parameters n=12 and p= .85

X may take values between 0 to 12

That is there may be o Americans who say yes or may be 2 .. 0r at the most 12 will say yes.

b. What is the probability that 10 out of 12 Americans say yes to the posed question?

P(X=10) = 12C10 *p10(1-p)12-10 = 12C10 *p10(1-p)2