STATISTIC & INFORMATION THEORY

(CSNB134)

MODULE 7APROBABILITY DISTRIBUTIONS FOR RANDOM VARIABLES (BINOMIAL

DISTRIBUTION)

Overview In Module 7, we will learn three types of

distributions for random variables, which are:- Binomial distribution - Module 7A- Poisson distribution - Module 7B- Normal distribution - Module 7C

This is a Sub-Module 7A, which includes lecture slides on Binomial Distribution.

The Binomial Random Variable Example of a binomial random variable:

coin-tossing experiment - P(Head) = ½. Toss a fair coin n = 3 times, x = number of heads.

We have seen in Module 6 that the probability distribution for the coin-tossing experiment, where a coin is toss for 3 times and x = number of heads is as follows.

x p(x)

0 1/81 3/82 3/83 1/8

The Binomial Random Variable (cont.) Many situations in real life resemble the coin toss, but with a

coin that might not be fair, i.e. P(H) 1/2.

Example: A lecturer samples 10 people and counts the number who of students who are from KL.

The analogy to coin tossing is as follows:

Student• Coin:• Head / H:• Tail / T:• Number of tosses:• P(H):

From KL

Not from KL

n = 10

P(From KL) = proportion of students in the population who

are from KL.

The Binomial Random Variable (cont.) Other examples:

1. Proportion of COIT students who are female.

2. Proportion of Malaysian population who watch Akademi Fantasia.

3. Proportion of people in Bangi who subscribe to ASTRO.

Female/Male

Watch/Don’t Watch

Subscribe/Don’t Subscribe

P(F) = .55 / P(M) = .45

P(W) = .35 / P(DW) = .65

P(S) = .85 / P(DS) = .15All exhibit the common characteristic of Binomial experiment!

Characteristic of a Binomial Experiment1. The experiment consists of n identical trials.2. Each trial results in one of two outcomes,

success (S) or failure (F) i.e: mutually exclusive.

1. The probability of success on a single trial is p and remains constant from trial to trial. The probability of failure is q = 1 – p.

2. The trials are independent (Note = please refer notes on Module 5 on ‘Independent Event’).

3. We are interested in x, the number of successes in n trials.

Binomial or Not? Example: Select two people from the U.S.

population (say 300 million), and suppose that 15% (45 million) of the population has the Alzheimer’s gene. For the first person:

p = P(gene) = 45,000,000/300,000,000 = .15

For the second person: p = P(gene) = 44,999,999/299,999,999

.15, even though one person has been removed from the population.

We assume that the removal of one person from the population have negligible effect.

The Binomial Probability Distribution For a binomial experiment with n trials and

probability p of success on a given trial, the probability of k successes in n trials is:

And the measures of center and spread are:

)!(!! Recall

.,...2,1,0for )!(!

!)(

knknC

nkqpknk

nqpCkxP

nk

knkknknk

npq

npq

np

:eviationStandard_d

:Variance

:Mean2

Exercise 1 A fair coin is flipped 6 times.

(1) What is the probability of obtaining exactly 3 heads?

For this problem, N =6, k=3, and p = .5, thus:

(2) Determine the probability of obtaining 3 or more successes.

P(x >=3) = P(3) + P(4) + P(5) + P(6) = pls find this out…..

Exercise 2

A rifle shooter hits a target 80% of the time. He fires five shots at the target. Assuming x = number of hits, what is the mean and standard deviation for x?

89.)2)(.8(.5

4)8(.5

npq

np

:deviation Standard

:Mean

Exercise 2 (cont.) Would it be unusual to find that none of the

shots hit the target?89.;4

• The value x = 0 lies

49.489.

40

xz

• more than 4 standard deviations below the mean. Very unusual!! (Note: Refer to z-score in Module 3)

n = p = x =success =

Exercise 2 (cont.)What is the probability that exactly 3 shots hit the target?

333)3()3( nn qpCPxP

5 .8hit # of hits

353 )2(.)8(.!2!3

!5

2048.)2(.)8(.10 23

Exercise 2 (cont.)What is the probability that more than 3 shots hit the target?

55555

45454)5()4()3( qpCqpCPPxP

0514 )2(.)8(.!0!5

!5)2(.)8(.!1!4

!5

7373.)8(.)2(.)8(.5 54

Example 2 (Cont.)For the same rifle shooter, if he fires 20 shots at the target. What is the probability that more than 5 shots hit the target?

0202020

416206

20

6

...)5( qpCqpCqpCxPk

kknnk

Effect of pTwo binomial distributions are shown below. Here, π is the same as p

Notice that for p = .5, the distribution is symmetric whereas for p = .3, the distribution is skewed right.

Probability Tables We can use the binomial probability

tables to find probabilities for selected binomial distributions.

They can either be Probability tables for each x = k or Cumulative probability tables

Probability Tables (cont.)

Probability table for each x = k (part of)

Cumulative Probability table(part of) (for n= 5)

Probability table for each x = k (for n = 5)

P(x<=3) = 0.263 (from the cumulative table) P(x<=3) = 0.000 + 0.006 + 0.051 + 0.205 (from the x=k table) = 0.262

Probability Tables (cont.)

Cumulative Probability TablesWe can use the cumulative probability tables to find probabilities for selected binomial distributions.Find the table for the correct value of n.

Find the column for the correct value of p.

The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Exercise 3A rifle shooter hits a target 80% of the time. He fires five shots at the target. What is the probability that exactly 3 shots hit the target?

Find the table for the correct value of n.

Find the column for the correct value of p.

Row “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)

Exercise 3 (cont.)k p = .800 .0001 .0072 .0583 .2634 .6725 1.000

P(P(xx = 3) = 3) = P(x 3) – P(x 2)= .263 - .058= .205

Check from formula: P(x = 3) = .2048

Exercise 2 (cont.)

k p = .800 .0001 .0072 .0583 .2634 .6725 1.000

What is the probability that more than 3 shots hit the target?

P(P(xx > 3) > 3) = 1 - P(x 3)= 1 - .263 = .737

Check from formula: P(x > 3) = .7373

STATISTIC & INFORMATION THEORY

(CSNB134)

PROBABILITY DISTRIBUTIONS OF RANDOM VARIABLES (BINOMIAL DISTRIBUTIONS)

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