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Lecture No.1

The Binomial Probability distribution

The Binomial Probability Distribution

Definition: A binomial experiment is one that has these five characteristics:

1. The experiment consists of n identical results.

2. Each trial results in one of two outcomes: one outcome is called a success, S, and

the other a failure, F.

3. The probability of success on a single trial is equal to p and remains the same from

trial to trial. The probability of failure is equal to (1 - p) = q.

4. The trials are independent.

5. We are interested in x, the number of successes observed during the n trials, for x =

0, 1, 2,…, n.

Example 1

Suppose there are approximately 1,000,000 adults in a country and an unknown

proportion p favor term limits for politicians. A sample of 1000 adults will be chosen

in such a way that every one of the 1,000,000 adults has an equal chance of being

selected, and each adult is asked whether he or she favors

term limits. (The ultimate objective of this survey is to estimate the unknown

proportion p, a problem that we will discuss in Chapter 8 ). Is this a binomial

experiment?

Solution

Does the experiment have the five binomial characteristics?

1. A “trial” is the choice of a single adult from the 1,000,000 adults in the country.

This sample consists of n = 1000 identical trials.

2. Since each adult will either favor or not favor term limits, there are two outcomes

that represent the “successes” and “failures” in the binomial experiment.

9

3. The probability of success, p, is the probability that an adult favors term

limits. Does this probability remain the same for each adult in the sample?

For all practical purposes, the answer is yes. For example, if 500,000

adults in the population favor term limits, then the probability of a

“success” when the first adult is chosen is 500,000 / 1,000,000 = 1/ 2.

When the second adult is chosen, the probability p changes slightly,

depending on the first choice. That is, there will be either 499,999 or

500,000 successes left among the 999,999 adults. In either case, p is still

approximately equal to 1/ 2.

4. The independence of the trials is guaranteed because of the large group of

adults from which the sample is chosen. The probability of an adult

favoring term limits does not change depending on the response of

previously chosen people.

5. The random variable x is the number of adults in the sample who favor

term limits.

Example No.2

A purchaser who has received a shipment consisting of 20 personal Computers,

want to sample 3 PCs to see whether they are in working order before accepting

the shipment. The nearest three Pcs are selected for testing and afterward are

declared either defective or non-defective unknown to the purchaser .Two of the

PC’s in the shipment of 20 are defective. Is this a binomial experiment?

Solution:

Does the experiment have the five binomial characteristics?

1. n=3 is the identical

2. Each trial results in one of two outcomes i.e. pc is defective or non-defective

i.e. S or F.

3. Probability of draw is a defective from 20 is 2/10.

4. The condition of indecencies between trials is not satisfied. For example, the

first trial results in a defective Pc, then there is only one defective left among

19 in the shipment there p/ defective in the remaining PC is =1/19

Rule of thumb:

If the sample size is large relative to the population size, in particular, if nN

≥ 0.05, then

the resulting experiment is not binomial.

There are two set.

Population set (utility of whole the observation)

Sample ( a subset of population)

Binomial probability distribution:

A binomial experiment consists of n identical trials with probability of success p on each

trial. The probability of k successes in n trials is

p ( x=k )=Cnk pk qn−k= n !

k ! (n−k )!pk qn−k for k=0,1,2 ;…,4

μ=np , σ2=npq

Example

Find P(x=2) for a binomial random variable with n=10 and p=0.1.

Solution:

μ=np=10 (0.1 )=1

σ 2=n . p . q=10 (0.1 ) (0.9 )=0.9

Population mean μ=∑ x i

Nwhere N represent the no . of element

Sample mean x=∑ x i

n

Populationvarience σ2=∑ ( x i−μ )2

N

Samplevarience S2=∑ ( xi−x )2

n

Where N=Number of elements∈a population

n=number of element ina sample