MATERIAL AND METHOD

BINOMIAL DISTRIBUTION

Table of Contents 1.0

2.0 3.1 3.2 3.2.1 3.3 3.3.1 3.3.2 4.0 5.0 6.0 7.0

Introduction Objectives

Definition and Properties of Binomial distribution Binomial Tables

Cummulative Binomial Probability Distribution P(x > x1) Measures in Binomial Distribution

Mean Variance Conclusion Summary

Tutor Marked Assignment References and Other Resources.

1.0 Introduction

This unit treats Binomial distribution, which is one of the most widely used probability distributions in statistics. Binomial distribution is a discrete random variable distribution in which one of the two possible mutually exclusive outcomes occurs in a trial or an experiment. Binomial distribution is derived from a process called Bernoulli trial. A Bernoulli trial results from experiment that has two possible mutually exclusive outcomes and one and only one of the outcomes can occur in a trial.

The unit will treat the properties of Binomial distribution, Binomial tables and measures of central tendency and dispersion in the distribution.

2.0 Objectives

At the end of the unit you should be able to:

• Know the definition and properties of Binomial distribution.

• Solve problems involving binomial distribution

• Read Binomial Tables

• Compute measures for a binomial distribution.

)

A Binomial distribution is derived from a process known as a Bernoulli trial.

A Bernoulli trial is said to be trial of some process or experiment that can result into one of the two possible mutually exclusive events such as

Male and Female Head and tail Present and absent Correct and incorrect Defective and not defective Acceptable or not acceptable

When there is a sequence of the trials we have a process. For instance a couple trying to have children, may decide to make five trials at births. These trials will now constitute a process. To achieve the process; the following conditions must be met (Daniel and Terrel, 1979).

(i) Each trial results in one of the two possible mutually exclusive

outcomes. We denote one of the two outcomes as a success and the other as a failure.

(ii) The probability of a success is arbitrarily denoted as P the probability of failure is 1 - P. which is denoted as q.

q=1-porP+q=1

(iii) The trials are independent, that is the outcome of a trial does not affect the outcome of any other trial.

In binomial distribution, there are trials, say n, where n = 1,2,3,4 --- N We also have success denoted as X where X = 0,1,2,3 --- N

For instance if you want to have four children and you regard the birth of a male as a success then the value of success will be 0,1,2,3,4. If there is no male it means there is 0 success. If there are four males, there are 4 successes.

On the basis of this, for n trials we have 0,1,2,3,4 successes. If it is possible to know the probability of a success, p, we can calculate the probability of Binomial distribution if we know the number of trials.

You shall know that if we have n trials where n = 1,2,3,4 --- n, we can have X successes where x=0,1,2,3 - - - - n. Suppose probability of a success is P, that of failure will be 1-P. Therefore the probability of X success will be

P(X1 = X2) =

(

^{X n}

^P

^{x qn-x}

2 = 5! = 5! = 5x4x3x2x1 =10 5 2!(5 - 2)! 2!3! 2xlx3x2x1

0.6² = 0.36 0.4³ = 0.064

)

n

X, you see

( )

( (

This is the general formula for binomial distribution. Previously in this course, you learned that

n

r n ! r! (n - r)!

( ^r )

5! = 5x4x3! =10 3!2! 3!x2x1

The formula is easy to learn. If you know the number of trials n and number of successor

n

x and multiplying it by probability of success raised is number of successor and multiply the product by probability of failure raised to the number of failure.

If there are n trials and x successors number of failure will be (n -x) failure.

Example 15.1

Suppose it is known that the probability that a person in a profession will be at home in particular time of the day is 0.6 or 60%. If a person calls on 5 persons in the profession at that particular time of the day, what is the probability of getting 2 of them at home?

Here number of trials is 5 = n Number of successes is 2 = x Probability of successor is 0.6 =

P. Probability of a failure is 1- 0.6 = 0.4 = q Number of failure is (5 - 2) = 3

Using the formula for calculating probability of binomial distribution p(x = 2) = 20.6²0.4³

(

5

p(x = 2)= 2

)

^0.620.4 = l0 x 0.36 x 0.064 = 0.2304

) (

Let us now solve the problem graphically. There are 5 trials and 2 successes.

Let S represent success and frequency a failure. We can have the possible arrangements for the 2 successes. This is equal to

5

2 5! = 5! = 5x4x3x2x1 = 10 2!(5 - 2) 2!3! 2xlx3x2x1

We will then have the following possible arrangements. The first arrangement is SSFFF.

The joint probability of this, if probability of success P and that of failure is q, is Pxpxgxq=p² g³.

The whole arrangement will be

Numbers of Arrangement Possible

Arrangement Probability

1 SSFFF Pxpxqxqxq=p² q³

2 SFSFF Pxqxpxqxq= p² q³

3 SFFSF Pxqxqxpxq= p² q³

4 SFFFS Pxqxqxqxp= p² q³

5 FFFSS qxqxqxpxp= p² q³

6 FFSSF qxqxpxpxq= p² q³

7 FSSFF qxpxpxgxq= p² q³

8 FFSFS qxgxpxqxq= p² q³

9 FSFFS qxpxqxqxp= p² q³

10 FSFSF qxpxqxpxq= p² q³

Total = 10 p² q³

The total is 10p²g³= l0x0.6²x0.4⁶ = 0.2304

You will realize that we also obtain the same value as we obtained before.

For n trials, X success and P as probability of success, we can have probability distribution for the binomial distribution.

For example if n=5,p=0.6, we can obtain probability values for 0,1,2,3,4, and 5 successes. The sequence of this value constitutes the distribution of interest.

For n=5 and p=0.6,

The probability distribution of the binomial distribution is given as

X = no of Success Probability

0 0.01024

1 0.0768

2 0.2304

3 0.3456

4 0.2592

5 0.07776

Total 1.00

With this distribution, we can make a lot of probability statements. For instance, what is the probability of meeting more than two of the persons at home at that particular time of the day?

P(x>2) =P(x=3)+P(x=4)+P(x=5)

=0.3456 x 0.2592 x 0.07776

=0.68256 Let us solve another,

What is the probability of getting at least a person at home?

P (at least a person) = P(x=1) + p(x=2) + P(x=3) + P(x=4) + P(x=5)

=1- P(x = 0) =1-0.01024

=0.98976

You should know that when we obtain probability distribution for all possible success, the sum of the probabilities is equal to 1.00

Student Assessment Exercise 15.1

Suppose the probability that a family will have a male child is given as 0.4. If a couple in the family decides to have five births, what is the probability of having?

(1) 3 male births

(ii) Less than 3 male births (iii) More than three male births (iv) At least a male birth

(v) At most 4 male births [prepare the probability distribution table first]

In document “Agricultural Economics and Rural Development realities and perspectives for Romania” (Page 68-72)