Download Binomial Distribution

Ch.4
DISCRETE PROBABILITY
DISTRIBUTION
Prepared by: M.S Nurzaman, S.E, MIDEc. ( deden )
[email protected]
081310315562
Introduction
We already know that a Discrete Random Variable is a variable
which can assume only integer values, such as, 7, 9, and so on. In
other words, a discrete random variable cannot take fractions as
value. Things such as people, cars, or defectives are things we can
count and are discrete items.

There are some types of Discrete Probability Distribution:
Binomial Distribution, Poisson Distribution, and Hypergeometric
Distribution. In this lecture we will discussing the first type

Probability Distribution:
A probability distribution is similar to the frequency distribution
of a quantitative population because both provide a long-run
frequency for outcomes. In other words, a probability distribution
is listing of all the possible values that a random variable can take
along with their probabilities. for example, suppose we want to
find out the probability distribution for the number of heads on
three tosses of a coin:


First toss.........T T T T H H H H

Second toss.....T T H H T T H H

Third toss........T H T H T H T H
the probability distribution of the above experiment is as
follows (columns 1, and 2 in the following table).
(Column 1).....................(Column 2)..............(Column 3)
Number of heads...............Probability.................(1)(2)
X.....................................P(X)..........................(X)P(X)
0......................................1/8................................0.0
1......................................3/8................................0.375
2......................................3/8................................0.75
3......................................1/8................................0.375
Total...................................................................1.5=E(X)
Mean, and Variance
of Discrete Random Variables
The equation for computing the mean, or expected value of discrete
random variables is as follows:
Mean = E(X) = Summation[X.P(X)]
where: E(X) = expected value, X = an event, and P(X) =
probability of the event
Note that in the above equation, the probability of each event is
used as the weight. For example, going back to the problem of
tossing a coin three times, the expected value is: E(X) =
[0(1/8)+1(3/8)+2(3/8)+3(1/8) = 1.5 (column 3 in the above table).
Thus, on the average, the number of heads showing face up in a
large number of tossing a coin is 1.5.
Example:
Suppose a charity organization is mailing printed
return-address stickers to over one million homes in
the U.S. Each recipient is asked to donate either $1,
$2, $5, $10, $15, or $20. Based on past experience,
the amount a person donates is believed to follow
the following probability distribution:
X:..... $1......$2........$5......$10.........$15......$20
P(X)....0.1.....0.2.......0.3.......0.2..........0.15.....0.05
The question is, what is expected that an average
donor to contribute, and what is the standard
devation.
The solution is as follows.
(1)......(2).......(3).............(4)..................(5)..........................................(6)
X......P(X)....X.P(X).......X - mean......[(X - mean)]squared...............(5)x(2)
1.......0.1......0.1...........- 6.25...............39.06........................................3.906
2.......0.2......0.4...........- 5.25...............27.56........................................5.512
5.......0.3......1.5...........- 2.25.................5.06........................................1.518
10.....0.2......2.0.............2.75.................7.56........................................1.512
15.....0.15....2.25...........7.75...............60.06........................................9.009
20.....0.05....1.0...........12.75.............162.56.........................................8.125
Total...........7.25 = E(X)....................................................................29.585
Thus, the expected value is $7.25, and standard deviation is the
square root of $29.585, which is equal to $5.55. In other words, an
average donor is expected to donate $7.25 with a standard deviation
of $5.55.
Binomial Distribution:
One of the most widely known of all discrete probability
distributions is the binomial distribution. Several characteristics
underlie the use of the binomial distribution.
Characteristics of the Binomial Distribution:
1. The experiment consists of n identical trials.
2. Each trial has only one of the two possible mutually exclusive
outcomes, success or a failure.
3. The probability of each outcome does not change from trial to
trial, and
4. The trials are independent, thus we must sample with
replacement.
Note that if the sample size, n, is less than 5% of the population,
the independence assumption is not of great concern. Therefore
the acceptable sample size for using the binomial distribution with
samples taken without replacement is [n<5% N] where n is equal
to the sample size, and N stands for the size of the population.
The birth of children (male or female), true-false or multiplechoice questions (correct or incorrect answers) are some examples
of the binomial distribution
Binomial Equation:
When using the binomial formula to solve problems, all that is
necessary is that we be able to identify three things: the number of
trials (n), the probability of a success on any one trial (p), and the
number of successes desired (X). The formulas used to compute
the probability, the mean, and the standard deviation of a binomial
distribution are as follows
where: n = the sample size or the number of trials, X = the number
of successes desired, p = probability of getting a success in one
trial, and q = (1 - p) = the probability of getting a failure in one
trial
Let's go back to previous lecture and solve the probability
problem of defective TVs by applying the binomial equation once
again. We said, suppose that 4% of all TVs made by Panasonic
Company in 1995 are defective. If eight of these TVs are
randomly selected from across the country and tested, what is the
probability that exactly three of them are defective? Assume that
each TV is made independently of the others

In this problem, n=8, X=3, p=0.04, and q=(1-p)=0.96. Plugging
these numbers into the binomial formula (see the above equation)
we get: P(X) = P(3) = 0.0003 or 0.03% which is the same answer
as in lecture number four. The mean is equal to (n) x (p) =
(8)(0.04)=0.32, the variance is equal to np (1 - p) = (0.32)(0.96) =
0.31, and the standard deviation is the square root of 0.31, which
is equal to 0.6
The Binomial Table:
Mathematicians constructed a set of binomial tables containing
presolved probabilities
The binomial tables are easy to use. Simply look up n and p, then
find X (located in the first column of each table), and read the
corresponding probability. The following table is the binomial
probabilities for n = 6. Note that the probabilities in each column
of the binomial table must add up to 1.0.
Binomial Probability Distribution Table (n = 6)
--------------------------------------------------------------------------------
Probability
X.....0.1........0.2.....0.3.....0.4.....0.5.....0.6.....0.7.....0.8.....0.9
-------------------------------------------------------------------------------
0.....0.531............0.118....................................................0.000
1.....0.354............0.303....................................................0.000
2.....0.098............0.324....................................................0.001
3.....0.015............0.185....................................................0.015
4.....0.001............0.060....................................................0.098
5.....0.000............0.010....................................................0.354
6.....0.000............0.001....................................................0.531
Example:
Suppose that an examination consists of six true and false
questions, and assume that a student has no knowledge of the
subject matter. The probability that the student will guess the
correct answer to the first question is 30%. Likewise, the
probability of guessing each of the remaining questions correctly
is also 30%. What is the probability of getting more than three
correct answers?
For the above problem, n = 6, p = 0.30, and X >3. In the above
table, search along the row of p values for 0.30. The problem is to
locate the P(X > 3). Thus, the answer involves summing the
probabilities for X = 4, 5, and 6. These values appear in the X
column at the intersection of each X value and p = 0.30, as
follows:
P (X > 3) = Summation of {P (X=4) + P(X=5) +P(X=6)} =
(0.060)+(0.010)+(0.001) = 0.071 or 7.1%
Thus, we may conclude that if 30% of the exam questions are
answered by guessing, the probability is 0.071 (or 7.1%) that more
than four of the questions are answered correctly by the student.
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