The characteristics of a binomial experiment are:
- There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter
nn denotes the number of trials. The nn trials are independent and are repeated using identical
conditions. Because the nn trials are independent, the outcome of one trial does not affect the
outcome of any other trial.
- There are only 2 possible outcomes, called "success" and, "failure" for each trial. The
letter pp denotes the probability of a success on one trial and qq denotes the probability of
a failure on one trial.
p+q=1p+q=1.
- For each individual trial, the probability, pp, of a success and probability, qq, of a failure
remain the same. For example, randomly guessing at a true - false statistics question has only
two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly.
Suppose Joe always guesses correctly on any statistics true - false question with probability p=0.6p0.6.
Then, q=0.4q0.4 .This means that for every true - false statistics question Joe answers,
his probability of success ( p=0.6p0.6) and his probability of failure ( q=0.4q0.4) remain the same.
The outcomes of a binomial experiment fit a binomial probability distribution. The
mean, μμ, and variance, σ2σ2, for the binomial probability distribution is
μ=np
μ
np
and
σ2=npq
σ2
npq
.
The standard deviation, σσ,
is then
σ=
npq
σ
npq
Any experiment that has characteristics 2 and 3 is called a Bernoulli Trial (named after
Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment
takes place when the number of successes are counted in one or more Bernoulli Trials.
At ABC College, the withdrawal rate from an elementary physics course is
30% for any given term. This implies that, for any given term, 70% of the students stay in the
class for the entire term. A "success" could be defined as an individual who withdrew. The
random variable is XX = the number of students who withdraw from the elementary physics
course per term.
Suppose you play a game that you can only either win or lose. The probability
that you win any game is 55% and the probability that you lose is 45%. If you play the game 20
times, what is the probability that you win 15 of the 20 games? Here, if you define
XX = the
number of wins, then
XX takes on the values
XX = 0, 1, 2, 3, ..., 20. The probability of a success
is
p=0.55
p
0.55
.
The probability of a failure is
q=0.45
q
0.45
.
The number of trials is
n=20
n
20
.
The probability question can be stated mathematically as
P
(X=15)P
(X
15).
A fair coin is flipped 15 times. What is the probability of getting more than 10
heads?
Let
XX = the number of heads in 15 flips of the fair coin.
XX takes on the values
xx = 0, 1, 2, 3, ..., 15. Since the coin is fair,
pp = 0.5 and
qq = 0.5. The number of trials is
nn = 15. The probability question can be stated mathematically as
P(
X>10)
P(
X
10)
.
Approximately 70% of statistics students do their homework in time for it to be
collected and graded. In a statistics class of 50 students, what is the probability that at least 40
will do their homework on time?
This is a binomial problem because there is only a success or a __________, there are a
definite number of trials, and the probability of a success is 0.70 for each trial.
If we are
interested in the number of students who do their homework, then how do we define
XX?
XX = the number of statistics students who do their homework on time
What values does
XX take on?
What is a "failure", in words?
Failure is a student who does not do his or her homework on time.
The probability of a success is
pp = 0.70. The number of trial is
nn = 50.
If
p+q=1
p+q
1
,
then what is
qq?
The words "at least"
translate as what kind in inequality?
greater than or equal to (≥)
The probability question is
P(
X≥40)
P(
X
40)
.
XX ~ B(
n,p)B(n,p)
Read this as "
XX is a random variable with a binomial distribution."
The parameters are
nn and
pp.
nn = number of trials
pp = probability of a
success on each trial
It has been stated that about 41% of adult workers have a high
school diploma but do not pursue any further education. If 20 adult workers are randomly
selected, find the probability that at most 12 of them have a high school diploma but do not
pursue any further education. How many adult workers do you expect to have a high school
diploma but do not pursue any further education?
Let
XX = the number of workers who have a high school diploma but do not pursue any further
education.
XX takes on the values 0, 1, 2, ..., 20 where
nn = 20 and
pp = 0.41.
qq = 1 - 0.41 = 0.59.
XX ~ B(
20,0.41)
B(20,0.41)
Find
P(
X≤12)
P(
X
12)
.
P(X≤12)=0.9738
P(X
12)
0.9738
.
(calculator or computer)
Using the TI-83+ or the TI-84 calculators, the calculations are as follows. Go into 2nd
DISTR. The syntax for the instructions are
To calculate (XX = value): binompdf(nn, pp, number)
If "number" is left out, the result is the binomial probability table.
To calculate
P(X≤value)
P(X
value)
:
binomcdf(nn, pp, number)
If "number" is left out, the result is the cumulative binomial probability table.
For this problem:
After you are in 2nd DISTR, arrow down to A:binomcdf. Press ENTER. Enter
20,.41,12). The result is
P(X≤12)=0.9738
P(X
12)
0.9738
.
If you want to find
P
(X=12)P
(X
12)
,
use the pdf (0:binompdf).
If you want to find
P
(X >
12)P
(X >
12),
use 1 - binomcdf(20,.41,12).
The probability at most 12 workers have a high school diploma but do not pursue any further
education is 0.9738
The graph of
XX ~ B(
20,0.41)
B(20,0.41) is:
The y-axis contains the
probability of XX, where
XX = the number of
workers who have only a
high school diploma.
The number of adult workers that you expect to have a high school diploma but not
pursue any further education is the mean, μ=np= (20)
(0.41)=
8.2μ=np=(20)(0.41)=8.2.
The formula for the variance is σ2=npq
σ2
npq. The standard deviation is
σ=npq
σ
npq.
σ=
(20)
(0.41)
(0.59)
=2.20
σ
(20)
(0.41)
(0.59)
2.20
.
The following example illustrates a problem that is not binomial.
It violates the condition of independence. ABC College has a student advisory
committee made up of 10 staff members and 6 students. The committee wishes to
choose a chairperson and a recorder. What is the probability that the
chairperson and recorder are both students? All names of the committee are put
into a box and two names are drawn without replacement. The first name
drawn determines the chairperson and the second name the recorder. There are
two trials. However, the trials are not independent because the outcome of the
first trial affects the outcome of the second trial. The probability of a student on
the first draw is 616616. The probability of a student on the second draw is 515515,
when the first draw produces a student. The probability is 615615 when the first
draw produces a staff member. The probability of drawing a student's name
changes for each of the trials and, therefore, violates the condition of
independence.
- Bernoulli Trials:
An experiment with the following characteristics:
- There are only 2 possible outcomes called “success” and “failure” for each trial.
- The probabilities pp of success and q = 1-pq=1-p for failure are the same for any trial.
- Binomial Distribution:
A discrete random variable (RV) which arises from the Bernoulli trials with the next additional requirements. There are fixed number, nn, of independent trials. “Independent” means that the result to any trial (for example, trial 1) in no way affects the answer to all the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV
XX size 12{X} {} is defined as the number of success in n trials. The notation is:
XX ~ B
(
n
,
p
)B(n,p); the domain is
the mean is μ=np
μ
np
, and the variance is
σ
2
=
df
σ
2
=df. The probability to have exactly xx successes in nn trials is
P
(
X
=
x
)
=
n
x
p
x
q
n
−
x
P(X=x)=
n
x
p
x
q
n
−
x
.
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