The binomial distribution is a discrete distribution displaying data that has only TWO OUTCOMES and each trial includes replacement. The binomial distribution approaches a normal distribution as the sample size increase and therefore this approximation is better for larger sample sizes.
Such as:
Assumptions
The following formula is used to compute the number of experimental outcomes resulting from x successes in n trials.
For example: 4! (4 factorial) = 4*3*2*1 = 24
The Binomial Distribution Probability Function is shown below:
A manufacturing process creates 3.4% defective parts.
A sample of 10 parts from the production process is selected. What is
the probability that the sample contains exactly 3 defective parts?
SOLUTION:
There are two outcomes: Defective / Not-Defective, therefore the Binomial Distribution equation is applied.
p = 0.034
n = 10
then q = 1-p = 1 - 0.034 = 0.966 (96.6%)
Need the probability that x = 3.
Substitute the values into the Binomial Probability Function and solve:
Forty-five percent of all registered voters in a
national election are female. A random sample of 8 voters is selected.
The probability that the sample contains 2 males is:
SOLUTION:
55% are male and there are two outcomes: MALE / FEMALE
n = 8
p = 0.55 (55%)
q = 1-p = 0.45
Need the probability that x = 2
79% percent of the students of a large class passed
the final exam. A random sample of 4 students are selected to be
analyzed by the school. What is the probability that the sample contains
fewer than 2 students that passed the exam?
There are two outcomes: PASS / FAIL
p = 0.79
n = 4
Solving for the probability that x < 2.
Note that we need to compute p(x<2). So, it is necessary to add the P(x=0) to the P(x=1).
Each P(x=0) and P(x=1) must be calculated separately and added.
p(x<2) = p(x=0) + p(x=1)
Substitute the following values into the binomial equation to find p(x=0):
n = 4
p = 0.79
q = 1 - p = 1 - 0.79 = 0.21
x = 0
PLUS
n = 4
p = 0.79
q = 1 - p = 1 - 0.79 = 0.21
x = 1
There is 3.12% chance of selecting fewer than 2 students that passed the
exam when randomly selecting 4 students when 79% of them passed.
There are also Binomial tables that can be used when the input variables are known (and found within the table).
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