subject: Binomial Distribution Standard Deviation [print this page] Introduction to Binomial Distribution Introduction to Binomial Distribution
Binomial distribution is a probability distribution of discrete random variables which results from an experiment called Bernoulli process, named after a seventeenth century Swiss mathematician Jacob Bernoulli. For an experiment to become a Bernoulli process, it has to be satisfy the following conditions
Each trail of the experiment should have exactly two outcomes like yes or no, pass or fail, heads or tails, male or female etc
The probability of each outcome should remain fixed over time.
The trails are statistically independent.
Some examples of Binomial Distributions are tossing of a coin, result of a student in an examination, sex of an applicant etc. In these cases, there are two outcomes with definite probabilities and each trail is independent of any other trail
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Mean and Standard Deviation of a Binomial Distribution
The mean of a probability distribution is the expected value of the distribution. This is the weighted average of each possible outcome multiplied by the probability of the outcome.
As an example, suppose a candidate has appeared for an examination and feels that the probability of his score as below
Score - Probability
80 - 0.2
90 - 0.7
95 - 0.1
His expected marks can be calculated as follows
80 'xx' 0.2 + 90 'xx' 0.7+ 95 'xx' 0.1 = 88.5
This does not mean that he will get 88.5. In the absence of any other information, this is the best value that can e assumed.
For a binomial distribution there are only two outcomes. If the probability of first outcome is p and the probability of the second outcome is then,
p+q=1
So, the mean of a binomial distribution can be calculated as
= np
where is the mean, n is the number of trials and p is the probability of success.
The standard deviation is given by
s = 'sqrt(npq)'
s = standard deviation
n = number of trails
p, q are the probabilities of each outcome
Before making use of these formulae, it must be ensured that the conditions of Bernoulli process are met
Let us illustrate this further in some examples
Ex 1: A packaging machine is observed produces 10 packaging defects for every 100 packets. If we take some samples, what is the mean and standard deviation?
Sol: The packaging is either good or defective so the distribution is binomial. The probability of defective packaging is 0.1 (='(10)/(100)'), so the probability of a good packet is 0.9 (=1-0.1).
The mean of the distribution is = np = 10 'xx' 0.1 = 1, One defective packaging
Standard deviation ='sqrt(npq)' ='sqrt(10 xx 0.1 xx 0.9)' = 0.9487 This is the average of the square of the deviations form the mean
Ex 2: Historically, result of a school in the tenth grade examination is 85 % First class. This year 400 students are appearing for the examination form the school. What is the average expected first class and what could be the standard deviation?
Ans: If a student appears for an examination, either the students gets first class or does not get it. So this is a binomial distribution
The mean of the distribution is = np = 400 'xx' 0.85 = 340 students
Standard deviation ='sqrt(npq)' ='sqrt(400 xx 0.85 xx 0.15)' = 7.141
Exercises on Binomial Distribution
Pro 1: In a factory of 350 workers there is 95% attendance on Mondays. If today is a Monday, what is the mean attendance and what could be the standard deviation?
Ans: Step 1 : The outcomes are the employee is present or absent, so the distribution could be binomial.
Step 2: The mean of the distribution is = np = 350 'xx' 0.95 = 332.5.workers
Standard deviation ='sqrt(npq)' ='sqrt(350 xx 0.95 xx 0.05)' = 7.06
Pro 2: In a toss of uniform coin, what is the mean and standard deviation in 100 trials?
Ans: Step 1 : The outcomes are heads or tails with a probability of 0.5 for each, so the distribution could be binomial.
Step 2: The mean of the distribution is = np = 100 'xx' 0.5 = 50.Heads or tails
Standard deviation ='sqrt(npq)' ='sqrt(100 xx 0.5 xx 0.5)' = 5
Pro 3: The weather forecast syas there is 60% probability of rains in this month of 30 days. What is the mean and standard deviation?
Ans: Step 1 : The outcomes are either it rains or does not, so the distribution could be binomial.
Step 2: The mean of the distribution is = np = 30 'xx' 0.6 = 18 days
Standard deviation ='sqrt(npq)' ='sqrt(30 xx 0.6 xx 0.4)' = 2.683
