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### Course: Statistics and probability>Unit 9

Lesson 6: Binomial mean and standard deviation formulas

# Mean and variance of Bernoulli distribution example

Sal calculates the mean and variance of a Bernoulli distribution (in this example the responses are either favorable or unfavorable). Created by Sal Khan.

## Want to join the conversation?

• What if the population had a third choice? Let's say that part of the population didn't have a clear opinion about it and didn't vote. How would that affect the example mentioned above?
• That would not be a 'Bernoulli distribution'. A Bernoulili distribution only consists of 2 options, failure or succes.
• what is the main differnce between Bernoulli and Binomial Distribution. For Bernoulli case, can I apply Binomial on it? I mean that for the flipping coin, there are also 2 options, head or tail, the same for Bernoulli with 2 options: yes or no, right?
• A Bernoulli distribution is a Binomial distribution with just 1 trial.

Or, a Binomial distribution is the sum of _n_ independent Bernoulli trials with the same probability of success.
• I thought the mean is a sum of numbers divided by the total number of data points. How can you use a mean that is not divided?
• When Sal says that 40% of the answers were unfavorable and 60% were favorable, that information is already calculated from the data points.
For example, suppose the population was 1000 people. Then to get 40% unfavorable, that means that 400 people answered unfavorable. Similarly, 600 people answered favorable. Then we could multiply 400*0 and add it to 600*1, then divide by 1000 to get 0.6.

If we know the percentage (or proportion) of the population in each category, that gives us enough information to calculate the mean even if we do not have access to the raw data. I can show you the algebra:
Let u be the number of people who answered unfavorable.
Let f be the number of people who answered favorable.
Let n be the number of people in the population.
We are given that u/n = 40% = 0.4 & f/n = 60% = 0.6
We calculate the mean:
mu = (u*0 + f*1)/n = (u*0)/n + (f*1)/n = (u/n)*0 + (f/n)*1 = 0.4 *0 + 0.6 * 1 = 0 + 0.6 = 0.6.
• how can you just decide to define u and f as 0 and 1?
why did you choose those numbers?
• once you defined one choice (favour, this case) as 1. the other must be 0 (unfavour) by definition of Bernoulli distribution

one of the conditions for binomial distribution is there must be 2 possible outcomes (success, failure)

you can treat Bernoulli distribution giving specific numbers (1 and 0) to two cases (success and failure) of binomial distribution
(1 vote)
• So a Bernoulli distribution is just a situation where there are only 2 options? Like Yes and No or Success and Failure or Positive and Negative? And do they have to be opposites from each other necessarily? So like if the question was: do you like chocolate or vanilla ice cream better, would the responses follow a Bernoulli distribution by definition, or no?
• As long as people had to choose chocolate or vanilla, then that would be a Bernoulli distribution (if they were able to say "neither", that would be a 3rd option and would not be Bernoulli).
• if mean and variance of bionominal distribution are 3 and 1.5 respectively, find the probablity of (1) at least one success (2) exactly 2 success.
• Nice problem!
If n represents the number of trials and p represents the success probability on each trial, the mean and variance are np and np(1 - p), respectively.
Therefore, we have np = 3 and np(1 - p) = 1.5.
Dividing the second equation by the first equation yields 1 - p = 1.5/3 = 0.5.
So p = 1 - 0.5 = 0.5, and n = 3/p = 3/0.5 = 6.

P(at least one success) = 1 - P(no successes) = 1 - (1 - p)^n = 1 - (0.5)^6 = 0.984375.
P(exactly 2 successes) = (n choose 2) p^2 (1-p)^(n-2) = [(6*5)/(1*2)] (0.5)^2 (0.5)^4 = 0.234375.

Have a blessed, wonderful day!
• At time , Sal says the distribution is skewed to the right. Isn't the distribution skewed left because the tail is to the left of the mean?
• The description of the distribution as skewed to the right might have been a slip in explanation. In the context provided, where the mean is 0.6 (60% favorable), and we're dealing with a binary outcome (favorable or unfavorable), the notion of "skewness" in the traditional sense isn't the most fitting descriptor. Skewness typically refers to the asymmetry of a distribution around its mean. Since this is a binary (Bernoulli) distribution, we don't have a long tail extending to the right or left as we would with continuous data. However, if the mean were closer to one of the extremities (0 or 1) and given we're assigning "0" to unfavorable and "1" to favorable, a higher mean suggests a concentration towards the favorable side, but not "skewness" in the classical sense.
(1 vote)
• What is the difference between the binomial and the Bernoulli distribution?
• •Bernoulli trial is a random experiment with only two possible outcomes.
•Binomial experiment is a sequence of Bernoulli trials performed independently.
• Sal how come you decided to define U and F as 0 and 1? If it's arbitrary, and you defined U to be 345 and F to be 3, couldn't you get a much different outcome?

In my class, we calculate variance as n*p*(1-p) ... I like your way better because it uses the same intuition as the analysis of random variables, but I don't understand the above.
(1 vote)
• When we define U as 0 and F as 1, then the sample mean of our data is an estimate of the proportion, p. Could we define these numbers differently? Sure, but there is no reason to do that, and we lose interpretability.