If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# 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?

• At Sal defines the variance as the probability weighted sum of the squared distances from the mean or the expected valued of the squared distances from the mean. What is the relation of this formula to what we learned in earlier videos about calculating variance as the sum of differences of a sample minus the mean of the sample squared divided by (n-1)?
(39 votes)
• You have all the right concepts in play, you just have to relate them. At the start of the video Sal remarks "[Imagine] we can survay every member of the population." This indicates we are doing the population variance not sample variance. In a previous video he stated that population variance is the sum of the squared distances from the mean divided by N. This is much like the population mean which is simply the sum of all the values divided by N. Now if you remember back to the video on expected value, we can express the population mean not only as the traditional formula (sum / N), but also as the sum of each value multiplied by its frequency (also called the weighted sum). This frees us from having a set size for N and we can take the expected value of an infinite set. Essentially what is happening here for the variance is the same process. Instead of dividing the square distances by N to arrive at the variance we are multiplying each by its weight (i.e. frequency, i.e. probability) in the distribution. With this method we can calculate the variance of an infinite population.
(75 votes)
• ( - ) So if you assigned the unfavorable as 1 and favorable as 0, you'd end up with a different mean...? How do you know what number to assign to each variable?
(25 votes)
• In fact, you could choose -1 for unfavorable and +1 for favorable. That way, a 0 mean would represent a neutral overall favorability rating, a negative number would yield a negative mean sentiment and a positive number would yield a positive mean sentiment. However, choosing a 0 to represent one of the values simplifies the math.
(16 votes)
• 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?
(7 votes)
• That would not be a 'Bernoulli distribution'. A Bernoulili distribution only consists of 2 options, failure or succes.
(18 votes)
• 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?
(7 votes)
• 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.
(13 votes)
• 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?
(4 votes)
• 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.
(14 votes)
• 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?
(3 votes)
• 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).
(8 votes)
• What happened to the (n-1) value in the denominator?
(2 votes)
• You have to divide by (n-1) if you want to calculate the sample variance (n-1 is a better approximation than just dividing by n), but here it's the variance of the whole population.
(9 votes)
• how can you just decide to define u and f as 0 and 1?
why did you choose those numbers?
(6 votes)
• 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)
• 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.
(3 votes)
• 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!
(3 votes)
• 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?
(3 votes)

## Video transcript

Let's say that I'm able to go out and survey every single member of a population, which we know is not normally practical, but I'm able to do it. And I ask each of them, what do you think of the president? And I ask them, and there's only two options, they can either have an unfavorable rating or they could have a favorable rating. And let's say after I survey every single member of this population, 40% have an unfavorable rating and 60% have a favorable rating. So if I were to draw the probability distribution, and it's going to be a discrete one because there's only two values that any person can take on. They could either have an unfavorable view or they could have a favorable view. And 40% have an unfavorable view, and let me color code this a little bit. So this is the 40% right over here, so 0.4 or maybe I'll just write 40% right over there. And then 60% have a favorable view. Let me color code this. 60% have a favorable view. And notice these two numbers add up to 100% because everyone had to pick between these two options. Now if I were to go and ask you to pick a random member of that population and say what is the expected favorability rating of that member, what would it be? Or another way to think about it is what is the mean of this distribution? And for a discrete distribution like this, your mean or you're expected value is just going to be the probability weighted sum of the different values that your distribution can take on. Now the way I've written it right here, you can't take a probability weighted sum of u and f-- you can't say 40% times u plus 60% times f, you won't get any type of a number. So what we're going to do is define u and f to be some type of value. So let's say that u is 0 and f is 1. And now the notion of taking a probability weighted sum makes some sense. So that mean, or you could say the mean, I'll say the mean of this distribution it's going to be 0.4-- that's this probability right here times 0 plus 0.6 times 1, which is going to be equal to-- this is just going to be 0.6 times 1 is 0.6. So clearly, no individual can take on the value of 0.6. No one can tell you I 60% am favorable and 40% am unfavorable. Everyone has to pick either favorable or unfavorable. So you will never actually find someone who has a 0.6 favorability value. It'll either be a 1 or a 0. So this is an interesting case where the mean or the expected value is not a value that the distribution can actually take on. It's a value some place over here that obviously cannot happen. But this is the mean, this is the expected value. And the reason why that makes sense is if you surveyed 100 people, you'd multiply 100 times this number, you would expect 60 people to say yes, or if you'd summed them all up, 60 would say yes, and then 40 would say 0. You sum them all up, you would get 60% saying yes, and that's exactly what our population distribution told us. Now what is the variance? What is the variance of this population right over here? So the variance-- let me write it over here, let me pick a new color-- the variance is just-- you could view it as the probability weighted sum of the squared distances from the mean, or the expected value of the squared distances from the mean. So what's that going to be? Well there's two different values that anything can take on. You can either have a 0 or you could either have a 1. The probability that you get a 0 is 0.4-- so there's a 0.4 probability that you get a 0. And if you get a 0 what's the distance from 0 to the mean? The distance from 0 to the mean is 0 minus 0.6, or I can even say 0.6 minus 0-- same thing because we're going to square it-- 0 minus 0.6 squared-- remember, the variance is the weighted sum of the squared distances. So this is the difference between 0 and the mean. And then plus, there's a 0.6 chance that you get a 1. And the difference between 1 and 0.6, 1 and our mean, 0.6, is that. And then we are also going to square this over here. Now what is this value going to be? This is going to be 0.4 times 0.6 squared-- this is 0.4 times point-- because 0 minus 0.6 is negative 0.6. If you square it you get positive 0.36. So this value right here-- I'm going to color code it. This value right here is times 0.36. And then this value right here-- let me do this in another-- so then we're going to have plus 0.6 times 1 minus 0.6 squared. Now 1 minus 0.6 is 0.4. 0.4 squared is 0.16. So let me do this. So this value right here is going to be 0.16. So let me get my calculator out to actually calculate these values. So this is going to be 0.4 times 0.36, plus 0.6 times 0.16, which is equal to 0.24. So our standard deviation of this distribution is 0.24. Or if you want to think about the variance of this distribution is 0.24 and the standard deviation of this distribution, which is just the square root of this, the standard deviation of this distribution is going to be the square root of 0.24, and let's calculate what that is. That is going to be-- let's take the square root of 0.24, which is equal to 0.48-- well I'll just round it up-- 0.49. So this is equal to 0.49. So if you were look at this distribution, the mean of this distribution is 0.6. So 0.6 is the mean. And the standard deviation is 0.5. So the standard deviation is-- so it's actually out here-- because if you go add one standard deviation you're almost getting to 1.1, so this is one standard deviation above, and then one standard deviation below gets you right about here. And that kind of makes sense. It's hard to kind of have a good intuition for a discrete distribution because you really can't take on those values, but it makes sense that the distribution is skewed to the right over here. Anyway, I did this example with particular numbers because I wanted to show you why this distribution is useful. In the next video I'll do these with just general numbers where this is going to be p, where this is the probability of success and this is 1 minus p, which is the probability of failure. And then we'll come up with general formulas for the mean and variance and standard deviation of this distribution, which is actually called the Bernoulli Distribution. It's the simplest case of the binomial distribution.