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## AP®︎/College Statistics

### Course: AP®︎/College Statistics>Unit 8

Lesson 6: Parameters for a binomial distribution

# Variance of a binomial variable

We can derive the variance of a binomial variable to be p(1-p), and the standard deviation is the square root of the variance.

## Want to join the conversation?

• I think 'X = sum Y = nY' is mislead people
since Y is a random variable
sum of Y can't be equal to n*Y(n is a constant)

to me its feel like Var(X) = Var(sum Y) = Var(nY) which is incorrect idea
'Var(sum Y) = sum Var(Y)' and 'Var(nY) = n^2*Var(Y)' so they can't be equal

it would be better to make clear 'X = sum Y' and 'Var(X) = Var(sum Y)'
thanks and sorry for my bad English
• Yes, Sal's terminology is a bit sloppy...

It would be clearer that 𝑋 is the sum of independent instances of 𝑌 if he had said
𝑋 = 𝑌(1) + 𝑌(2) + 𝑌(3) + ... + 𝑌(𝑛) = ∑𝑌(𝑖)

Then,
𝐸(𝑋) = 𝐸(∑(𝑌(𝑖))) = ∑𝐸(𝑌(𝑖)) = 𝑛 ∙ 𝐸(𝑌)

Similarly,
Var(𝑋) = 𝑛 ∙ Var(𝑌)
• Variance of aY is a^2 variance(Y). How would that justify with the derivation
(1 vote)
• This was exactly my question!... if X = nY, then Var(X) = Var(nY) = n^2Var(Y).... so Sal needs to explain his steps...
• Couldn't this be simplified by saying the Variance of a binomial variable is the variance of a Bernoulli distribution multiplied by n trials? The reason being the variance addition property.

@ time Sal says "indeed the variance for a binomial variable" I think he means the variance of a Bernoulli distribution?
• What does the variance of a binomial variable describe? Why is the variance of a binomial variable important?
(1 vote)
• Would this derivation of the variance = p(1-p) work if Sal started by using p(0-p)^2 + (1-p)(1-p)^2? I can't seem to derive the same result if I try calculate it this way. I guess my question is why did Sal use p(1-p)^2 as the first term and not p(0-p)^2? Shouldn't we arrive at the same result?
(1 vote)
• i need help it is to h
• X and Y actually are two sets of data
Therefore Var(X)=Var(Y+Y+Y...) = nVar(Y)

Proof:
Var(X+Y) = Var(X)+Var(Y)+2Cov(X,Y)
If X and Y are independent of each other, then Cov(X,Y) = 0