# Binomial Distribution

How to find the expected value, variance, and standard deviation of a binomial distribution: formula, 3 examples, and their solutions.

## Formula

For a binomial experiment B(n, p),

if n is big,

then B(n, p) shows a binomial distribution.

Then you can find E(X), V(X), and σ(X)

of the binomial distribution.

E(X) = np

Expected Value

V(X) = npq

Variance

σ(X) = √npq

Standard Deviation

n: Number of the binomial experiment

p: Probability of the wanted event

q = 1 - p

Probability: Not A

## ExampleE(X)

First, write the given condition as

B(n, p).

Getting a '2' of a fair die

isn't affected by the previous trial.

So this is an independent event.

This is repeated.

So this is a binomial experiment.

n = 180

So write

B(180.

The probability of getting a '2' of a die is

1/6.

So p = 1/6.

So write

1/6).

So the given binomial experiment is

B(180, 1/6).

B(180, 1/6)

n = 180

p = 1/6

Then the expected value E(X) is

180⋅[1/6].

180⋅[1/6] = 30

So E(X) = 30.

## ExampleV(X)

You just found that

B(180, 1/6).

B(180, 1/6)

p = 1/6

Then q = 1 - 1/6 = 5/6.

B(180, 1/6)

n = 180

p = 1/6

q = 5/6

Then the variance V(X) is

180⋅[1/6]⋅[5/6].

180⋅[1/6] = 30

Cancel the denominator 6

and reduce 30 to, 30/6, 5.

5⋅5 = 25

So V(X) = 25.

## Exampleσ(X)

The condition is different

from the previous examples.

So write the given condition as

B(n, p).

The probability of

getting 'the multiple of 3', 3 or 6, is

2/6 = 1/3.

So p = 1/3.

Getting 'the multiple of 3' of a fair die

isn't affected by the previous trial.

So this is an independent event.

This is repeated.

So this is a binomial experiment.

n = 180

p = 1/3

So write

B(180, 1/3).

B(180, 1/3)

p = 1/3

Then q = 1 - 1/3 = 2/3.

B(180, 1/6)

n = 180

p = 1/3

q = 2/3

Then the standard deviation σ(X) is

√180⋅[1/3]⋅[2/3].

180⋅[1/3] = 60

Cancel the denominator 3

and reduce 60 to, 60/3, 20.

20

= 4⋅5

= 2^{2}⋅5

Prime Factorization

√2^{2}⋅5⋅2

= 2√5⋅2

= 2√10

Simplify a Radical

So σ(X) = 2√10.