# Binomial Distribution

How to find the expected value, variance, standard deviation, and the probability from the binomial distribution data: formulas, examples, and their solutions.

## Formulas

If a binomial experiment is repeated *n* times,

then the expected value of this experiment

shows a binomial distribution.

The expected value, E(*X*), is *np*.

The variance, V(*X*), is *npq*.

And the standard deviation, σ(*X*) is √*npq*.*p* is the probability of [want].

And *q* = 1 - *p*.

## Example 1: Expected Value

Recall that

tossing a die is a binomial experiment.

Binomial experiment

So this example shows a binomial distribution.

So, to find the expected value,

use E(*X*) = *np*.

A fair die is tossed 180 times.

So *n* = 180.

And for a single toss,

the probability of getting a '1' is 1/6.

So *p* = 1/6.

*n* = 180*p* = 1/6

So E(*X*) = 180⋅(1/6).

Expected value

180⋅(1/6) = 30

So E(*X*) = 30.

## Example 2: Variance

Write *n* = 180 and *p* = 1/6.

*p* = 1/6

So *q* = 1 - 1/6

= 5/6.

Probability of (not *A*, Complement)

*n* = 180*p* = 1/6*q* = 5/6

So V(*X*) = 180⋅(1/6)⋅(5/6).

180⋅(1/6) = 30

Cancel the denominator 6

and reduce the numerator 30 to 5.

5⋅5 = 25

So V(*X*) = 25.

## Example 3: Standard Deviation

Write *n* = 180, *p* = 1/6, and *q* = 5/6.

*n* = 180*p* = 1/6*q* = 5/6

So σ(*X*) = √180⋅(1/6)⋅(5/6).

180⋅(1/6)⋅(5/6) = 25

So σ(*X*) = √25.

You can also start from V(*X*) = 25.

σ(*X*) = √V(*X*)

So σ(*X*) = √25.

√25 = 5

The standard deviation is (+).

So you don't have to write the [±] sign.

So σ(*X*) = 5.

## Example 4: Normal Approximation

Write *n* = 180, *p* = 1/6, and *q* = 5/6.

*n* = 180*p* = 1/6*q* = 5/6.

So E(*X*) = 180⋅(1/6)

= 30.

And σ(*X*) = √180⋅(1/6)⋅(5/6)

= 5.

*n* = 180 is a very large number.

If *n* is a very large number,

then the shape of the binomial distribution curve

becomes close to a normal distribution curve.

Then you can assume the binomial distribution

as a normal distribution.

This assumption is called

the normal approximation.

The normal distribution satisfies

E(*X*) = 30 and σ(*X*) = 5.

And it says

to find P(at least 25 of '1').

This is P(*X* ≥ 25).

Normal distribution

Find the z-score of 25.

E(*X*) = 30

σ(*X*) = 5

So *Z* = (25 - 30)/5.

Z-score

25 - 30 = -5

-5/5 = -1

So, for 25 of '1', *Z* = -1.

*Z* = -1 for 25 of '1'.

So P(*X* ≥ 25)

= P(*Z* ≥ -1).

Draw the normal distribution curve.

The percentage area of [-1 ≤ *Z* ≤ 0]

is 34%.

And the right half is 50%.

P(*Z* ≥ -1) is the colored area.

0.34 + 0.50 = 0.84

So P(*Z* ≥ -1) = 0.84.

This is the answer.