 # 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.