# Binomial Standard Deviation

When you flip a coin, there is a chance of getting heads or tails.The heads and tails each have the same probability  `1/2` .  There are also the situations in which the coin is biased, so that heads and tails have different probabilities.The distributions for which there are only two outcomes that are possible with fixed probability summing to one are called binomial distributions.

In the calculation of probability, the first outcome is `p ` and that of the nextoutcome is given as,`1-p=q` .

Binomial standard deviation is useful in main ways for calculations involved in mathematics. Binomial standard deviation in many aspects in accordance with various maths topics.

When `p` is equal to ` 0.5`, the binomial distribution will be symmetric. When ` p ` is not equal to ` 0.5` , the binomial distribution will not be symmetric.If `p` is closer to ` 0.5` and the number of trials is `n` , then the distribution is more symmetric.

## Binomial Standard Deviation:

There are two types of standard deviations. One is Binomial Standard Deviation and second one is Statistical Standard Deviation.

Two events with two outcomes is for the calculation of binomial standard deviation. The binomial distribution is used in the popular binomial test of statistical significance.

Formulas:

The mean (expected value) `= np` .

The variance `(sigma^2) =np(1-p` ).

The standard deviation (standard error) `=sigma=sqrt(sigma^2)=sqrt(np(1-p))`.

Where` p=` probability of first outcome.

`q=1-p=` probability of second outcome.

## The formula for binomial distribution is

` P(x)=((N!)/(x!(N-x)!))(pi^x)(1-pi)^(N-x).`

where `P(x)` is the probability of` x` successes out of `N` trials, `N ` is the number of trials, and `pi ` is the probability of success on a given trial.

## Problems

Simple Example:

If a coin is tossed twice then it has four possible outcomes.Each outcome has fixed probability.The four possible outcomes are head and head, head and tail, tail and head, tail and tail.The four outcomes each have the same probability of `1/4. `

Hence,

The probability for outcomes((head and head) or (tail and tail)) = product of `P(H)` and `P(H)` .

`=1/2xx1/2`

`=1/4.`

Similarly,

Where` P(H) =` probability for outcome(head or tail).

Therefore, the probability of getting one or more heads `= 1/4+1/4=1/2` .

Practice problems:

Prob1 : If a coin is tossed twice, what is the probability of getting one or more heads?

Sol:   Formula is `P(x)=((N!)/(x!(N-x)!))(pi^x)(1-pi)^(N-x).`

`P(0)=((2!)/(0!(2-0)!))(0.5^0)(1-0.5)^(2-0)=(2/2)(1)(0.25)=0.25` .

`P(1)=((2!)/(1!(2-1)!))(0.5^1)(1-0.5)^(2-1)=(2/1)(0.5)(0.5)=0.5.`

`P(2)=((2!)/(2!(2-2)!))(0.5^2)(1-0.5)^(2-2)=(2/2)(0.25)(1)=0.75` .

The probability of getting one head is `0.50 ` and the probability of getting two heads is `0.25.` Then the probability of getting one or more heads is `0.50 + 0.25 = 0.75` .

Prob2 : We toss a coin `12 ` times. What is the probability that we get from `0` to `3` heads?

### Sol:       Formula is `P(x)=((N!)/(x!(N-x)!))(pi^x)(1-pi)^(N-x).`

`P(0)=((12!)/(0!(12-0)!))(0.5^0)(1-0.5)^(12-0)=(12/12)(1)(0.0002)=0.0002.`

`P(1)=((12!)/(1!(12-1)!))(0.5^1)(1-0.5)^(12-1)=(12/1)(0.5)(0.00048)=0.0028`

` P(2)=((12!)/(2!(12-2)!))(0.5^2)(1-0.5)^(12-2)=(6xx11)(0.25)(0.00097)=0.0161` .

`P(3)=((12!)/(3!(12-3)!))(0.5^3)(1-0.5)^(12-3)=(2xx11xx10)(0.75)(0.0019)=0.052`

The probability of getting `3` heads `= P(0)+P(1)+P(2)+P(3)`

`=0.0002+0.0028+0.0161+0.052`

`=0.0713.`

Therefore we are using binomial calculator to make it easy to calculate these probabilities.