Mean and Variance of Binomial Distribution is a topic in probability theory and statistics. Mean and Variance is properties of Binomial Distribution. In this article, we will study Mean and Variance of Binomial Distribution, how to find Mean and Variance of Binomial Distribution, formula, derivation with proof, solved examples, mean & variance of negative binomial distribution and FAQs

The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes-no question, and each with its own Boolean-valued outcome: success or failure. Mean and Variance is the properties of Binomial Distribution. The concept of mean and variance is also seen in standard deviation. Binomial Distribution is a topic of statistics. Mean deviation is also a useful topic of probability.

Mean is the expected value of Binomial Distribution. It is calculated by multiplying the number of trials (n) by the probability of successes (p), or n x p.

Variance is a measure of dispersion that takes into account the spread of all data points in a data set. The variance is the mean squared difference between each data point and the centre of the distribution measured by the mean.

The mean of the distribution\( (μ_x)\) is equal to np. The variance \((σ^2_x)\) is \(n\times{p}\times( 1 – p )\). The standard deviation \((σ_x)\) is \(\sqrt{n\times{p}\times( 1 – p )}\) When p > 0.5, the distribution is skewed to the left. When p < 0.5, the distribution is skewed to the right

Bernoulli distribution is a discrete probability distribution for a Bernoulli trial. It is a random experiment that has only two outcomes, usually either a Success or a Failure.

Example: The probability of getting a head i.e a success while flipping a coin is 0.5. The probability of “failure” is 1 – P (1 minus the probability of success, which also equals 0.5 for a coin toss). It is a special case of the binomial distribution for n = 1. In other words, it is a binomial distribution with a single trial (e.g. a single coin toss). The expected value for a random variable, X, for a Bernoulli distribution is E[X] = p.

For example, if p = 0.4, then E[X] = 0.4.

The variance of a Bernoulli random variable is:

Var[X] = p(1 – p).

Now let’s see the derivation of how the formulae of the Mean and Variance are derived.

\(\begin{equation} \text{If } P(x)= \binom{n}{x} p^x(1-p)^{n-x} \text{then}\\ \mathop{\mathbb{E}[X] = \sum_{x=0}^{n} x \cdot \binom{n}{x} p^x(1-p)^{n-x}}\\ =\sum_{x=0}^{n}{n!\over{(n-x)!x!}}p^x(1-p)^{n-x}\\ =\sum_{x=1}^{n}{n!\over{(n-x)!(x-1)!}}p^x(1-p)^{n-x}\\ \mathop{\mathbb{E}[X] = \sum_{x=1}^{n}{n(n-1)!\over{(n-x)!(x-1)!}}p.p^{x-1}(1-p)^{n-x}}\\ =np\sum_{x=1}^{n}{(n-1)!\over{(n-x)!(x-1)!}}.p^{x-1}(1-p)^{n-x}\\ =np\sum_{x=1}^{n}{(n-1)!\over{[(n-1)-(x-1)]!(x-1)!}}.p^{x-1}(1-p)^{n-x}\\ =np\sum_{x=1}^{n} x \cdot \binom{n-1}{x-1}p^{x-1}(1-p)^{n-x}\\ \text{We put 1 – p = q}\\ =np\sum_{x=1}^{n} x \cdot \binom{n-1}{x-1}p^{x-1}q^{n-x}\\ =np[^{n-1}C_0q^{n-1}+^{n-1}C_1pq^{n-2}+^{n-1}C_2p^2q^{n-3}+ … + ^{n-1}C_{n-1}p^{n-1}]\\ [^{n-1}C_0q^{n-1}+^{n-1}C_1pq^{n-2}+^{n-1}C_2p^2q^{n-3}+ … + ^{n-1}C_{n-1}p^{n-1}] \\ =\text{Binomial Expansion of} (p+q)^{n-1}\\ \mathop{\mathbb{E}[X]} = np(p+q)^{n-1}\\ \text{But we know that p + q = 1}\\ \mathop{\mathbb{E}[X]} = np(1)^{n-1} = np\\ \text{This the mean of the binomial distribution.}\\ \text{Now,} Var(X) = \mathop{\mathbb{E}[X^2] – [{\mathop{\mathbb{E}[X]}]}}^2\\ \mathop{\mathbb{E}[X^2]} = \sum_{x=0}^{n} x^2 \cdot \binom{n}{x} p^xq^{n-x}\\ = {\sum_{x=0}^{n} [x(x-1)+x] \cdot \binom{n}{x} p^xq^{n-x}} + \sum_{x=0}^{n} x \cdot \binom{n}{x} p^xq^{n-x}\\ = \sum_{x=2}^{n} {\frac{x(x-1)n!}{(n-x)!x(x-1)(x-2)!}} p^x q^{n-x} + np\\ = n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{(n-x)!(x-2)!}} p^{x-2}q^{n-x}\\ = n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{[(n-2)-(x-2)]!(x-2)!}} p^{x-2}q^{n-x}\\ = n(n-1)p^2\sum_{x=2}^{n} \binom{n-2}{x–2}p^{x-2}q^{n-x}\\ = n(n-1)p^2 [^{n-2}C_0q^{n-2}+^{n-2}C_1pq^{n-3}+^{n-2}C_2p^2q^{n-4}+ … + ^{n-2}C_{n-2}p^{n-2}] + np\\ = n(n-1)p^2[(p+q)^{n-2}]+np\\ \text{Since p + q =1, we have} \\ \mathop{\mathbb{E}[X^2]} = n(n-1)p^2+np\\ \text{Using this,} \\ Var(X) = n(n-1)p^2+np -(np)^2\\ = n^2p^2 – np^2 + np – n^2p^2\\ = np(1-p)\\ = npq\\ \text{Hence the variance of the binomial distribution is npq.}\\ \end{equation}\)

Q. What is the mean of a binomial random variable with n = 18 and p = 0.4?

Ans: The mean of a binomial random variable X is represented by the symbol \(\mu\)

\(\mu\) \(\mu\) = np Here, n = 18 and p = 0.4, so \(\mu\) = 18 x 0.4 = 7.2

Q. What is the standard deviation of a binomial distribution with n = 18 and p = 0.4? Round your answer to two decimal places.

Ans: The standard deviation of X is represented by \(\sigma\) and represents the square root of the variance. If X has a binomial distribution, the formula for the standard deviation is \(\begin{matrix} \sigma=\sqrt{npq} \sigma=\sqrt{18\times0.4(1-0.4)} \end{matrix}\)

Q. An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years:

Ans: Case 1: If all 5 people are living \(\begin{matrix} B(5, \frac{2}{3}) p = \frac{2}{3} 1 – p = \frac{1}{3}\\ p(X = 5) = \binom{5}{5} (\frac{2}{3})^5 = 0.132 \end{matrix}\)

Case 2: At least three people are still living.

\(\begin{matrix} p(X \geq 3) = p (X = 3) + p (X = 4) + p (X = 5)\\ = \binom {5}{3} (\frac{2}{3})^3 (\frac{1}{3})^2 + \binom {5}{4} (\frac{2}{3})^4 (\frac{1}{3}) + \binom{5}{5} (\frac{2}{3})^5 = 0.791 \end{matrix}\)

Case 3: Exactly two people are still living.

\(p(X = 2) = \binom{5}{2} (\frac{2}{3})^2 (\frac{1}{3})^3 = 0.164\)

Q. If from six to seven in the evening one telephone line in every five is engaged in a conversation: what is the probability that when 10 telephone numbers are chosen at random, only two are in use?

Ans. : B(10, 1/5)p = 1/51 − p = 4/5 \(\begin{matrix} B(10, \frac{1}{5}) p = \frac{1}{5} 1 – p = \frac{4}{5}\\ p(X = 2) = \binom {10}{2} (\frac{1}{5})^2 \cdot (\frac{4}{5})^8 = 0.3020 \end{matrix}\)

Q. A pharmaceutical lab states that a drug causes negative side effects in 3 of every 100 patients. To confirm this affirmation, another laboratory chooses 5 people at random who have consumed the drug. What is the probability of the following events?

Ans: Case: 1 None of the five patients experience side effects.

B(100, 0.03) p = 0.03 q = 0.97 p(X = 0) = \binom {5}{0} 0.97^5 = 0.8687[/latex]᠎᠎

Case 2: At least two experience side effects.

\(\begin{matrix} p(X \geq 2) = 1 – p (X 0.5, the distribution is skewed to the left. When p < 0.5, the distribution is skewed to the right.Q.3 What is the mean and variance of a Bernoulli binomial distributionAns.3 he expected value for a random variable, X, for a Bernoulli distribution is: E[X] = p. For example, if p = 0.4, then E[X] = 0.4. The variance of a Bernoulli random variable is: Var[X] = p(1 – p).Q.4 Is in binomial distribution mean always greater than variance?Ans.4 Since p and q are numerically less than or equal to 1, npq < np. The variance of a binomial variable is always less than its mean.Q.5 What are the applications of Binomial Distribution?Ans.5 The manufacturing company uses binomial distribution to detect defective goods or items. In clinical trial binomial trial is used to detect the effectiveness of the drug. Moreover, binomial trial is used in various fields such as market research. In a manufacturing context, the number of faulty items in a batch of products might follow a binomial distribution, if the probability of failures is constant. Continue Reading in AppCreate Your Free Account to Continue Reading

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