Answer to Question #173484 in Statistics and Probability for ANJU JAYACHANDRAN

Question

10 (b) Find the mean and variance of binomial distribution.

Answer 1

Solution:

Let X be the random variable following binomial distribution with parameters n and p.

$X\sim B(n,p)$

First, we find the mean or expectation of X.

From the definition of expectation:

$\mathrm{E}(X)=\sum x \operatorname{Pr}(X=x)$

Thus:

$\mathrm{E}(X)=\sum_{k=0}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right) p^{k} q^{n-k}$ [Definition of Binomial Distribution, with $p+q=1$ ]$=\sum_{k=1}^{n} k\left(\begin{array}{l} n \\ k \end{array}\right)$ $p^{k} q^{n-k} \quad [\because k=0, k\left(\begin{array}{l}n \\ k \end{array}\right) p^{k} q^{n-k}=0]$

$=\sum_{k=1}^{n} n\left(\begin{array}{l} n-1 \\ k-1 \end{array}\right) p^{k} q^{n-k}$

$[\text{Factors of Binomial Coefficient}: k\left(\begin{array}{l}n \\ k\end{array}\right)=n\left(\begin{array}{l}n-1 \\ k-1\end{array}\right)]$

$\begin{array}{l} =n p \sum_{k=1}^{n}\left(\begin{array}{l} n-1 \\ k-1 \end{array}\right) p^{k-1} q^{(n-1)-(k-1)} \quad \text { taking out } n p \text { and using }(n-1) \\ =n p \sum_{j=0}^{m}\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j} \quad \text { putting } m=n-1, j=k-1 \end{array}$

$=n p$ [Binomial Theorem and $p+q=1$]

Now, we find variance.

$\mathrm{E}(X^2)=\sum x^2 \operatorname{Pr}(X=x)$

$\begin{aligned} \mathrm{E}\left(X^{2}\right) &=\sum_{k \geq 0}^{n} k^{2}\left(\begin{array}{l} n \\ k \end{array}\right) p^{k} q^{n-k} \\ &=\sum_{k=0}^{n} k n\left(\begin{array}{c} n-1 \\ k-1 \end{array}\right) p^{k} q^{n-k} \\ &=n p \sum_{k=1}^{n} k\left(\begin{array}{c} n-1 \\ k-1 \end{array}\right) p^{k-1} q^{(n-1)-(k-1)} \\ &=n p \sum_{j=0}^{m}(j+1)\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j} \\ &=n p\left(\sum_{j=0}^{m} j\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j}+\sum_{j=0}^{m}\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j}\right) \\ &=n p\left(\sum_{j=0}^{m} m\left(\begin{array}{c} m-1 \\ j-1 \end{array}\right) p^{j} q^{m-j}+\sum_{j=0}^{m}\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j}\right) \\ &=n p\left((n-1) p \sum_{j=1}^{m}\left(\begin{array}{c} m-1 \\ j-1 \end{array}\right) p^{j-1} q^{(m-1)-(j-1)}+\sum_{j=0}^{m}\left(\begin{array}{c} m \\ j \end{array}\right) p^{j} q^{m-j}\right) \\ &=n p((n-1) p+1) \\ &=n^{2} p^{2}+n p(1-p) \end{aligned}$

Now, $var(X)=E(X^2)-(E(X))^2$

$=n^2p^2+np(1-p)-(np)^2 \\ =n^2p^2+np(1-p)-n^2p^2 \\=np(1-p)$

Hence, mean = np, variance = np(1-p).