Probability
15.0 Mean and variance of binomial distribution
15.0 Mean and variance of binomial distribution
Let $X$ be a binomial variable with parameters $n$ and $p$.
$$P(X = r) = {\;^n}{C_r}{p^r}{q^{n - r}}\;\;,\;\;\;\;\;\;\;\;r = 0,1,2,...,n$$
Comparing with probability distribution,
$${p_i} = P(X = {x_i})$$
we get,
$${x_i} = r$$
and
$${p_i} = P(X = r)$$
$$Mean = \sum\limits_{}^{} {{p_i}{x_i}} = \sum\limits_{}^{} {P(X = {x_i}){x_i}} $$
Thus,
$$\begin{equation} \begin{aligned} Mean = \sum\limits_{r = 0}^n {P(X = r)r} = \sum\limits_{r = 0}^n {r \times {\;^n}{C_r}{p^r}{q^{n - r}}} \\ = \sum\limits_{r = 0}^n {r \times {n \over r}{\;^{n - 1}}{C_{r - 1}}{p^{r - 1}}{q^{n - r}}p} \\ = np\sum\limits_{r = 0}^n {^{n - 1}{C_{r - 1}}{p^{r - 1}}{q^{(n - 1) - (r - 1)}}} \\ = np{(q + p)^{n - 1}} \\ Since,\;p + q = 1 \\\end{aligned} \end{equation} $$
$$Mean = np$$
$$\begin{equation} \begin{aligned} Var = \left( {\sum\limits_{r = 0}^n {P(X = r){r^2}} } \right) - {(mean)^2} \\ = \left( {\sum\limits_{r = 0}^n {{r^2} \times {\;^n}{C_r}{p^r}{q^{n - r}}} } \right) - {\left( {np} \right)^2} \\ = \left( {\sum\limits_{r = 0}^n {\left( {{r^2} + r - r} \right) \times {\;^n}{C_r}{p^r}{q^{n - r}}} } \right) - {\left( {np} \right)^2} \\ = \left( {\sum\limits_{r = 0}^n {\left( {r(r - 1) + r} \right) \times {\;^n}{C_r}{p^r}{q^{n - r}}} } \right) - {\left( {np} \right)^2} \\ = \left( {\sum\limits_{r = 0}^n {\left( {r(r - 1){ \times ^n}{C_r}{p^r}{q^{n - r}}} \right) + \;\sum\limits_{r = 0}^n {r{ \times ^n}{C_r}{p^r}{q^{n - r}}} } } \right) - {\left( {np} \right)^2} \\ = \left( {n \times (n - 1){p^2}\sum\limits_{r = 0}^n {\left( {{\;^{n - 2}}{C_{r - 2}}{p^{r - 2}}{q^{n - r}}} \right) + \;np} } \right) - {\left( {np} \right)^2} \\ = \left( {n \times (n - 1){p^2} \times {{(q + p)}^{n - 2}} + np} \right) - {\left( {np} \right)^2} \\ = {n^2}{p^2} - n{p^2} + np - {n^2}{p^2} \\ = np - n{p^2} = np(1 - p) = npq \\\end{aligned} \end{equation} $$
$$Var = npq$$
Note: Mean> Variance
Illustration 51. If the sum of mean and variance of a binomial distribution for $5$ trials is $1.8$, find the distribution.
Solution: Let $n$ and $p$ be the parameters of the distribution.
Then,
$$\begin{equation} \begin{aligned} Mean = np \\ Var = npq \\\end{aligned} \end{equation} $$
Given that,
$n=5$
Mean + Variance = $1.8$
$$\begin{equation} \begin{aligned} \Rightarrow np + npq = 1.8 \\ \Rightarrow np(1 + q) = 1.8 \\ \Rightarrow 5p(1 + 1 - p) = 1.8 \\ \Rightarrow - 5{p^2} + 10p - 1.8 = 0 \\ \Rightarrow {p^2} - 2p + 0.36 = 0 \\ \Rightarrow (p - 0.2)(p - 1.8) = 0 \\\end{aligned} \end{equation} $$
Since, $p < 1$
Thus, we have,
$$n = 5,\;p = 0.2\;\;and\;\;q = 0.8$$
Therefore, if $X$ denotes binomial variate,
then, the binomial distribution is given by,
$$P(X = r) = {\;^5}{C_r}{(0.2)^r}{(0.8)^{5 - r}}\;\;,\;\;\;\;\;\;\;\;r = 0,1,2,3,4,5$$
Illustration 52. The sum of mean and variance of a binomial distribution is $15$ and the sum of their squares is $117$. Determine the distribution.
Solution: Let $n$ and $p$ be the parameters of the distribution. Then,
$$Mean = np\;\;and\;\;Var = npq$$
Given that,
$$\begin{equation} \begin{aligned} Mean\; + \;Variance = 15 \\ {\left( {Mean} \right)^2} + {\left( {Variance} \right)^2} = 117 \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} np + npq = 15 \\ np(1 + q) = 15 \\ {n^2}{p^2}{(1 + q)^2} = 225 \\ {n^2}{p^2}(1 + {q^2}) = 117 \\ {{{n^2}{p^2}{{(1 + q)}^2}} \over {{n^2}{p^2}(1 + {q^2})}} = {{225} \over {117}} \\ {{1 + {q^2} + 2q} \over {1 + {q^2}}} = {{225} \over {117}} \\ 1 + {{2q} \over {1 + {q^2}}} = {{225} \over {117}} \\ {{2q} \over {1 + {q^2}}} = {{108} \over {117}} = {{12} \over {13}} \\ 13 \times 2q = 12 \times (1 + {q^2}) \\ 13q = 6 + 6{q^2} \\ 6{q^2} - 13q + 6 = 0 \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} q = {{13 \pm \sqrt {{{13}^2} - 4 \times 6 \times 6} } \over {2 \times 6}} \\ q = {{13 \pm \sqrt {169 - 144} } \over {12}} = {{13 \pm 5} \over {12}} \\ q = {8 \over {12}},{{18} \over {12}} \\\end{aligned} \end{equation} $$
Since, $q < 1$
$$q = {8 \over {12}} = {2 \over 3}\,\;and\;p = {1 \over 3}\,$$
Now,
$$\begin{equation} \begin{aligned} Mean\; + \;Variance = 15 \\ np(1 + q) = 15 \\ {n \over 3}\left( {1 + {2 \over 3}{\mkern 1mu} } \right) = 15 \\ n\left( {{5 \over 3}{\mkern 1mu} } \right) = 45 \\ n = 27 \\\end{aligned} \end{equation} $$
Hence,
$$n = 27,\;\;q = {2 \over 3}\,\;and\;p = {1 \over 3}\,$$
Therefore,
the binomial distribution is,
$$P(X = r) = {\;^{27}}{C_r}{\left( {{1 \over 3}} \right)^r}{\left( {{2 \over 3}} \right)^{27 - r}}\;\;,\;\;\;\;\;\;\;\;r = 0,1,2,.....,27$$
Illustration 53. If $X$ follows binomial distribution with mean $4$ and variance $2$, find $P\left( {\left| {x - 4} \right| \le 2} \right)$.
Solution: Let $n$ and $p$ be the parameters of the binomial distribution.
$$\begin{equation} \begin{aligned} Mean = np = 4 \\ Variance = npq = 2 \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} {{np} \over {npq}} = {4 \over 2} = 2 \\ q = {1 \over 2}\;and\;p = {1 \over 2} \\ n = 8 \\\end{aligned} \end{equation} $$
The binomial distribution is,
$$P(X = r) = {\;^8}{C_r}{\left( {{1 \over 2}} \right)^r}{\left( {{1 \over 2}} \right)^{8 - r}}\;\;,\;\;\;\;\;\;\;\;r = 0,1,2,....,8$$
$$\begin{equation} \begin{aligned} P\left( {\left| {X - 4} \right| \le 2} \right) = P( - 2 \le X - 4 \le 2) \\ = P(2 \le X \le 6) \\ = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \\ { = ^8}{C_2}{\left( {{1 \over 2}} \right)^8}{ + ^8}{C_3}{\left( {{1 \over 2}} \right)^8}{ + ^8}{C_4}{\left( {{1 \over 2}} \right)^8}{ + ^8}{C_5}{\left( {{1 \over 2}} \right)^8}{ + ^8}{C_6}{\left( {{1 \over 2}} \right)^8} \\ = {{\left( {28 + 56 + 70 + 56 + 28} \right)} \over {256}} = {{238} \over {256}} = {{119} \over {128}} \\\end{aligned} \end{equation} $$
