# What are the mean and standard deviation of a binomial probability distribution with n=145  and p=7/11 ?

Apr 13, 2016

$\mu = n p \left[p + \left(1 - p\right)\right] = n p = 145 \cdot \frac{7}{11}$
$\sigma = \sqrt{n p \left(1 - p\right)} = \sqrt{145 \cdot \frac{7}{11} \cdot \frac{4}{11}}$

#### Explanation:

Given: Binomial PDF, B(x;n,p) = ((n),(x))p^x(1-p)^(n-x)
n = 145; p = 7/11

Required : Mean, $\mu$ and standard deviation $\sigma$

Solution strategy:

1. Use the formula for mean, standard deviation:
1.1) $\mu = {\Sigma}_{i = x}^{n} x \textcolor{red}{P D F \left(X\right)}$
${\Sigma}_{x = 0}^{n} x \textcolor{red}{\left(\begin{matrix}n \\ x\end{matrix}\right) {p}^{x} {\left(1 - p\right)}^{n - x}}$
 = Sigma_(x=0)^n x [(n!)/((x)!(n-x)!) p^x(1-p)^(n-x)]

 = npSigma_(x=1)^n cancelx/cancelx [(n-1)!)/[((n-1)-(x-1))!] p^(x-1)(1-p)^[(n-1)-(x-1)]

 = npSigma_(x=1)^n [(n-1)!)/[((n-1)-(x-1))!] p^(x-1)(1-p)^[(n-1)-(x-1)]

$= n p {\Sigma}_{x = 1}^{n} \left[\begin{matrix}n - 1 \\ x - 1\end{matrix}\right] {p}^{x - 1} {\left(1 - p\right)}^{\left(n - 1\right) - \left(x - 1\right)}$

Let m = n-1; k = x-1

$\mu = n p {\Sigma}_{k = 0}^{m} \left[\begin{matrix}m \\ k\end{matrix}\right] {p}^{k} {\left(1 - p\right)}^{\left(m - k\right)}$

Now let's use "The Binomial Theorem". We know from BT:

${\left(a + b\right)}^{m} = {\sum}_{k = 0}^{n} \left[\begin{matrix}m \\ k\end{matrix}\right] {a}^{k} {b}^{m - k}$

Let a=p and b = (1-p) thus:

$\mu = n p \left[p + \left(1 - p\right)\right] = n p$

1.2) It can be shown that Variance, v $E \left[{\left(X - \mu\right)}^{2}\right] = E \left({X}^{2}\right) - {E}^{2} \left(X\right)$
We know E^2(X), just square above, thus we need to derive $E \left({X}^{2}\right)$?

$E \left({X}^{2}\right) = {\Sigma}_{x = 0}^{n} {x}^{2} \textcolor{red}{\left(\begin{matrix}n \\ x\end{matrix}\right) {p}^{x} {\left(1 - p\right)}^{n - x}}$
Reduce by $x$, which means you have to also reduce by $n$
 = Sigma_(x=1)^n xn [(n-1), (x-1)] p^x(1-p)^[(x-n))]
$= n p {\Sigma}_{x = 1}^{n} x \left[\begin{matrix}n - 1 \\ x - 1\end{matrix}\right] {p}^{x - 1} {\left(1 - p\right)}^{\left(n - 1\right) - \left(x - 1\right)}$

Let color(red)m= n-1; color(blue)k=(x-1)

$= n p {\Sigma}_{k = 0}^{m} \textcolor{red}{\left(k + 1\right)} \left[\begin{matrix}\textcolor{red}{m} \\ \textcolor{b l u e}{k}\end{matrix}\right] {p}^{\textcolor{red}{k}} {\left(1 - p\right)}^{\textcolor{red}{m} - \textcolor{b l u e}{k}}$

Apply Linearity

$= n p {\Sigma}_{k = 0}^{m} \textcolor{red}{k} \left[\begin{matrix}\textcolor{red}{m} \\ \textcolor{b l u e}{k}\end{matrix}\right] {p}^{\textcolor{red}{k}} {\left(1 - p\right)}^{\textcolor{red}{m} - \textcolor{b l u e}{k}} + n p {\Sigma}_{x = 0}^{m} \left[\begin{matrix}\textcolor{red}{m} \\ \textcolor{b l u e}{k}\end{matrix}\right] {p}^{\textcolor{red}{k}} {\left(1 - p\right)}^{\textcolor{red}{m} - \textcolor{b l u e}{k}}$

$= {S}_{1} + {S}_{2}$
we saw in 1.1) ${S}_{2} = n p$
$\textcolor{red}{k} \left[\begin{matrix}\textcolor{red}{m} \\ \textcolor{b l u e}{k}\end{matrix}\right] = m \left[\begin{matrix}m - 1 \\ k - 1\end{matrix}\right]$

${S}_{1} = n p \left\{{\Sigma}_{k = 0}^{m} m \left[\begin{matrix}m - 1 \\ k - 1\end{matrix}\right] {p}^{k} {\left(1 - p\right)}^{m - k}\right\}$

${S}_{1} = n p \left\{\left(n - 1\right) p {\Sigma}_{k = 1}^{m} \left[\begin{matrix}m - 1 \\ k - 1\end{matrix}\right] {p}^{k - 1} {\left(1 - p\right)}^{\left(m - 1\right) - \left(k - 1\right)}\right\}$
We saw in 1.1) that the term in the sum is a binomial ${\left(a + b\right)}^{m}$ with
a=p; b=(1-p) thus the sum is $1 \mathmr{and} {S}_{1} = n p \left(n - 1\right) p$
$E \left({X}^{2}\right) = {S}_{1} + {S}_{2} = n p \left(n - 1\right) p + n p$
$E \left[{\left(X - \mu\right)}^{2}\right] = E \left({X}^{2}\right) - {E}^{2} \left(X\right) = n p \left(n - 1\right) p + n p - {\left(n p\right)}^{2}$
$V a r = n p \left(1 - p\right)$
and standard deviation is:
$\sigma = \sqrt{n p \left(1 - p\right)} = \sqrt{145 \cdot \frac{7}{11} \cdot \frac{4}{11}}$