Question

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

N=126 p=0.27

Answer 1

A random variable following a binomial distribution over
`n` trials with success probability
`p` has PMF


`f_X(x)=\dbinom nxp^x(1-p)^(n-x)`

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.


`\displaystyle\sum_xf_X(x)=\sum_(x=0)^n\binom nxp^x(1-p)^(n-x)=1`

The mean is given by the expected value of the distribution,


`\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_(x=0)^nx\binom nxp^x(1-p)^(n-x)`

`\mathbb E(X)=\displaystyle\sum_(x=1)^nx(n!)/(x!(n-x)!)p^x(1-p)^(n-x)`

`\mathbb E(X)=\displaystyle\sum_(x=1)^n(n!)/((x-1)!(n-x)!)p^x(1-p)^(n-x)`

`\mathbb E(X)=\displaystyle np\sum_(x=1)^n((n-1)!)/((x-1)!((n-1)-(x-1))!)p^(x-1)(1-p)^((n-1)-(x-1))`

`\mathbb E(X)=\displaystyle np\sum_(x=0)^n((n-1)!)/(x!((n-1)-x)!)p^x(1-p)^((n-1)-x)`

`\mathbb E(X)=\displaystyle np\sum_(x=0)^n\binom{n-1}xp^x(1-p)^((n-1)-x)`

`\mathbb E(X)=\displaystyle np\sum_(x=0)^(n-1)\binom{n-1}xp^x(1-p)^((n-1)-x)`

The remaining sum has a summand which is the PMF of yet another binomial distribution with
`n-1` trials and the same success probability, so the sum is 1 and you're left with


`\mathbb E(x)=np=126*0.27=34.02`

You can similarly derive the variance by computing
`\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2`, but I'll leave that as an exercise for you. You would find that
`\mathbb V(X)=np(1-p)`, so the variance here would be


`\mathbb V(X)=125*0.27*0.73=24.8346`

The standard deviation is just the square root of the variance, which is


`√(\mathbb V(X))=√(24.3846)\approx4.9834`