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:
- Use the formula for mean, standard deviation:
1.1) mu = Sigma_(i=x)^n x color(red)(PDF(X))
Sigma_(x=0)^n xcolor(red)[((n),(x))p^x(1-p)^(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)]
= npSigma_(x=1)^n [(n-1), (x-1)] p^(x-1)(1-p)^[(n-1)-(x-1)]
Let m = n-1; k = x-1
mu = npSigma_(k=0)^m [(m), (k)] p^(k)(1-p)^[(m-k)]
Now let's use "The Binomial Theorem". We know from BT:
(a+b)^m = sum_(k=0)^n [(m), (k)]a^kb^(m-k)
Let a=p and b = (1-p) thus:
mu = np [p+(1-p)] = np
1.2) It can be shown that Variance, v E[(X-mu)^2]= E(X^2)-E^2(X)
We know E^2(X), just square above, thus we need to derive E(X^2)?
E(X^2) = Sigma_(x=0)^n x^2color(red)[((n),(x))p^x(1-p)^(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))]
= npSigma_(x=1)^n x [(n-1), (x-1)] p^(x-1)(1-p)^[(n-1)-(x-1)]
Let color(red)m= n-1; color(blue)k=(x-1)
= npSigma_(k=0)^m color(red)((k+1)) [(color(red)m), (color(blue)k)] p^(color(red)k)(1-p)^[color(red)(m)-color(blue)(k)]
Apply Linearity
= npSigma_(k=0)^m color(red)(k)[(color(red)m), (color(blue)k)] p^(color(red)k)(1-p)^[color(red)(m)-color(blue)(k)] +npSigma_(x=0)^m [(color(red)m), (color(blue)k)] p^(color(red)k)(1-p)^[color(red)(m)-color(blue)(k)]
= S_1 + S_2
we saw in 1.1) S_2 = np
color(red)(k)[(color(red)m), (color(blue)k)] = m[(m-1),(k-1) ]
S_1 = np{Sigma_(k=0)^m m[(m-1),(k-1)] p^k(1-p)^(m-k)}
S_1 = np{(n-1)pSigma_(k=1)^m [(m-1),(k-1)] p^(k-1)(1-p)^((m-1)-(k-1))}
We saw in 1.1) that the term in the sum is a binomial (a+b)^m with
a=p; b=(1-p) thus the sum is 1 and S_1=np(n-1)p
E(X^2)= S_1 + S_2= np(n-1)p + np
E[(X-mu)^2]= E(X^2)-E^2(X)= np(n-1)p + np - (np)^2
Var = np(1-p)
and standard deviation is:
sigma= sqrt(np(1-p)) =sqrt(145*7/11 *4/11 )
