**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 )#