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Find moment generating function of Binomial distribution & hence find mean and variance.
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By definition, moment generating function m.g.f. about origion is

$ M_0(t) = E(e^tX) = ∑p_i e^{tx_i}$

=$ ∑(^nC_x) p^xq^{n-x}.e^{tx}$

= $∑ (^nC_x)q^{n- x}(pe^t)^x = (q +pe^t)^n $

[Note = ∑$(^nC_x) a^xb^{n-x} = (a+b)^n $]

Differentiating M$_0$(t) and putting t =0 ,we get the required moments.

Now, $[\frac{\mathrm{d} }{\mathrm{d} t} M_0(t)] = n(q +pe^t)^{n-1}. pe^t $

$[\frac{\mathrm{d} }{\mathrm{d} t} M_0(t)]_{t=0}$ = np.(q+p)

∴ mean = μ$_1'$= np , since p+q =1

$[\frac{\mathrm{d^2} }{\mathrm{d} t^2} M_0(t)] = np.[ e^{t(n-1)} (q +pe^t)^{n-2}.pe^t + (q +pe^t)^{n-1}.e^t]$

$[\frac{\mathrm{d^2} }{\mathrm{d} t^2} M_0(t)]_{t=0}$ = npq + n$^2$p$^2$ = μ$_2'$

Therefore variance = μ$_2'$ - [μ$_1'$]$^2$ = npq +n$^2$p$^2$ - n$^2$p$^2$ = npq

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