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**Answer**

Mean of Binomial Distribution Proof

The MGF of Binomial Distribution is given by; ${M}_{X}\left(t\right)={(P{e}^{t}+q)}^{n}$

To get the Mean using the MGF Technique, we use the first derivative;

${M}_{X}^{I}\left(t\right)=n{(P{e}^{t}+q)}^{n-1}\times P{e}^{t}nP{e}^{t}{(P{e}^{t}+q)}^{n-1}$

$Replacethetwith0\phantom{\rule{0ex}{0ex}}$

${M}_{X}^{I}\left(0\right)=n{(P{e}^{0}+q)}^{n-1}\times P{e}^{0}$

$=np{(p+q)}^{n-1}Rememberthatp+q=1$

$And1raisedtoanypower=1$

=np

$ThustheMeanofBinomialDistribution=np$

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