# Mean of binomial distribution

Prove that mean of binomial distribution is:

$$E(X)$$ = $$np$$

Where n=number of trials and p=probability of success.

Since I could not think of how to put it as an mcq i decided to put it in a note.

Enjoy solving....!

Note by Mahimn Bhatt
3 years, 3 months ago

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Please do help me in this one .

- 3 years, 3 months ago

$$E(X)=\displaystyle \sum_{x=1}^{n} x\binom{n}{x} p^{x}(1-p)^{n-x}$$

$$= \displaystyle \dfrac{n!}{x!\times (n-x)!}xp^{x}(1-p)^{n-x}$$

$$= \displaystyle np\dfrac{n!}{(x-1)!\times (n-x)!}p^{x-1}(1-p)^{n-x} \quad x=t+1 ,n=m+1$$

$$=\displaystyle \dfrac{m!}{t!\times (m-t)!}xp^{x}(1-p)^{m-y} \\= np\sum p_{i}$$

$$= \displaystyle np \quad \because \sum p_{i}=1$$

P.S. I think you understand why I didn't start the sum with x=0 .

- 3 years, 3 months ago

haha yes i understand. I know how to derive i was just posting it as a problem.......:)

- 3 years, 3 months ago

Ok .

- 3 years, 3 months ago