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....!

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....!

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TopNewestPlease do help me in this one . – Gaurav Jain · 2 years ago

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

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

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

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

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

P.S. I think you understand why I didn't start the sum with x=0 . – Azhaghu Roopesh M · 2 years ago

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– Mahimn Bhatt · 2 years ago

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

– Azhaghu Roopesh M · 2 years ago

Ok .Log in to reply

@Azhaghu Roopesh M @Karan Siwach @Deepanshu Gupta @Ronak Agarwal @Mvs Saketh @Sandeep Bhardwaj – Mahimn Bhatt · 2 years ago

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